Common Factors Definition, Examples | How to Find Common Factors?

Common Factors

A factor is a number that is the exact multiplicand of another number. Every number factor is less than or equal to the given number but it cannot be greater than the given number. Every number has at least 2 factors. Common factors are also the factors that are common to two or more numbers. Fet the detailed steps to find the common factors of 2 or more numbers, solved examples, and applications in the below sections.

What are Common Factors?

Common Factors are defined as the factors that are common to two or more numbers. You can also say that a common factor is a number with which a set of two or more numbers will be divided exactly.

To find the common factors of two numbers, you need to list the factors of each number separately and then compare them. Now write the factors which are common and those are called common factors for the given numbers.

How to find Common Factors?

Factors are the numbers that divide the original number. Here are the steps to check whether two or more numbers have common factors or not.

  • Get the factors of each number separately.
  • Compare the factors of two numbers.
  • If you find common numbers then those are common factors between two numbers.

Example:

Common Factors of 4, 12

Find the factors of given numbers

4 = 1, 2, 4

12 = 1, 2, 3, 4, 6, 12

The common factors between 4 and 12 are 1, 2, 4.

Read More Articles,

Common Factors Examples

Example 1:

Find the common factors of 2, 16?

Solution:

The given numbers are 2, 16

Factors of 2 = 1, 2

Factors of 16 = 1, 2, 4, 8, and 16

Therefore, common factors of 2 and 16 = 1, 2.

Example 2:

Calculate the common factors of 14, 21?

Solution:

The given numbers are 14, 21

Factors of 14 = 1, 2, 7, 14

Factors of 21 = 1, 3, 7, 21

Therefore, common factors of 14 and 21 = 1, 7.

Example 3:

Find the common factors of 15, 45?

Solution:

The given numbers are 15, 45

Factors of 15 = 1, 3, 5, 15

Factors of 45 = 1, 3, 15, 5, 9, 45

Therefore, common factors of 15 and 45 = 1, 3, 5, 15.

Example 4:

Find the common factors of 36 and 63.

Solution:

The given numbers are 36, 63

The factors of 36 are

1 × 36 = 36

2 × 18 = 36

3 × 12 = 36

4 × 9 = 36

6 × 6 = 36

Stop here, since the number 6 is repeated.

So, 1, 2, 3, 4, 6, 9, 12, 18, and 36 are factors of 36.

The factors of 63 are

1 × 63 = 63

3 × 21 = 63

7 × 9 = 63

9 × 7 = 63

Stop here, since the numbers 7 and 9 are repeated.

So, 1, 3, 7, 9, 21, and 63 are factors of 63.

1, 3, and 9 are common in both lists.

Hence, the common factors of 36 and 63 are 1, 3, 9.

Angle of Elevation Definition, Formulas, and Examples | How to find the Angle of Elevation?

Angle of Elevation

If two straight lines coincide at a point, then it is called the angle. The angle of elevation is defined as the angle between the horizontal plane and oblique line from the observer’s eye to an object which is located above his eye. Finally, the angle formed above the surface. It is a widely used concept and it is related to height and distance. The different terms used in the angle of elevation concept, trigonometric formulas, and solved examples are mentioned below for the sake of students.

Angle of Elevation – Definition

The angle of elevation of an object seen by the observer is the angle formed between the horizontal and the line from the object to the observer’s eye. The line in which the observer’s eye is there is known as the line of sight.

Angle of Elevation

From the above image, you can see that the observer is looking at the object from the ground making an angle θ with the line of sight and horizontal line. By joining the imaginary line between the object and the end of the horizontal line, it forms a right-angle triangle. Thus we can use the trigonometry concept to find the distance from the observer to the object.

Example:

A student named Rakesh is looking at a bird sitting at the building, then there is an angle formed, which is inclined towards the bird from the Rakesh eye. This elevation angle is helpful to find the distance, height of the building with the trigonometric function like sine, cosine, and tangent.

Terms used in Elevation Angle

The three most important terms used in this concept are angle, horizontal line, and line of sight. Let us discuss each of them in detail in the following sections.

Horizontal Line:

A straight line on the flat surface where all points lie on a line and those points will have the same y-coordinate value. The angle and horizontal lines combine to form the angle of elevation.

Angle:

If two line segments meet at an endpoint, then the point is called the vertex. When two straight lines meet at a common point form an angle. It is also defined as the gap between two limes that connect on one side. It is measured in degrees or radians.

Line of Sight:

The line which is drawn from the observer’s eye to the object is known as the line of sight. In the angle of elevation, the object is kept above the line of sight of the observer. If you know the elevation angle, then it is easy to determine the distance and altitude.

Also, Check:

Angle of Elevation Formulas

Trigonometric ratios are used to find the unknown measures of the opposite, adjacent, and hypotenuse sides of the right angle.

Angle of Elevation

The trigonometric ratios are

sin θ = \(\frac { Opposite side }{ Hypotenuse } =\frac { y }{ h } \)

cos θ = \(\frac { Hypotenuse }{ Adjacent side } =\frac { h }{ x } \)

tan θ = \(\frac { Opposite side }{ Adjacent side } =\frac { x }{ y } \)

Another type of problem is along the lines.

Angle of Elevation 2

PQ = y, TQ = SR = I, PT = (y – l), QR = ST = x, PS = h

The formulas are

sin θ = \(\frac { PT }{ PS } =\frac { y – l }{ h } \)

cos θ = \(\frac { QR }{ PS } =\frac { x }{ h } \)

tan θ = \(\frac { PT }{ QR } =\frac { y – l }{ x } \)

Angle of Elevation Problems

Example 1:

A tower stands vertically on the ground. From a point on the ground, which is 48 m away from the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower.

Solution:

Let AB be the height of the tower.

Angle of Elevation 3

Take AB = h and BC is the distance between the tower and the point C. In the right-angled trinagle ABC, ∠ACB = 30°

tan θ = \(\frac { AB }{ BC } \)

tan 30 = \(\frac { h }{ 48 } \)

\(\frac { 1 }{ √3 } \) = \(\frac { h }{ 48 } \)

cross multiply the fractions

48 = h√3

h = \(\frac { 48 }{ √3 } \)

h = 16√3

Therefore, the height of the tower is 16√3 m

Example 2:

A ladder of length 20 m is kept in against a wall of length 15 m such that their topmost point is in contact with one another and their bottom point is at a certain distance as shown in the figure. Find the angle subtended by the ladder on the floor.

Angle of Elevation 4

Solution:

The length of the ladder is PQ = 20 m

The height of the wall is PR = 15 m

We have to find angle PQR = angle subtended by the ladder on the floor.

Let angle PQR = θ

We know that,

sin θ = \(\frac { PR }{ PQ } \)

sin θ = \(\frac { 15 }{ 20 } \)

θ = sin-1 (\(\frac { 15 }{ 20 } \))

θ = sin-1 (\(\frac { 3 }{ 4 } \))

θ = 48.59°

The angle subtended by the ladder on the floor is 48.59°.

Example 3:

A man of height 1 m 30 cm is standing in front of a tree of height 30 m. Find the angle of elevation to be made by the man’s eyes to look at the topmost point of the tree if the man is standing at a distance of 5 m from the tree.

Solution:

Angle of Elevation 5

Here, PQ is the height of the tree = 30m

TR is the height of man = 1 m 30 cm = 1.30 m

RQ is the distance between the foot of the man and the tree = ST = 5 m

We have to find the angle of elevation, θ =?

We know that,

tan θ = \(\frac { (30 – 1.30) }{ 5 } \)

tan θ = 5.74

θ = tan-1 (5.74)

θ = 80.117°

The angle of elevation to be made by the man’s eye is 80.117°.

Example 4:

When the angle of elevation of the Sum is 45°, the shadow of a coconut tree is 15 m in length. What is the height of the coconut tree?

Solution:

Let AB denote the height of the coconut tree and BC denotes the length of the shadow.

Therefore, according to the problem ∠ACB = 45°, BC = 18 m.

Let the height of the coconut tree AB = x meters.

Now, tan 45° = \(\frac { AB }{ BC } \)

1 = \(\frac { x }{ 18 } \)

x = 18

Therefore, the height of the coconut tree is 18 meters.

FAQs on Angle of Elevation

1. What is the angle of elevation in trigonometry?

The angle of elevation is the angle between the horizontal line of the sight and the line of sight up to an object.

2. Differences between the angle of elevation and depression?

The angle of depression is opposite to the angle of elevation. In the angle of depression, the observer is standing at the height and the object is placed below the line of sight of the observer.

3. What is the formula for Angle of Elevation?

The formula for Angle of Elevation is given by

sin θ = \(\frac { Opposite side }{ Hypotenuse } =\frac { y }{ h } \)

cos θ = \(\frac { Hypotenuse }{ Adjacent side } =\frac { h }{ x } \)

tan θ = \(\frac { Opposite side }{ Adjacent side } =\frac { x }{ y } \)

Round off to Nearest 100 – Definition, Rules, Examples| How to Round Numbers to Nearest Hundred?

Round off to Nearest 100

Rounding is making a number simple but keeping its value nearest to what it was. The result of round-off is less accurate and easier to use especially while doing arithmetical operations. Likewise, rounding a number to the nearest hundred means making the units and tens place zeros and either increasing or decreasing the remaining part of the number by 1. Students can get the solved examples questions, definitions, and round-off rules in the below-mentioned sections of this page.

Round off to Nearest 100

Round off to the Nearest 100 is a process where you need to convert the given number into an easy form for various reasons. The obtained easy number is not the actual value but is an approximate value of the original number. Rounding to the nearest hundred means, you need to convert the given number to the nearest 100.

The purpose of rounding the numbers is it makes the numbers easier to understand and remember, the calculations become easier. The application of rounding is when you want to estimate an answer or try to find the most sensible guess, then rounding is used.

Rules for Rounding Numbers to Nearest 100

  • Rule 1: While round off to nearest 100, if the digit in the tens place is between 0 to 4 or <5, then the tens place is replaced by 0.
  • Rule 2: If the digit in the unit’s place is equal to 5 or greater than 5, then the tens place is replaced by 0 and the hundreds place is increased by 1.

How to Round Numbers to Nearest Hundred?

Get the detailed steps to rounding to the nearest 100. They are along the lines

  • Get the number we want to round.
  • Identify the digit in the tens place.
  • If the digit in the tens place is less than 5, then place zero’s in the tens, units place of the number.
  • If the digit in the tens place is more than or equal to 5, then place zero’s in the tens and units place, increase the hundredth place digit by 1.
  • Now, write the obtained number.

Also, Read: Rounding Decimals to Nearest Whole Number

Examples on Rounding Numbers to Nearest 100

Example 1:

Round the following numbers to the nearest 100.

(i) 148

(ii) 5520

(iii) 95

Solution:

(i) 148

Given number is 148

We see the digit in the tens place is 4 means that is less than 5. So, we round to the nearest multiple of a hundred which is less than the number. Keep zeros in the units and tens place.

Therefore, rounding of 148 to the nearest 100 is 100.

(ii) 5520

Given number is 5520

We can see that the digit in tens place is 2 which is less than 2. So, keep zero’s in the tens, units place, and write the remaining digits as it is.

Therefore, rounding of 5520 to the nearest 100 is 5500.

(iii) 95

Given number is 95

The tens digit of the given number is 9 that is more than 5. So, we need to place zeros in the tens, units place and increase the hundreds digit by 1. So, the obtained number is 100.

Example 2:

Round the following numbers to the nearest hundred.

(i) 696

(ii) 1,00,678

(iii) 12,05,896

Solution:

(i)696

Given number is 696

The digit in the tens place of the original number is 9. So, increase the hundreds digit by 1 and place zeros in the units, tens place.

Therefore, the obtained number is 700.

(ii) 1,00,678

Given number is 1,00,678

We can identify that the digit in the tens place of the number is 7 i.e > 5. Therefore, increase the digit in the hundreds place of the number by 1 and put zeros in the units, tens places.

Hence, the rounding off 1,00,678 to the nearest hundred is 1,00,700.

(iii) 12,05,896

Given number is 12,05,896

We see the digit in the tens place is 8, we round to the nearest multiple of hundred which is greater than the number. Hence, 12,05,896 is nearer to 12,05,900 than 12,05,800.

Example 3:

Round of To Nearest 100.

(i) 50

(ii) 255

(iii) 510

Solution:

(i) 50

Given number is 50

The digit in the tens position of the given number is 5. So, increase the hundreds position of the number by 1, place zeros in the tens, units place.

So, the round-off 50 to the nearest 100 is 100.

(ii) 255

Given number is 255

We choose the two multiple of 100 just greater than and just less than 255.

The nearest hundreds of 255 are 200, 300

255 – 200 = 55

300 – 255 = 45

As 300 has the lowest difference value.

So, 300 is the nearest 100 for 255.

(iii) 510

Given number is 510

We see the digit in the tens place is 1, we round to the nearest multiple of hundred which is less than the number. Hence, 510 is nearer to 500 than 600.

Pairs of Angles Definition, Examples | Different Types of Angle Pairs

Pairs of Angles

Angles are formed when two lines intersect each other at a point. The pair of angles are nothing but two angles. The angle pairs can relate to each other in various ways. Those are complementary angles, supplementary angles, vertical angles, adjacent angles, alternate interior angles, alternate exterior angles, and corresponding angles.

Pairs of Angles – Definition

The region between two infinitely long lines pointing a certain direction from a common vertex is called an angle. Which is the amount of turn is measured by an angle. The pairs of angles mean two angles. If there is one common line for two angles, it is known as angle pairs. Get the definitions and examples of all pairs of angles in the following section.

1. Complementary Angles

Two angles whose sum is 90° are called complementary angles and one angle is the complement of another angle.

Pairs of Angles 1

Here, ∠AOB = 20°, ∠BOC = 70°

So, ∠AOB + ∠BOC = 20° + 70° = 90°

Therefore, ∠AOB and ∠BOC are called complementary angles.

∠AOB is a complement of ∠BOC and ∠BOC is the complement of ∠AOB.

Example:

(i) Angles of measure 50° and 40° are complementary angles because 50° + 40° = 90°.

Thus, the complementary angle of 50° is the angle measure 40°. The complementary angle of 40° is the angle measure 50°.

(ii) Complementary of 60° is 90° – 60° = 30°

(iii) Complementary of 45° is 90° – 45°= 45°

(vi) Complementary of 25° is 90 – 25° = 65°

Working rule: To find the complementary angle of a given angle subtracts the measure of an angle from 90°.

So, the complementary angle = 90° – given angle

Also, Read:

2. Supplementary Angles

The pair of angles whose sum is 180° is called the supplementary angles and one angle is called the supplement of the other angle.

pairs of angles3

Here, ∠AOC = 120°, ∠COB = 60°

∠AOC + ∠COB = 120° + 60° = 180°

Therefore, ∠AOC, ∠COB are called supplementary angles.

∠AOC is the supplement of ∠COB, ∠COB is a supplement of ∠AOC.

Example:

(i) Angles of measure 90° and 90° are supplementary angles because 90°+ 90° = 180°

Thus, the supplementary angle of 90° is the angle of measure 90°.

(ii) Supplement of 100° is 180° – 100° = 80°

(iii) Supplement of 50° is 180° – 50° = 130°

(iv) Supplement of 95° is 180° – 95°= 85°

(v) Supplement of 140° is 180°- 140° = 40°

Working rule: To find the supplementary angle of the given angle, subtract the measure of angle from 180°.

So, the supplementary angle = 180° – given angle

3. Adjacent Angles

Two non-overlapping angles are said to be adjacent angles if they have a common vertex, common arm, and other two arms lying on the opposite side of this common arm so that their interiors do not overlap.

Pairs of Angles 4

In the above figure, ∠DBC and ∠CBA are non-overlapping, have BC as the common arm and B as the common vertex. The other arms BD, AB of the angles ∠DBC and ∠CBA are opposite sides of the common arm BC.

Hence, the arm ∠DBC and ∠CBA form a pair of adjacent angles.

4. Linear Pair of Angles

The angles are called liner pairs of angles when they are adjacent to each other after the intersection of two lines. Two adjacent angles are said to form a linear pair if their sum is 180°. The types of linear pairs of angles are alternate exterior angles, alternate interior angles, and corresponding angles.

Alternate interior angles

Two angles in the interior of the parallel lines and on opposite sides of the transversal. Alternate interior angles are non-adjacent and congruent.

pairs of angles 5

Alternate exterior angles

Two angles in the exterior of the parallel lines, and on the opposite sides of the transversal. Alternate exterior angles are non-adjacent and congruent.

Pairs of angles 6

Corresponding angles

The pair of angles, one in the interior and another in the exterior that is on the same side of the transversal. Corresponding angles are non-adjacent and congruent.

Pairs of angles 7

5. Vertical Angles

Two angles formed by two intersecting lines having no common arm are called the vertically opposite angles.

Pairs of angles 8

When two lines intersect, then vertically opposite angles are always equal.

∠1 = ∠3

∠2 = ∠4

Pair of Angles Examples

Example 1:

Suppose two angles ∠AOC and ∠ BOC form a linear pair at point O in a line segment AB. If the difference between the two angles is 40°. Then find both the angles.

Solution:

Given that,

∠AOC and ∠BOC form a linear pair

so, ∠AOC + ∠BOC = 180° —- (i)

∠AOC – ∠BOC = 40° —- (ii)

Add both equations

∠AOC + ∠BOC + ∠AOC – ∠BOC = 180° + 40°

2∠AOC = 220°

∠AOC = 220° / 2

∠AOC = 110°

Now, substitute ∠AOC in (i)

110° + ∠ BOC = 180°

∠BOC = 180° – 110°

∠BOC = 70°

Therefore, two angles are 70°, 110°.

Example 2:

Find the values of the angles x, y, and z in the following figure.

Pairs of angles 9

Solution:

From the given figure,

lines AD and EC intersect each other and ∠DOC and ∠AOE are vertically opposite angles

When two lines intersect, then vertically opposite angles are always equal.

So, ∠DOC = ∠AOE

Therefore, z = 40°

AD is a line

∠DOE and ∠AOE are adjacent angles. The sum of adjacent angles are 180°

So, ∠DOE + ∠AOE = 180°

y + 40° = 180°

y = 180°- 40°

y = 140°

And, lines AD and CE intersect

∠DOE, ∠COA are vertically opposite.

When two lines intersect, then vertically opposite angles are always equal.

So, ∠DOE = ∠COA

y = ∠COB + ∠BOAA

140° = x + 25°

140° – 25° = x

x = 115°

Hence, x = 115°, y = 140°, z = 40°

Example 3:

Identify the five pairs of adjacent angles in the following figure.

Pairs of angles 10

Solution:

Adjacent angles are the angles that have a common side, vertex, and no overlap.

So, (i) ∠AOD, ∠AOE are the adjacent angles

The common side is AO, the common vertex is O and OE, OD is not overlapping.

(ii) ∠AOD, ∠DOB is the adjacent angles

The common vertex is O, the common side is OD and OA, OB is not overlapping.

(iii) ∠DOB, ∠BOC is the adjacent angles

The common side is OB, the common vertex is O, and OD, OC are not overlapping.

(iv) ∠COE, ∠BOC are adjacent angles

The common side is OC, the common vertex is O, and OE, OB are not overlapping.

(v) ∠COE, ∠EOA are adjacent angles

The common vertex is O, the common side is OE, and OC, OA is not overlapping.

Converting Fractions to Decimals Examples | How to Convert Fractions to Decimals?

Converting Fractions to Decimals

Converting Fractions to Decimals means you need to press the fractional numbers in the form of exact decimals. You can use different methods like just divide the numerator by the denominator of the fraction apply the long division method or any other simple method. Check out those different forms of converting a fraction into a decimal and apply the best one to get the answers quickly and effortlessly.

What is meant by a Fraction?

The fraction is a part of the whole number and it is also expressed as a ratio of two numbers. The form of the fraction is a/b where ais the numerator, b is the denominator and it is not equal to 0. The different types of fractions are mixed, proper, improper fractions. A fraction is also known as the rational number.

Examples: 2/5, 10/8, 120/150.

What is meant by a Decimal?

A decimal is a number that has a dot or point in between the digits. In simple words, we can say that decimals are fractions with denominators as multiples of 10.

Examples: 5.26, 1.25, 0.06

Methods of Converting Fractions to Decimals

The three different ways of converting fractions to decimals are mentioned here.

Method 1: Divide the numerator by denominator

  • Get the fraction or mixed fraction
  • Convert the mixed fraction into a fraction
  • Just divide the numerator of the fraction by the denominator
  • While dividing add a dot to the quotient when you reach a point where the remainder is lesser than the divisor.

Example:

Convert 1/4 into decimal form

1/4 = 0.25

Method 2: Multiply both numerators, denominators by the same number 

  • Identify the fraction, you need to convert to the decimal number.
  • Multiply the denominator of the fraction by some number to get the denominator as the multiple of 10.
  • Find that number and multiply both the numerator and denominator by the same number.
  • Then, you will get the denominator as the multiple of 10.
  • After that, mark the decimal point after one place or two places or three places from right towards left if the given fraction’s denominator is 10 or 100 or 1000 respectively.

Example:

Convert 1/4 into the decimal form?

Given fraction is 1/4

1/4 = (1 x 25)/(4 x 25) = 25/100

= 0.25

Method 3: Using Long Division

  • Find the fractional number.
  • Divide the numerator of the fraction by denominator using the long division.
  • Then, you will get the decimal number as a quotient.

Example:

Convert 1/4 into the decimal form?

Given fraction is 1/4

Divide 1/4 using long division.

Converting Fractions To Decimals

So, 1/4 = 0.25

Have a look at the related articles

Example Questions on Fractions to Decimal Conversion

Example 1:

Convert the following fractions to decimals.

(i) 9(1/2)

(ii) 14/5

(iii) 125/6

Solution:

(i) The given mied fraction is 9(1/2)

= 19/2

= 9.5 using the division method

So, 9(1/2) = 9.5

(ii) The given fraction is 14/5

Multiply both numerator and denominator by 2

= (14 x 2)/(5 x 2)

= 28/10

= 2.8

So, 14/5 = 2.8

(iii) The given fraction is 125/6

Using the long division method

Converting Fractions To Decimals 1

So, 125/6 = 20.83333

Example 2:

Convert the following fractions to decimals.

(i) 3/10

(ii) 101/100

(iii) 856/9

Solution:

(i) The given fraction is 3/10

As the denominator is the multiple of 10, add point to the numerator.

So, 3/10 = 0.3

(ii) The given fraction is 101/100

As the denomintaor is 100, add point after two digits from the left side

So, 101/00 = 1.01

(iii) The given fraction is 856/9

Using the long division method

Converting Fractions To Decimals 2

So, 856/9 = 95.11111

Example 3:

Convert the following fractions to decimals.

(i) 18/7

(ii) 2/5

(iii) 3/5

Solution:

(i) The given fraction is 18/7

Using the long division method

Converting Fractions To Decimals 3

So, 18/7 = 2.571428571

(ii) The given fraction is 2/5

Multiply both numerator and denominator by 2

2/5 = (2 x 2)/(5 x 2)

= 4/10 = 0.4

So, 2/5 = 0.4

(iii) The given fraction is 3/5

Multiply both numerator and denominator by 2

3/5 = (3 x 2)/(5 x 2)

= 6/10 = 0.6

So, 3/5 = 0.6

Convex and Concave Polygons – Definitions, Properties, Formulas, and Examples

Convex and Concave Polygons

Polygon is a two-dimensional geometric figure which has a finite number of sides. Each side of the polygon is a straight line and all line segments form a closed figure. Convex and Concave Polygons are the two different types of polygons. A polygon that has all the interior angles lesser than 180°, then it is called the convex polygon. If the polygon contains one or more interior angles greater than 180°, then it is called the concave polygon. Check out the definitions, properties, and examples in the remaining sections of this article.

Convex and Concave Polygons

A polygon is said to be a convex polygon when the measures of the interior angles are lesser than 180 degrees. The vertices of the convex polygon always point outwards.

Convex and Concave Polygons 1

A polygon that measures at least one angle greater than 180 degrees is called the concave polygon. The vertices of this are inwards and outwards also. Convex, concave polygons are opposite to each other.

Convex and Concave Polygons 2

Other Types of Polygons

The various types of polygons are described here with their definitions.

  • Regular polygons: All sides of the polygons measure equal dimensions.
  • Irregular Polygons: All sides of the polygons measure unequal dimensions.
  • Quadrilateral Polygons: It is a four-sided quadrilateral
  • Convex Polygons: Examples are rhombus, parallelogram, square, rectangle, pentagon, etc
  • Concave Polygons

Also, read Regular and Irregular Polygons

Regular Convex Polygons

A regular convex polygon has all sides of equal length, all interior angles are the same and below 180 degrees. The distance between the center point to all vertices is equal. An example is square.

Irregular, Regular Concave Polygons

All the concave polygons are irregular concave polygons. Because concave polygons must have at least one angle greater than 180 degrees and irregular polygons measure interior angles differently. There is no concave polygon that is regular.

Convex and Concave Polygons – Formulas

The perimeter of a convex and concave polygon is the sum of all sides or the total region coved around the boundary.

Perimeter = Sum of all sides

Area of convex polygon is given as A = ½ | (x₁ • y₂ – x₂ • y₁) + (x₂ • y₃ – x₃ • y₂) + . . . + (xn • y₁ – x₁ • yn) |

Where (x₁, y₁), (x₂, y₂), (x₃, y₃), . . . (xn, yn) are the vertices of the convex polygon

Area of a concave polygon

Like regular polygons, there is no formula to calculate the area of the irregular concave polygons. Here, on each side, interior angles will be of different lengths. So, we need to split the polygons into triangles or other shapes to find the area.

Properties of Concave and Convex Polygons

  • The interior angles of convex polygons have to be less than 180 degrees. The concave polygon should have at least one reflex angle.
  • A concave polygon must have one vertex that points inwards to give the concave shape.
  • Sum of all interior angles of the convex polygon of n sides = (n – 2) x 180°
  • If a line segment is drawn crossing the concave polygon, then it will intersect the boundary more than two times.
  • Concave polygons have more than one diagonal that lie outside the boundary.
  • Concave polygons need to have one pair of sides joining a vertex that goes outside the vertex.

Interior, Exterior Angles

The interior angles are defined as the angles inside the polygon. The number of interior and exterior angles in a polygon are the same. According to the convex and concave polygon sum formula, for any n sided polygon, the sum of interior angles is (n – 2)180°. By knowing the sum, divide the sum by a total number of sides to get each interior angle measurement.

The exterior angle is defined as the angle formed by extending the side of the polygon. We already know that the sum of exterior angles of any polygon is 360°. Divide the sum by the total number of sides to get the measure of each exterior angle of Convex and Concave Polygons.

Examples of Convex and Concave Polygon

Example 1:

Find the area of the convex polygon with three sides, and vertices are (5, 7), (8, 4), (-2, 1).

Solution:

Given that,

Vertices of the polygon are (5, 7), (8, 4), (-2, 1).

(x₁, y₁) = (5, 7), (x₂, y₂) = (8, 4), (x₃, y₃) = (-2, 1)

The formula to calculate the area of a convex polygon is A = ½ | (x₁ • y₂ – x₂ • y₁) + (x₂ • y₃ – x₃ • y₂) + (x₃ • y₁ – x₁ • y₃) |

= ½ | (5 • 4 – 8 • 7) + (8 • 1 – (-2) • 4) + (-2 • 7 – 5 • 1) |

= ½ | (20 – 56) + (8 + 8) + (-14 – 5) |

= ½ | -36 + 16 – 19 |

= ½ | -55 + 16 |

= ½ | -39 |

= 19.5

Therefore, the area is 19.5 sq units.

Example 2:

Calculate the perimeter and area of the pentagon has a side length of 4 cm.

Solution:

Given that,

The side length of pentagon s = 4 cm

n = 5

The regular polygon area formula is A = n x s² x cot(π/n) / 4

A = 5 x 4² x cot(π/5) / 4

= 5 x 16 x 1.3763 / 4

= 110.104 / 4

= 27.526 cm²

Perimeter p = Sum of all sides

= 4 + 4 + 4 + 4 + 4

= 20 cm

Hence the area, perimeter of the pentagon is 27.526 cm², 20 cm.

Example 3:

Find the perimeter, area of the below given concave polygon.

Convex and Concave Polygons 3

Solution:

The perimeter of the polygon = sum of all sides

= 4 + 5 + 2 + 2 + 5

= 18 cm

To find the area divide the given polygon into a square and two triangles by drawing a dotted line.

Convex and Concave Polygons 4

Area of square = side²

= 4² = 16 cm²

Area of triangle = 1/2 x base x height

hypotenuese² = base² + height²

2² =1² + height²

4 = 1 + h²

h² = 4 – 1

h² = 3

h = √3

So, area = 1/2 x 1 x √3

= √3/2 cm²

So, area of polygon = area of square + 2 x area of the triangle

= 16 + 2(√3/2)

= 16 + √3

Therefore, the perimeter, area of the concave polygon is 18 cm, (16 + √3) cm²

Example 4:

Find the measure of each interior, exterior angle of a regular octagon.

Solution:

Octagon has 8 sides

The sum of interior angles of polygon = (n – 2)180°

= (8 – 2) 180°

= 6 x 180°

= 1080°

The measure of each interior angle of an octagon = sum of interior angles/ number of sides

= 1080°/8

= 135°

The sum of exterior angles = 360°

The measure of each exterior angle of an octagon = sum of exterior angles/ number of sides

= 360/8

= 45°

So, the measure of each interior angle, exterior angle of the octagon is 135°, 45°. Hence, it is a convex polygon.

FAQs on Convex and Concave Polygons

1. Write the differences between concave and convex polygons?

Each and every polygon is either convex or concave. The main difference between them is angles. For a polygon to be convex, it must have all interior angles lower than 180 degrees. If any of the interior angles is greater than 180 degrees, then it is a concave polygon.

2. How do you know if a polygon is convex?

Get the measure of each interior angle of the polygon. If all the angles are below 180 degrees, then you can say it is a convex polygon.

Roman Numerals Chart | Rules, Conversion from Roman Numerals to Numbers and Vice Versa, Examples

Roman Numerals

Roman Numerals are the numeral system which is originated in ancient Rome. It is a decimal or base 10 number system. It is an additive or subtractive system in which letters are used to denote some base numbers and arbitrary numbers in the number system and denoted using a different combination of symbols. Here, students can check the Roman Numerals Chart, how to convert roman numerals to numbers, and solved examples in the following sections.

Roman Numerals List

Here provided are some basic roman numerals.

Roman Numerals 1

Roman Numerals Chart:

Roman numerals Chart has particular roman numbers for the decimal numbers from 1 to 1000.

Roman Numerals 2    Roman Numerals 3    Roman Numerals 4

Rules for formation of Numbers

We have three different rules to form the roman numerals as numbers. They are multiplication rule, addition rule, and subtraction rule.

Multiplication Rule:

When a roman symbol is repeated in sequence, then we have to multiply the value of the numeral by the number of times it is repeated. A symbol cannot be repeated more than three times in series.

Example:

II = 1 x 2 = 2

XX = 10 x 2 = 20

III = 1 x 3 = 3

We cannot express 4 as IIII. Because the symbol cannot be repeated more than three times.

Addition Rule:

If a smaller number is located to the right of a larger number, then you need to add numbers.

Example:

VII = 5 + 1 + 1 = 7

VIII = 5 + 1 + 1 + 1 = 8

XXII = 10 + 10 + 1 + 1 = 22

Subtraction Rule:

Writing a smaller number to the left side of a larger number means that the smaller number has to be subtracted from the larger number. The symbol I can be subtracted from V and X and X can be subtracted from L and C.

Example:

IV = 5 – 1 = 4

IL = 50 – 1 = 49

We do not repeat V twice to get 10. We already have a symbol for 10. So VV for writing 10 is not correct.

We do not subtract 5 from any symbol. VX is not correct.

Roman Numerals Questions with Solutions

Example 1:

Write the Roman Numerals for 57.

Solution:

The given decimal number is 57

Break up the number into Tens and Ones.

57 = 50 + 7

The symbol for 50 is L

7 = 5 + 2 = 5 + 1 + 1

The symbol for 7 is VII.

57 convert roman numerals as LVII.

Example 2:

Write the number for XXIV.

Solution:

Given roman numeral is XXIV

V = 5

IV = 5 – 1 = 4

X = 10

XX = 10 + 10 = 20

Therefore, XXIV = 20 + 4

XXIV = 24.

Example 3:

Write the Roman Numerals for 18.

Solution:

The given decimal number is 18.

Break up the number into Tens and Ones.

18 = 10 + 8

The symbol for 10 is X.

8 = 5 + 1 + 1 + 1

The symbol for 5 is V, 1 is I

So, the symbol for 8 is VIII

18 convert roman numerals as XVIII.

Example 4:

Write the number for XXXIX.

Solution:

Given roman numeral is XXXIX

X = 10

IX = 10 – 1 = 9

XXX = 10 x 3 = 30

Therefore, XXXIX = 30 + 9 = 39

FAQs on Roman Numerals

1. How do you write roman numerals?
Roman numerals can be written by using seven different alphabets. They are I for 1, V for 5, X for 10, L for 50, C for 100, D for 500, M for 1000. By using these symbols, you can write all roman numerals easily.

2. How to evaluate 550 in roman numerals?

550 = 500 + 50

500 = D, 50 = L

Therefore, 550 = DL.

3. What is the use of roman numerals?

Roman numerals can be used for labeling the name or position of a person or an object. Examples are Kiran came IInd in the class. Prince Charles III, Schools have Classes from VIth to Xth.

4. Write roman numerals from 1 to 10?

The roman numerals from 1 to 10 are I, II, III, IV, V, VI, VII, VIII, IX, and X.

Correct to Two Decimal Places Definition, Rules, Examples | Rounding Decimals to 2 Places

Correct to Two Decimal Places

Looking for help on Correcting Decimals to 2 Places? If so, you have arrived at the right place and get complete knowledge regarding the Rounding Off Decimals to Hundredths Place. Here you will find how to round off to hundredths place or Correct to Two Decimal Places. Get the definition, rules, detailed step-by-step process of rounding numbers to the nearest hundredths place. Solved Examples on Rounding Decimals to 2 Places help you understand the concept clearly and learn the approach used to solve them.

Correct to Two Decimal Places – Definition

The process of Correct to Two Decimal Places is nothing but rounding the to hundredths. Rounding off to decimal places is a technique used to find the approximate values of the decimal number. Here, the decimal numbers are rounded off to 2 decimal places to make them easy to read, understand, and manageable instead of having lengthy string decimal places. We will keep the original 1st decimal number as it is, may change the second decimal number, and eliminate the remaining decimal numbers.

Rules for Rounding off to Two Decimal Places

For rounding the given numbers to the nearest hundredths, you need to check the below-provided rules.

  • Rule 1: If the digit in the thousands place is less than 5, then remove the following digits or place 0 in place of them
  • Rule 2: When the digit in the thousands place is equal to or more than 5, then the digit in the hundredths place is increased by 1 and the following digits are replaced by 0.

Also, check:

How to Correct to Two Decimal Places?

Follow the below-listed procedure to learn Rounding Decimals to Two Places. They are along the lines

  • Identify the number for which you need to round off to two decimal places.
  • See the digit in the thousands place of the given number.
  • If the digit is equal to or greater than 5, then add 1 to the digit in the hundredth place of the number and remove the following digits.
  • When the digit is smaller than 5, then replace the following digits with 0.

Solved Examples on Correcting Decimals to 2 Places

Example 1:

Round off the following to the two decimal places.

(i) 186. 256

(ii) 25. 532

Solution:

(i) 186. 256

The given number is 186. 256

We can see the digit in the thousandths place is 6 then round it to the nearest hundredths which is greater than the given decimal number. Since 6 > 5 then the decimal number is rounded to 186.260.

Therefore, the solution is 186.260.

(ii) 25. 532

The given number is 25. 53256

We can identify the digit in the thousandths place is 2 then round it to the nearest hundredths which is smaller than the given decimal number. Since 2 < 5 then the decimal number is rounded to 25.5300.

Therefore, the solution is 25.53.

Example 2:

(i) 120.085

(ii) 12,856.558

Solution:

(i) 120.085

The given number is 120.085

The digit in the thousands place is 5 which is equal to 5.

So, increase the hundreds place digit by 1 i.e 8 + 1 = 9, and remove the following digits.

Therefore, the solution is 120.09.

(ii) 12,856.558

The given number is 12,856.558

We can see the digit in the thousandths place is 5 then round it to the nearest hundredths which is greater than the given decimal number. Since 5 = 5 then the decimal number is rounded to 12,856.56.

Therefore, the solution is 12,856.56.

Example 3:

(i) 0.042

(ii) 12.567

Solution:

(i) 0.042

The given number is 0.042

The digit in the thousand’s place of the given number is 2

Since 2 is less than 5, then replace the following digits with 0.

Therefore, the solution is 0.04.

(ii) 12.567

The given number is 12.567

The digit in the thousand’s place of the given number is 7

Since 2 > 5 add 1 to the hundredths place of the number and remove the following digits.

So, the rounds off number is 12.57.

Hexadecimal Addition and Subtraction | How to Add & Subtract Hexadecimal Numbers? | Hexadecimal Arithmetic Examples

Hexadecimal Addition and Subtraction

In the hexadecimal number system, the numbers are expressed with the base 16. Hexadecimal is also called as Hex. It is like decimal, binary or octal numbers. The list of 16 hexadecimal numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F. Here, we are going to learn how to perform addition, subtraction operations in between two hexadecimal numbers with the examples for the better understanding of the concept. So, interested people can check out this complete page.

What is Hexadecimal Number System?

A hexadecimal number is a number having a base of 16. These numbers are also called the 16 number system. It has 16 different symbols, 0 to 9 represent the binary values, A, B, C, D, E, F represent 10 to 15 values respectively. Each position in the hexadecimal number represents 0 power of the base 16. The last position in the number represents an x power of base 16.

Examples of Hexadecimal Numbers:

1. B84F

The decimal value of B84F is 11 x 16 + 8 x 16 + 4 x 16 + 16 x 16

= 176 + 128 + 64 + 256

= 624

2. Convert 857 to hexadecimal

857 ÷ 16 = 53.5625

0.5625 x 16 = 9 (Remainder 9)

53 ÷ 16 = 3.3125

0.3125 x 16 = 5 (Remainder 5)

3 ÷ 16 = 0.1875

0.1875 x 16 = 3 (Remainder 3)

Read remainders from bottom to top

So, 857 = (359)16

Also, Read: Binary Subtraction

Procedure for Adding & Subtracting in Hexadecimal

Below provided are the simple steps that are helpful to compute the addtion and subtraction of two hexadecimal numbers.

1. Hexadecimal Numbers Addition

  • Write two hexadecimal numbers one after another in two different lines
  • Begin adding from the rightmost digits.
  • If the digit is in the form of an alphabet then convert it to the respective decimal number to make the process easy
  • Add those digits and convert the sum to the hexadecimal
  • If you got the carry, then represent it on the top of the first number next digit and result on the bottom of the second number added digit.
  • Continue the process until you left nothing on the left side.

We can also add two hexadecimal numbers by following this table.

Hexadecimal Addition and Subtraction

For Example:

926 + 1A2

9 2 6

(+) 1 A 2

=    A C 8

So, 926 + 1A2 = AC8

2. Hexadecimal Numbers Subtraction

  • Write two hexadecimal numbers in different lines
  • Subtraction starts from the rightmost digits of the numbers.
  • Convert the alphabets into decimals and subtract two digits and again convert the difference value as hexadecimal.
  • In case the first number digit is smaller than the second number digit, then barrow from the left side digit.
  • The borrowed value is always 16 as its base is 16. Then add borrowed value and first number digit and subtract.
  • Don’t forget to mention the borrowed value on the top of the first number digit.
  • After borrowing, the left side digit decreased by 1.
  • Repeat the process till you have nothing remaining on the left side.

For Example:

938 – 1A2

8 – 2 = 6

3 – A(10) = (16 + 3) – 10 = 19 – 10 = 9

(9 – 1) – 1 = 8 – 1 = 7

So,  938 – 1A2 = 796.

Hexadecimal Addition and Subtraction Examples

Example 1:

Evaluate (1AB2)16 + (2198)16

Solution:

Given expression is (1AB2)16 + (2198)16

From the table,

2 + 8 = A

B + 9 = 4 and 1 is carry

1 + A + 1 = C

1 + 2 = 3

Therefore, (1AB2)16 + (2198)16 = 3C4A

Example 2:

Find subtraction of (B84F)16 and (A53)16.

Solution:

F means 15. F – 3 = 15 – 3 = 12 = C

4 + 16 = 20 – 5 = 15 = F

8 – 1 = 7

7 + 16 = 23 – A = 23 – 10 = 13 = D

8 – 1 = 7
23
10 7 20

B 8 4 F

(-) 0 A 5 3

= 7 D F C

So, (B84F)16 – (A53)16 = (7DFC)16

Example 3:

Find the addition, subtraction of (AB53)16, (155)16

Solution:

The addition of numbers is (AB53)16 + (155)16

3 + 5 = 8

5 + 5 = 10 = A

B + 1 = 11 + 1 = 12 = C

A + 0 = 10 + 0 = 10 = A

So, (AB53)16 + (155)16 = (ACA8)16

Subtraction of numbers is (AB53)16 – (155)16

(3 + 16) – 5 = 19 – 5 = 14 = E

(5 – 1) – 5 = 4 – 5

(4 + 16) – 5 = 20 – 5 = 15 = F

(B – 1) – 1 = (11 – 1) – 1 = 10 – 1 = 9

A – 0 = A

A B 5 3

(-) 0 1 5 5

= A 9 F E

So, (AB53)16 – (155)16 = (A9FE)16

Example 4:

(i) Calculate (9AB)16 + (12C)16

(ii) Compute (CB5)16 – (223)16

Solution:

(i) (9AB)16 + (12C)16

B + C = 11 + 12 = 23 = 7 and 1 is carry

1 + A + 2 = 3 + 10 = 13 = D

9 + 1 = 10 = A

So, (9AB)16 + (12C)16 = (AD7)16

(ii) (CB5)16 – (223)16

5 – 3 = 2

B – 2 = 11 – 2 = 9

C – 2 = 12 – 2 = A

So, (CB5)16 – (223)16 = (A92)16

Trigonometrical Ratios of 90 Degree Minus Theta | Relation between Quadrants | Solved Examples

Trigonometrical Ratios of 90 Degree Minus Theta

Are you looking for any material to know the relation between all Trigonometrical Ratios of 90 Degree Minus Theta? Then, you can relax now. On this page, we have enclosed the detailed information on the relation between Trigonometric Ratios of (90° – θ) along with the proofs. Get the example questions and step-by-step solutions in the following sections of this article. Check out the simple formula to memorize the Trigonometric Functions.

How to Determine the Trigonometric Ratios of 90 Degree Minus Theta?

Here, you can see Trigonometrical Functions of 90 Degree Minus Theta can be determined. As per the ASTC “All Silver Tea Cups” or “All Students Take Calculus”

A means All, S means “Sinθ, Cosecθ”, T means “Tanθ, Cotθ”, C means Cosθ, Secθ.

The pictorial representation of the ASTC formula is as follows:

ASTC Formula

From the above picture, (90° – θ) falls in the first quadrant.

sin (90° – θ) = cos θ

cos (90° – θ) = sin θ

tan (90° – θ) = cot θ

cosec (90° – θ) = sec θ

sec (90° – θ) = cosec θ

cot (90° – θ) = tan θ

Evaluate Trigonometrical Ratios of 90 Degree Minus Theta

1. Evaluate Sin(90° – θ)?

To evaluate sin (90° – θ), we have to consider the following important points.

  •  (90° – θ) will fall in the 1st quadrant.
  •  When we have 90°, “sin” will become “cos”.
  •  In the 1st quadrant, the sign of “sin” is positive.

Considering the above points, we have

Sin (90° – θ) = Cos θ

2. Evaluate Cos(90° – θ)?

To evaluate cos (90° – θ), we have to consider the following important points.

  • (90° – θ) will fall in the 1st quadrant.
  •  When we have 90°, “cos” will become “sin”.
  •  In the 1st quadrant, the sign of “cos” is positive.

Considering the above points, we have

Cos (90° – θ) = Sin θ

3. Evaluate Tan(90° – θ)?

To evaluate tan (90° – θ), we have to consider the following important points.

  •  (90° – θ) will fall in the 1st quadrant.
  •  When we have 90°, “tan” will become “cot”.
  •  In the 1st quadrant, the sign of “tan” is positive.

Considering the above points, we have

Tan (90° – θ) = Cot θ

4. Evaluate Cot(90° – θ)?

To evaluate cot (90° – θ), we have to consider the following important points.

  •  (90° – θ) will fall in the 1st quadrant.
  •  When we have 90°, “cot” will become “tan”
  •  In the 1st quadrant, the sign of “cot” is positive.

Considering the above points, we have

Cot (90° – θ) = Tan θ

5. Evaluate Cosec(90° – θ)?

To evaluate Cosec (90° – θ), we have to consider the following important points.

  • (90° – θ) will fall in the 1st quadrant.
  •  When we have 90°, “Cosec” will become “sec”.
  •  In the 1st quadrant, the sign of “Cosec” is positive.

Considering the above points, we have

Cosec (90° – θ) = Sec θ

6. Evaluate Sec(90° – θ)?

To evaluate sec (90° – θ), we have to consider the following important points.

  •  (90° – θ) will fall in the 1st quadrant.
  •  When we have 90°, “sec” will become “cosec”.
  •  In the 1st quadrant, the sign of “sec” is positive.

Considering the above points, we have

Sec (90° – θ) = Cosec θ

More Related Articles:

Solved Examples on Trigonometric Ratios of 90° – θ

Example 1:

Find the value of Tan 45°?

Solution:

Tan 45° = Tan (90° – 45°)

We know that Tan (90° – θ) = Cot θ

So, Tan 45° = Cot 45°

= 1  [Cot 45° = 1]

Therefore, Tan 45° = 1.

Example 2:

Find the value of Sin 65°?

Solution:

Sin 65° = Sin (90° – 25°)

We know that Sin (90° – θ) = Cos θ

So, Sin 65° = Cos 25°

= 0.906308 [ Cos 25° = 0.906308]

Therefore, Sin 65° = 0.906308

Example 3:

Find the value of Cot 80°?

Solution:

Cot 80° = Cot (90° – 10°)

We know that Cot (90° – θ) = Tan θ

So, Cot 80° = Tan 10°

= 0.17633 [ Tan 10° = 0.17633]

Therefore, Cot 80° = 0.17633

Example 4:

Find the value of Cos 50°?

Solution:

Cos 50° = Cos (90° – 40°)

We know that Cos (90° – θ) = Sin θ

So, Cos 50° = Sin(40°)

= 0.64278760968 [ Sin(40°) = 0.64278760968]

Therefore, Cos 50° = 0.64278760968.

What is Place Value? – Definition, Properties of Place, Face Value, Place Value, Examples

Place Value

Place value is nothing but the position or place of a digit in the decimal number. In a number, every digit has someplace. The positions of digits in number starts from one’s place. The order of place value of digits of a number of right to left is units, tens, hundreds, thousands, ten thousand’s, and so on. Generally, a number is formed by grouping digits together.

Place, Place Value and Face Value Definitions

In a number, every digit has a fixed position known as the digit place. And each digit has a value depending on its place called the place value of the digit. Face value of a digit for any place in the value of the digit itself.

Place Value of a digit = (face value of the digit) x (value of the place)

Place Value Table

Place value chart is helpful to ensure that the digits are in the correct places. Place value tells us how much each digit stands for. The place value table is mentioned here.

place value 1

Properties of Place Value

  • The place value of every one-digit number is the same as the face value.

The place value and face value of 1, 2, 3, 4, 5, 6, 7, 8, and 9 are 1, 2, 3, 4, 5, 6, 7, 8, and 9. The place value of 0 is always 0. It may present at any place in the number, its value is always zero.

Example:

(i) In the numbers 105, 270, 1025 the place of value 0 is 0.

(ii) The place of 1 in 251 is 1, 7 in 8567 is 7.

  • In a two-digit number, the place value of the ten-place digit is equal to 10 times of the digit.

Example:

(i) The place value of 6 in 67 is 6 x 10 = 60

(ii) The place value of 2 in 526 is 2 x 10 = 20

  • In the number 567, the digit 7 is at one’s place, the digit 6 is at ten’s place and the digit 5 is at the hundred’s place.

So, the place value of 7 is 7, 6 is 6 x 10 = 60, and 5 is 5 x 100 = 500

Thus, for the place value of a digit, the digit is multiplied by the place value of 1 it has to be that place.

Example:

In the number 286,

The place value of 6 is 6 x 1 = 6

The place value of 8 is 8 x 10 = 80

The place value of 2 is 2 x 100 = 200

  • Now it is the general law that the digit possesses its place value as the product of the digit and place value of one to be at that position.

Examples:

(i) In the number 3578,

The place value of 8 is 8 x 1 = 8 because 8 is at the unit’s place.

The place value of 7 is 7 x 10 = 70 because 7 is at ten’s place.

The place value of 5 is 5 x 100 = 500 because 5 is at hundred’s place

The place value of 3 is 3 x 1000 = 3000 because 3 is at thousand’s place.

(ii) In the number 58762, the place value of each digit is as follows

2 is 2 x 1 = 2

6 is 6 x 10 = 60

7 is 7 x 100 = 700

8 is 8 x 1000 = 8000

5 is 5 x 10000 = 50000

(iii) Find the place value of digits 30589

place value 2

Also, check out

Example Questions on Place Value

Example 1:

Write the place value of the given numbers.

(i) 5 in 38956

(ii) 2 in 2587

(iii) 6 in 6845321

Solution:

(i) The given number is 38956

The place value of 5 in 38956 is 5 x 10 = 50. Because 5 is at tens place.

(ii) The given number is 2587

The place value of 2 in 2587 is 2 x 1000 = 2000. Because 2 is at thousands place.

(iii) The given number is 6845321

The place value of 6 in 6845321 is 6 x 1000000 = 60,00,000. Because 6 is at ten lakhs place.

Example 2:

Write the place value of a highlighted digit in the given numbers

(i) 2589

(ii) 67525

(iii) 2515963

Solution:

(i) The given number is 2589

The place value of highlighted 8 in 2589 is 8 x 10 = 80 as it is located at the tens position.

(ii) The given number is 67525

The place value of highlighted digit 5 in 67525 is 5 x 100 = 500 as it is located at the hundred’s position.

(iii) The given number is 2515963

The place value of highlighted digit 1 in 2515963 is 1 x 10000 = 10000 as it is located at the ten thousand’s position.

Example 3:

Circle the following.

(i) Digit at the hundreds place in 5289.

(ii) Digit at the lakhs place in 2563891

(iii) Digit at the units place in 5280

Solution:

(i) In the number 5289,

9 is at units place, 8 is at tens place, 2 is at hundreds place and 5 is at thousand’s place

So, Digit at the hundreds place in 5289 is 2.

(ii) In the number 2563891,

1 is at units place, 9 is at tens place, 8 is at hundreds place, 3 is at thousand’s place, 6 is at ten thousand’s place, 5 is at lakhs place, 2 is at ten lakhs place.

Therefore, Digit at the lakhs place in 2563891 is 5.

(iii) In the number 5280,

0 is at the unit’s place, 8 is at the ten’s place, 2 is at the hundred’s place and 5 is at the thousand’s place.

Hence, the digit at the units place in 5280 is 0.

Example 4:

Find the place value of 5 in the given numbers.

(i) 6,00,521

(ii) 5,23,168

(iii) 2,05,387

Solution:

(i) The given number is 6,00,521

The place value of 5 is 5 x 100 = 500.

(ii) The given number is 5,23,168

The place value of 5 is 5 x 10,000 = 50,000.

(iii) The given number is 2,05,387

The place value of 5 is 5 x 1000 = 5000.

FAQs on Place Value

1. Write the differences between place value and face value?

Place value means the position of a particular digit in the number but face value represents the exact value of a digit in that number. For example in the number 2556, the place value of 2 is thousands but the face value is 2.

2. Define place value with an example?

The place value is the position of a digit in a number. The place values of digits are represented as ones, tens, hundreds, thousands, ten thousand, and so on. The example is the place value of 8 in 589 is 8 tens i.e 80.

Cuboid – Definition, Properties, Formulas and Examples

Cuboid

A cuboid is a three-dimensional shape having three axes. It has 3 faces which are convex polyhedrons, 12 edges, 12 vertices. Find the different formulas of the cuboid like perimeter, total surface area, lateral surface area, base surface area, and diagonal in the following sections of this page. One can also find the cuboid definition, properties, and solved example questions here.

Cuboid Definition

The cuboid is a closed three-dimensional geometric figure having 6 rectangular regions. Each rectangular region is called the face. The point of intersection of three edges in the cuboid is called the vertices or corners. The sides of all rectangular faces are called the edges of the cuboid. Some of the examples for the objects in the cuboid shape are matchbox, shoebox, bricks, matrices.

The shape of the cuboid is shown here.

 

cuboid

Faces, Edges, Vertices of the Cuboid

The cuboid is made up of 6 rectangular faces, 12 edges, 8 corners. The faces, edges, and corners for the above-mentioned cuboid image are as follows:

  • The six faces are ABCD, BDEF, ABGF, AGCH, CDHE, EFGH.
  • Eight corners or vertices are A, B, C, D, E, F, G, H.
  • Twelve edges and opposite sides of the rectangle are AB = CD = EH = GF, AC = BD = EF = GH, DE = BF = CH = AG.

Cuboid Formulas

The cuboid formulas are provided-below. Get the total surface area, diagonal, perimeter, and volume of the cuboid. Let us consider l, b, h are the length, breadth, and height of the cuboid respectively.

Cuboid Surface Area

The surface area is nothing but the total region covered by all the faces. Generally, the cuboid surface area is classified into two types they are lateral surface area and total surface area. It is also defined as the sum of areas of six faces of the cuboid.

Lateral Surface Area

The lateral surface area is the sum of the areas of all faces except the top and bottom faces.

Cuboid Lateral Surface Area = (Area of ABCD + Area of BDEF + Area of EFGH + Area of AGCH)
= (b × h) + (b × h) + (l × h) + (l × h)

= 2h(l + b)

Total Surface Area

Cuboid Total Surface Area is the sum of the faces.

Total Surface Area of the Cuboid = (Area of ABCD + Area of BDEF + Area of EFGH + Area of AGCH + Area of ABGF + Area of CDHE)

= (l × b) + (l × b) + (b × h) + (b × h) + (l × h) + (l × h)

= 2lb + 2bh + 2hl

= 2(lb + bh + lh)

Cuboid Diagonal

The length of the diagonal of a cuboid is along the lines.

Diagonal = √(l² + b² + h²)

Cuboid Perimeter

The perimeter is the sum of the edges of all edges.

The perimeter of the Cuboid = AB + BF + FE + BD + DE + CD + CH + AC + GH + AG + GF + HE

= l + l + l + l + b + b + b + b + h + h + h + h

= 4(l + b + h)

Cuboid Volume

Cuboid volume is the product of the base area and height.

Volume = length x breadth x height

= lbh

Read More Related Articles:

Properties of Cuboid

Here is the list of Properties of Cuboids such as faces, edges, vertices, angles, etc. They are as follows

  • A cuboid has 6 faces, twelve edges, and 8 vertices.
  • It has rectangular-shaped faces.
  • The angles are plane and at a right angle.
  • Opposite edges are parallel to each other.

Solved Examples on Cuboid

Example 1:

Calculate the volume, diagonal, surface area, and perimeter of the cuboid, if the cuboid length is 8 cm, width is 5 cm, and height is 10 cm.

Solution:

Given that,

Cuboid length l = 8 cm

breadth b = 5 cm

height h = 10 cm

Cuboid volume v = lbh

= 8 x 5 x 10

= 400 cm³

Diagonal of the cuboid = √(l² + b² + h²)

= √(8² + 5² + 10²)

= √(64 + 25 + 100)

= √(189)

= 3√(21) cm

Perimeter = 4(l + b +h)

= 4(8 + 5 + 10)

= 4(23)

= 92 cm

Total Surface Area = 2(lb + bh + lh)

= 2(8 x 5 + 5 x 10 + 8 x 10)

= 2(40 + 50 +80)

= 2(170)

= 340 cm²

∴ Perimeter, diagonal, total surface area, and volume of the cuboid are 92 cm, 3√(21) cm, 340 cm² & 400 cm³.

Example 2:

The lunchbox measures 20 cm long, 10 cm wide, and 5 cm high. what is the total surface area, the lateral surface area of the box?

Solution:

Given that,

Lunchbox length l = 20 cm

Width b = 10 cm

Height h = 5 cm

Total Surface Area of the lunchbox = 2(lb + bh + lh)

= 2(20 x 10 + 10 x 5 + 20 x 5)

= 2(200 + 50 + 100)

= 2(350)

= 700 cm²

Lateral Surface Area of the lunchbox = 2h(l + b)

= 2 x 5(20 + 10)

= 10(30)

= 300 cm²

∴ The total surface area of the lunchbox is 700 cm², lateral surface area is 300 cm².

Example 3:

If the cuboid volume is ∛(126) m³, breadth is 2 m and height is 3 m. Find its length?

Solution:

Given that,

The volume of the cuboid = ∛(126) m³

lbh = ∛(126) m³

l x 2 x 3 = ∛(126)

l x 6 = ∛(126)

l = ∛(126) / 6

l = 21 m

∴ The cuboid length is 21 m.

Example 4:

If the volume of a room is 792 m³ and the area of the floor is 132 m², find the height of the room.

Solution:

Given that,

The volume of a room = 792 m³

Area of the floor = 132 m²

Height of the room = Volume of a room / Area of the floor

= 792/132

= 6 m

Therefore, the height of the room is 6 m.

Conversion of Minutes into Hours Definition, Formula, Examples | How to Convert Min to Hr?

Conversion of Minutes into Hours

Here you can convert minutes to hours easily in a fraction of seconds. We are providing three simple and easy steps that are helpful for the students during their calculations. We already know that 1 hour means 60 minutes. So, divide the number of minutes by 60 to get hours. Get the solved example questions on Conversion Of Minutes Into Hours in the following sections. The formula to convert minutes into hours is the number of minutes/60.

Minutes to Hours Conversion Definition

Conversion of minutes into hours means you need express minutes measurement of time as hours measurement. As we know, both minutes and hours are the unit measurement of the time. One minute is 1/60th of the hour. So, multiply the minutes by (1/60) to get the hours version.

60 minutes = 1 hour

So, 1 minute = 1/60 hour

Ways to Convert Minutes into Hours

We have 3 different and simple ways for conversion of minutes into hours. Those methods with detailed step by step explanation is mentioned below.

Method 1:

  • Make a note of the number of minutes from the question.
  • Multiply the given number of minutes by (1 hour /60 minutes)
  • After performing the multiplication, the product is answer.

Method 2:

  • Observe from the questions, how many minutes are to be converted to hours.
  • We know that 1 hour = 60 minutes, 1 minute = 1/60 hour
  • So, divide minutes by 60 to check the hour conversion of minutes.

Method 3:

  • Get an idea of how many minutes are to be converted into hours.
  • And use the calculator to get the solution.

See More Related Articles:

Solved Examples on Minutes to Hours Conversions

Example 1:

Convert 125 minutes into hours.

Solution:

Given that,

Number of minutes = 125

The formula to convert minutes into hours = Number of minutes/60

So, 125 minutes = 125/60

= 25/12 hour

Therefore, 125 minutes = 25/12 hours.

Example 2:

Convert 60 minutes into hours

Solution:

Given that,

Number of minutes = 60 minutes

The formula to convert minutes into hours = Number of minutes/60

So, 60 minutes = 60/60

= 1 hour
Therefore, 60 minutes = 1 hour

Example 3:

Convert 45 minutes into hours.

Solution:

Given that,

Number of minutes = 45 minutes

We know that,

60 minutes = 1 hour

1 minute = 1/60 hour

So, 45 minutes = 45/60

= 3/4

Therefore, 45 minutes = 3/4 hour

FAQs on Conversion Of Minutes Into Hours

1. How do you convert minutes into hours and minutes?

Just simply divide the number of minutes by 60 to get the solution. The division gives minutes in the form of hours.

2. What is the rule for converting hours to minutes?

To convert the hours into minutes, multiply hours by 60. The time measurement in minutes is equal to hours measurement when it is multiplied by 60.

3. What is the formula to convert minutes into hours?

The simple formula to convert minutes measurement into hours measurement is the number of minutes/60. Because 1 hour = 60 minutes. So, 1 minute = 1/60 hour.

Signed Magnitude Representation of Binary Numbers with Examples | How to Represent Sign Magnitude Form?

Signed Magnitude Representation

In mathematics, every number has a sign. Negative numbers with any base are denoted by a prefix symbol ‘-‘. Binary numbers are the numbers that are expressed in base 2 having two symbols 0 and 1. Generally, the computer uses binary numbers. The sign of numbers in binary form is represented by the first binary digit in the number. Get the definition of signed magnitude representation and convert binary numbers into decimal and solved examples in the below sections. You can also find the difference between signed binary numbers and two’s complement numbers on this page.

Signed Magnitude Representation – Introduction

Every 8-bit binary number has magnitude and symbol which is used to indicate either the magnitude is positive or negative. The symbol defines the magnitude of the number. The sign bit is the left-most bit in the binary number. It is also known as the most significant bit. If the sign bit is 1, then it is a negative number, if the sign bit is 0, then it is a positive number.

The 8-bit binary numbers range from -127 to +127. To know the magnitude of the binary number, convert it into a decimal by following the below-mentioned rules and guidelines.

How to Convert Binary Numbers to Decimal Numbers?

Follow the below-listed steps to convert from a binary number to a decimal number. They are along the lines

  • Let us take any 8-bit binary number.
  • By checking the MSB digit, identify whether the number is positive or negative.
  • Now, take the remaining 7 digits into consideration.
  • Write the number 2 to the power 0 to n from the right side
  • And multiply the 2 power number by the binary number
  • Add all the terms to get the decimal number.

Also, Read:

Signed Magnitude Representation Examples

Example 1:

(10101011)2

Solution:

Given binary number is (10101011)2

The most significant digit is 1. So, the sign is negative

Decimal number = 26 x 0 + 25 x 1 + 24 x 0 + 23 x 1 + 22 x 0 + 21 x 1 + 20 x 1

= 64 x 0 + 32 x 1 + 16 x 0 + 8 x 1 + 4 x 0 + 2 x 1 + 1 x 1
= 32 + 8 + 2 + 1
= 43

Therefore, (10101011)2 = -(43)10

Example 2:

(01111111)2

Solution:

Given binary number is (01111111)2

The most significant digit is 0. So, the sign of the number is positive.

Decimal number = 26 x 1 + 25 x 1 + 24 x 1 + 23 x 1 + 22 x 1 + 21 x 1 + 20 x 1

= 64 x 1 + 32 x 1 + 16 x 1 + 8 x 1 + 4 x 1 + 2 x 1 + 1 x 1

= 64 + 32 + 16 + 8 + 4 + 2 + 1

= 127

Therefore, (01111111) = +(127)10

Example 3:

(01001101)2

Solution:

Given binary number is (01001101)2

The most significant digit is 0. So, the sign of the number is positive.

The decimal form of the number = 26 x 1 + 25 x 0 + 24 x 0 + 23 x 1 + 22 x 1 + 21 x 0 + 20 x 1

= 64 x 1 + 32 x 0 + 16 x 0 + 8 x 1 + 4 x 1 + 2 x 0 + 1 x 1

= 64 + 0 + 0 + 8 + 4 + 0 + 1

= 77

Therefore, (01001101) = +(77)10

Example 4:

(10010000)2

Solution:

Given binary number is (10010000)2

The most significant digit is 1. So, the sign is negative.

The decimal form of the number = 26 x 0 + 25 x 0 + 24 x 1 + 23 x 0 + 22 x 0 + 21 x 0 + 20 x 0

= 64 x 0 + 32 x 0 + 16 x 1 + 8 x 0 + 4 x 0 + 2 x 0 + 1 x 0

= 0 + 0 + 16 + 0 + 0 + 0 + 0

= 16

Therefore, (10010000) = -(16)10

FAQs on Signed Magnitude Representation

1. What is signed magnitude representation?

Sign and magnitude are two parts of a binary 8-bit number. The sign of numbers can be either positive or negative. The left-most digit of the number decides the sign of a decimal number. If the digit is 1, then negative sign, otherwise positive sign.

2. What is MSB?

MSB full form is the most significant bit. In a binary number, the bit at the left most corner is called the MSB, and the bit furthest to the right side is called the least significant bit.

3. Which digit in the binary number represents the sign?

The left-most digit in the binary number represents the sign of the decimal number. So, it is not included in the magnitude part of the number.

Big Ideas Math Answers Grade 8 Chapter 2 Transformations

Big Ideas Math Answers Grade 8 Chapter 2

Big Ideas Math Answers Grade 8 Chapter 2 Transformations provided helps students to learn the associated lessons easily. All the Solutions provided in BIM 8th Grade Chapter 2 Transformations are aligned according to the Transformation Textbook Chapter. Students can identify the knowledge gap accordingly and improvise on them and get a good hold of the concepts. Be on the right track and achieve success in your journey of learning math. Download the Big Ideas Math Grade 8 Answers Chapter 2 Transformations for free of cost and clear your queries.

Big Ideas Math Book 8th Grade Answer Key Chapter 2 Transformations

Avail the handy Big Ideas Math Grade 8 Chapter 2 Transformations Solution Key over here and solve any kind of problem easily. 8th Grade Big Ideas Math Ch 2 Transformations Answers are given by experts after doing extensive research. Simply access them by clicking on the direct links available and prepare accordingly. Enhance your Problem Solving Ability and math skills with regular practice and attempt the exams with confidence.

Performance

Lesson: 1 Translations

Lesson: 2 Reflections

Lesson: 3 Rotations

Lesson: 4 Congruent Figures

Lesson: 5 Dilations

Lesson: 6 Similar Figures

Lesson: 7 Perimeters and Areas of Similar Figures

Chapter 2: Transformations 

Transformations STEAM Video/Performance

STEAM Video

Shadow Puppets

Some puppets are controlled using strings or wires. How else can a puppet be controlled?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 1.1

Watch the STEAM Video “Shadow Puppets.” Then answer the following questions.

Question 1.
Tory and Robert are using a light source to display puppets on a screen. Tory wants to show the pig jumping from the floor to the window. Should she use a translation, reflection, rotation, or dilation? Explain.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 1.2

Answer:
In the situation given, if we translate first, we move the pre-image closer to the center of dilation than if we translate second. That will result in a different image.

Question 2.
How can Tory show the pig getting smaller as it jumps out the window?

Performance Task

Master Puppeteer

After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given the coordinates of a kite being used bya puppeteer.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 1.3
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 1.4
You will be asked to identify transformations for given movements of the kite. When might a puppeteer want to use a reflection?

Transformations Getting Ready for Chapter 2

Getting Ready for Chapter 2

Chapter Exploration
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 2

Question 1.
Work with a partner. Form each triangle on a geoboard.

  • Which of the triangles are congruent to the triangle at the right?
  • Measure the sides of each triangle with a ruler. Record your results in a table.
  • Write a conclusion about the side lengths of triangles that are congruent.

Big Ideas Math Answers Grade 8 Chapter 2 Transformations 3
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 4

Answer: option d

When two triangles are congruent they will have exactly the same three sides and exactly the same three angles. The equal sides and angles may not be in the same position

Vocabulary

The following vocabulary terms are defined in this chapter. Think about what the terms might mean and record your thoughts.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 5

Lesson 2.1 Translations

EXPLORATION 1
Work with a partner.
a. For each figure below, draw the figurein a coordinate plane. Then copy the figureonto a piece of transparent paper and slide the copy to a new location in the coordinate plane. Describe the location of the copy compared to the location of the original.

  • point
  • triangle
  • line segment
  • rectangle
  • line

Big Ideas Math Answers Grade 8 Chapter 2 Transformations 6
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 6.1
b. When you slide figures, what do you notice about sides, angles, and parallel lines?
c. Describe the location of each point below compared to the point A(x, y).
B(x + 1, y + 2)
C(x – 3, y + 4)
D(x – 2, y + 3)
E(x + 4, y – 1)
d. You copy a point with coordinates (x, y) and slide it horizontally a units and vertically b units. What are the coordinates of the copy?

2.1 Lesson

Try It

Tell whether the blue figure is a translation of the red figure.

Question 1.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 7

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same but the size is different. The red figure slide to form a blue figure but it is not the same size. So blue figure is not the translation of red figure.

Question 2.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 8

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same and also the size is the same. The red figure slide to form exactly blue figure. So blue figure is the translation of red figure.

Try It

Question 3.
WHAT IF?
The red triangle is translated 4 units left and 2 units up. What are the coordinates of the image?

Answer:
Big Ideas Math Grade 8 Chapter 2 Answer Key img_6
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(-2,1) B(2,5), C(1,2) and a = -4, b = 2
A'(-2+a, -2+b) = A'(-2-4, 1+2) = A'(-6,3)
B'(2+a, 5+b) = B'(2-4, 5+2) = B'(-2,7)
A'(1+a, 2+b) = C'(1-4, 2+2) = C'(-3,4)
Hence the coordinate of image are A'(-6,3), B'(-2,7), C'(-3,4).

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

IDENTIFYING A TRANSLATION
Tell whether the blue figure is a translation of the red figure.

Question 4.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 9

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same but the size is different. The red figure slide to form a blue figure but it is not the same size. So blue figure is not the translation of red figure.

Question 5.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 10

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same, and also the size is the same. The red figure slide to form the exactly blue figure. So blue figure is the translation of red figure.

Question 6.
The vertices of a triangle are A(2, 2), B (0, 2), and C (3, 0). Translate the triangle 1 unit left and 2 units up. What are the coordinates of the image?

Answer:
Bigideas Math Answers Grade 8 Chapter 2 img_7
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(2,2) B(0,2), C(3,0) and a = -1, b = 2
A'(2+a, 2+b) = A'(2-1, 2+2) = A'(1,4)
B'(0+a, 2+b) = B'(0-1, 2+2) = B'(-1,4)
A'(3+a, 0+b) = C'(3-1, 0+2) = C'(2,2)
Hence the coordinate of image are A'(1,4), B'(-1,4), C'(2,2).

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
A neighborhood planner uses a coordinate plane to design a new neighborhood. The coordinates A(1, -1), B(1, -2), and C (2, -1) represent House A, House B, and House C. The planner decides to place a playground centered at the origin, and moves the houses to make space. House A is now located at A'(3, -4). What are the new coordinates of House B and House C when each house is moved using the same translation? Justify your answer.

Answer:
Big Ideas Math Answers 8th Grade Chapter 2 img_8
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(1,-1) B(1,-2), C(2,1) and A'(3,-4)
A'(1+a, -1+b) = A'(3,-4) So, a = 2, b = -3
New coordinates of the houses are
B'(1+a, -2+b) = B'(1+2, -2-3) = B'(3,-5)
C'(2+a, -1+b) = C'(2+2, -1-3) = C'(4,-4)
Hence the coordinate of image are B'(3,-5),C'(4,-4)

Question 8.
The locations of a quarterback and a wide receiver on a football field are represented in a coordinate plane. The quarterback throws the football to the point (6, -2). Use a translation to describe a path the wide receiver can take to catch the pass.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 11

Answer:
Coordinate of Receiver: (1,3) and football point:(-6,-2)
Horizontal shift: a = x2-x1 = 6 – 1 = 5
Verrical shift: b = y2 – y1 = -2 – 3 = -5
Hence the path which receiver will take 5 unit right and 5 unit down.

Translations Homework & Practice 2.1

Review & Refresh

Solve the equation for y.

Question 1.
6x + y = 12

Answer:
Given
6x + y = 12
Subtract 6x from both sides
y = 12 – 6x
Now arranging the terms
y = -6x + 12
y = 6 (-x + 2)
Thus y = 6(-x + 2)

Question 2.
9 = x + 3y

Answer:
Given,
9 = x + 3y
3y = 9 – x
Dividing by 3 on both sides
y = (9 – x)/3
y =  \(\frac{9}{3}\) – \(\frac{x}{3}\)
Now arranging the terms
y = – \(\frac{x}{3}\) + 3
Thus y = – \(\frac{x}{3}\) + 3

Question 3.
\(\frac{1}{3}\)x + 2y = 8

Answer:
Given,
\(\frac{1}{3}\)x + 2y = 8
Subtracting x/3 from both sides
2y = 8 – \(\frac{x}{3}\)
Now arranging the terms
2y = – \(\frac{x}{3}\) + 8
y = – \(\frac{x}{6}\) + 4

Question 4.
You put $550 in an account that earns 4.4% simple interest per year. How much interest do you earn in 6 months?
A. $1.21
B. $12.10
C. $121.00
D. $145.20

Answer: $12.10

Explanation:
Given:
You put $550 in an account that earns 4.4% simple interest per year.
Principal amount: P = $550
Rate of Interest: r = 4.4%
Time: t = 6 months = 0.5 year
We know that formula for Simple Interest is SI = prt/100
SI = (550 × 4.4 × 0.5)/100
SI = 1210/100
SI = 12.10
Hence the simple interest is $12.10
Thus the correct answer is option B.

Concepts, Skills, & Problem Solving

DESCRIBING RELATIONSHIPS
For each figure, describe the location of the blue figurerelative to the location of the red figure. (See Exploration 1, p. 43.)

Question 5.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 12

Answer: The path which the receiver will take is 6 units right and 3 units down.

Explanation:
Coordinate of Point A: (-3,2) and Point A’: (3,-5)
Horizontal shift: a = x2 – x1 = 3 – (-3) = 3 + 3 = 6
Vertical shift: b = y2 – y1 = -5 – (-2) = -5 + 2 = -3
Hence, The path which the receiver will take is 6 units right and 3 units down.

Question 6.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 13

Answer: The path which receive will take is 5 units left and 2 units down.

Explanation:
Coordinate of point A: (3,-2) and point A’: (-2,-4)
Horizontal shift: a = x2 – x1 = -2 – (3) = -2 – 3 = -5
Vertical shift: b = y2 – y1 = -4 – (-2) = -4 + 2 = -2
Hence, The path which receive will take is 5 unit left and 2 unit down.

IDENTIFYING A TRANSLATION
Tell whether the blue figure is a translation of the red figure.

Question 7.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 14

Answer:
Blue figure is the translation of red figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figure are the same. Also, the orientation of the blue figure is the same as the red figure. This means that the red figure sides to form the blue figure. So, the blue figure slides to form the red figure.

Question 8.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 15

Answer:
Blue figure is not the translation of the red figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figure are the same. But the orientation of the blue figure is different from the red figure. This means that the blue figure is not the translation of the red figure.

Question 9.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 16

Answer:
Blue figure is not the translation of the red figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figure are the same. But the orientation of the blue figure is different from the red figure. This means that the blue figure is not the translation of the red figure. The blue figure is the mirror image of the red figure.

Question 10.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 17

Answer:
Blue figure is the translation of red-figure.

Here in the given figure, we can see that the shape of both blue and red-figure are the same, and also the size of both the figure are the same. Also, the orientation of the blue figure is the same as the red figure. This means that the red figure slides to form the blue figure. so the figure slides to form the red figure.

Question 11.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 18

Answer:
Blue figure is the translation of red-figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figures are the same. Also, the orientation of the both figure is the same as the red figure. This means that the red figure slides to form the blue figure. So blue figure slides to form the red figure.

Question 12.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 19

Answer: Blue figure is not the translation of the red-figure.

Explanation:
Here in the given figure we can see that the shape of both blue and red figure are same and the size of both the figure are not same. The red figure are smaller as compared to the blue figure. This means that blue figure is not the translation of red figure.

TRANSLATE A FIGURE
The vertices of a triangle are L(0, 1), M(1, -2), and N(-2, 1). Draw the figure and its image after the translation.

Question 13.
1 unit left and 6 units up

Answer:
Big Ideas Math Answers Grade 8 Chapter 2 Transformations img_1

explanation:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to the x-coordinate and ‘b’ is added to the y-coordinates of the vertices.
A(x,y) – A'(x+a, y+b)
The value ‘a’ and ‘b’ will be positive if the shift is Right and Vertical up and the value of ‘a’ and ‘b’ will be negative if the shift is left and vertical down.
Given:L(0,1),M(1,-2),N(-2,1)anda=-1,b=6
L'(0+a,1+b)=l'(0-1,1+6)=L'(-1,7)
M'(1+a,-2+b)=M'(1-1,1+6)=M'(0,4)
N'(-2+a,1+b)=N'(-3-1,1+6)=N'(-4,7)
Hence,the coordinate of image are L'(-1,7),M'(0,4),N'(-4,7)

Question 14.
5 units right

Answer:
We know that to translate a figure ‘a’ units horizontal and ‘b’ units vertically in the coordinate plane, ‘a’ is added to X-coordinate and ‘b’ is added to Y-coordinate of the vertices.
a(x,y),=A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if the shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given:L(0,1),M(1,-2),N(-2,1)and a=5,b=0
L(0+a,1+b)=L'(0+5,1+0)=L'(5,1)
M'(1+a,-2=b)=M'(1+5,-2+0)=M'(6,-2)
N'(-2+a,1+b)=N'(-2+5,1+0)=N'(3,1)
Hence the coordinate of image are L'(5,1),M'(6,-2),N'(3,1).

Question 15.
(x + 2, y + 3)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: L(0,1), M(1,-2), N(-2,1) and (x+2,y+3)
So the value of: a = 2 and b = 3
L'(o + a, 1 + b) = L'(0 + 2,1 + 3) = L'(2,4)
M'(1+a, -2 + b) = M'(1 + 2, -2 + 3) = M'(3, 1)
N'(-2 + a, 1 + b) = N'(-2 + 2, 1 + 3) = N'(0, 4)
Hence the coordinate of the image is L'(2,4), M'(3,1), N'(0,4)

Question 16.
(x – 3, y – 4)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: L(0,1), M(1,-2), N(-2,1) and (x-3, y-4)
So the value of a = -3 and b = -4
L'(0 + a, 1 + b) = L'(0-3,1-4) = L'(-3,-3)
M'(1 + a, -2 + b) = M'(1 – 3, -2-4) = M'(-2,-6)
N'(-2 + a, 1 + b) = N'(-2 – 3, 1 – 4) = N'(-5, -3)
Hence the coordinate of the image are L'(-3,-3), M'(-2,-6), N'(-5, -3)

Question 17.
YOU BE THE TEACHER
Your friend translates point A 2 units down and 1 unit right. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 20

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given points,
A(3, 1) and a = 1, b = -2
A'(3+a, 1+b) = A'(3+1, 1-2) = A'(4, -1)
So, the point A’ translated by my friend is wrong. He has reversed the x and y coordinate for translation.
Hence the correct translate point is A'(4,-1)

Question 18.
TRANSLATING A FIGURE
Translate the triangle 4 units right and 3 units down. What are the coordinates of the image?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 21

Answer:
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations img_3
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given points from graph: J(-1,3), K(-1,1), L(-4,1)
and a = 4, b = -3
J'(-1 + a, 3 + b) = J'(-1+4,3-3) = J'(3,0)
K'(-1 + a, 1 + b) = K'(-1 + 4, 1 – 3) = K'(3,-2)
L'(-4 + a, 1 + b) = L'(-4 + 4, 1 – 3) = L'(0,-2)
Hence the coordinate of image is J'(3,0), K'(3,-2), L'(0,-2)

Question 19.
TRANSLATING A FIGURE
Translate the figure 2 units left and 4 units down. What are the coordinates of the image?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 22

Answer:
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations img_4
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: A(-1,4), B(2,3), C(3,0), D(-1,-1)
and a = -2, b = -4
A'(-1+a, 4+b) = A'(-1-2, 4-4) = A'(-3,0)
B'(2+a, 3+b) = B'(2-2, 3-4) = B'(0,-1)
C'(3+a, 0+b) = C'(3-2, 0-4) = C'(1,-4)
D'(-1+a, -1+b) = D'(-1-2, -1-4) = D'(-3,-5)
The coordinate of image are A'(-3,0), B'(0,-1), C'(1,-4), D'(-3,-5)

DESCRIBING A TRANSLATION
Describe the translation of the point to its image.

Question 20.
(3, 2) → (1,0)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Coordinate of the point A:(3,-2) and the image point A’:(1,0)
Horizontal shift: a = x2 – x1 = 1 – 3 = -2
Vertical shift: b = y2 – y1 = 0 – (-2) = 0 + 2 = 2
Hence the translation path will be 2 units left and 2 units up.

Question 21.
(-8, -4) → (-3, 5)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Coordinate of the point A:(-8,-4) and the image point A’:(-3,5)
Horizontal shift: a = x2 – x1 = -3 – (-8) = 5
Vertical shift: b = y2 – y1 = 5 – (-4) = 9
Hence the translation path will be 5 units left and 9 units up.

Question 22.
REASONING
You can click and drag an icon on a computer’s desktop. Is this an example of a translation? Explain.

Answer:
Yes, the dragging of an icon on a computer’s desktop is an example of translation.
Because when dragging an icon on desktop the icon directly slides and is stored in its new position. While dragging the icon there is no change in shape and size of the icon, thus fulfilling the criteria of translation.

Question 23.
MODELING REAL LIFE
The proposed location for a new oil platform is represented in a coordinate plane by a rectangle with vertices A(1, 3), B(1, 4), C(4, 4), and D(4, -3). An inspector recommends moving the oil platform 4 units right and 2 units down. Find the coordinates of the image. Then draw the original figureand the image in the coordinate plane.

Answer:
BIM 8th Grade Answers Chapter 2 Transformations img_5
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: A(1,-3), B(1,4), C(4,4), D(4,-3) and a = 4, b = -2
A'(1+a, -3+b) = A'(1+4, -3-2) = A'(5,-5)
B'(1+a, 4+b) = B'(1+4, 4-2) = B'(5,2)
C'(4+a, 4+b) = C'(4+4, 4-2) = C'(8,2)
D'(4+a, -3+b) = D'(4+4, -3-2) = D'(8,-5)
Hence the coordinate of image are A'(5,-5), B'(5,2), C'(8,2), D'(8,-5)

Question 24.
PROBLEM SOLVING
A school of fish translates from point F to point D.
a. Describe the translation of the school of fish.
b. Can the fishing boat make the same translation? Explain.
c. Describe a translation the fishing boat could make to get to point D.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 22.1

Answer:
a. Coordinate of the point F: (-3,2) and the point D: (2,3)
Horizontal shift: a = x2 – x1 = 2 – (-3) = 2 + 5 = 7
Vertical shift: b = y2 – y1 = 3 – 2 = 1
Hence the path of translation is 5 unit Right and 1 unit Up.
b. No, the fishing boat (point B) cannot make the same translation as by fish (point F). Because in path between from point B to point D the is an island which will interrupt the translation of fishing boat.
c. Coordinate of the point B:(-2,-1) and the point D: (2,3)
Horizontal shift: a = x2 – x1 = 2 – (-2) = 2 + 2 = 4
Verical shift: b = y2 – y1 = 3 -(-1) = 3 + 1 = 4
Hence the path of translation is 4 unit Right and 4 unit Up.

Question 25.
REASONING
The vertices of a triangle are A(0, -3), B(2, -1), and C(3, -3). You translate the triangle 5 units right and 2 units down. Then you translate the image 3 units left and 8 units down. Is the original triangle identical to the final image? Explain your reasoning.

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: A(0,-3), B(2,-1), C(3,-3) and a1 = 5, b1 = -2
A'(0+a, -3+b) = A'(0+5, -3-2) = A'(5,-5)
B'(2+a, -1+b) = B'(2+5, -1-2) = B'(7,-3)
C'(3+a, -3+b) = C'(3+5,-3-2) = C'(8,-5)
Hence the coordinate of the first image are A'(5,-5), B'(7,-3), C'(8,-5)
Given: A'(5,-5), B'(7,-3), C'(8,-5) and a2 = -3, b2 = -8
A”(5+a, -5+b) = A”(5-3, -5-8) = A”(2,-13)
B”(7+a, -3+b) = B”(7-3, -3-8) = B”(4,-11)
C”(8+a, -5+b) = C”(8-3,-5-8) = C”(5,-13)
Hence the coordinate of the first image are A”(2,-13), B”(4,-11), C”(5,-13)
a = a1+a2 = 5 – 3 = 2, and b = b1 + b2 = -2 – 8 = -10
A'(0+a, -3+b) = A'(0+2, -3-10) = A'(2,-13)
B'(2+a, -1+b) = B'(2+2, -1-10) = B'(4,-11)
C'(3+a, -3+b) = C'(3+2,-3-10) = C'(5,-13)
Hence the original triangle is identical to the final image. This is because we can use both the translation by finding the resultant translation.
For final translation we can use:(x+2, y-10)

Question 26.
DIG DEEPER!
In chess, a knight can move only in an L-shaped pattern:

  • two vertical squares, then one horizontal square;
  • two horizontal squares, then one vertical square;
  • one vertical square, then two horizontal squares; or
  • one horizontal square, then two vertical squares.

Write a series of translations to move the knight from g8 to g5.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 23

Answer:
The series of translation to move knight from g8 to g5
1. Move 1 units Right to h8 and then 2 units Down to h6
2. Move 2 units Left to f6 and then 1 unit Up to f7
3. Move 2 units Down to f5 and then 1 unit Right to g5

Lesson 2.2 Reflections

Reflecting Figures

Work with a partner.
a. For each figure below, draw the figure in the coordinate plane. Then copy the axes and the figure onto a piece of transparent paper. Flip the transparent paper and align the origin and the axes with the coordinate plane. For each pair of figures, describe the line of symmetry.

  • point
  • triangle
  • line segment
  • rectangle
  • line

Big Ideas Math Answers Grade 8 Chapter 2 Transformations 24
b. When you reflect figures, what do you notice about sides, angles, and parallel lines?
c. Describe the relationship between each point below and the point A(4, 7) in terms of reflections.
d. A point with coordinates (x, y) is reflected in the x-axis. What are the coordinates of the image?
e. Repeat part(d) when the point is reflected in the y-axis

2.2 Lesson

Try It

Tell whether the blue figure is a reflection of the red figure.

Question 1.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 25

Answer: Blue figure is not the reflection of the red figure

Explanation:
By seeing the above figure we can say that the blue figure is not the mirror image of the red figure. Thus Blue figure is not the reflection of the red figure.

Question 2.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 26

Answer: Blue figure is the reflection of the red figure

Explanation:
By seeing the above figure we can say that the blue figure is the mirror image of the red figure. If the red figure is flipped it would form the shape of the blue figure. Thus Blue figure is the reflection of the red figure

Try It

Question 3.
The vertices of a rectangle are A(-4, -3), B(-4, -1), C(-1, -1), and D(-1, -3). Draw the figure and its reflection in (a) the x-axis and (b) the y-axis.

Answer:
Given,
The vertices of a rectangle are A(-4, -3), B(-4, -1), C(-1, -1), and D(-1, -3).
Reflection about the x-axis:
A(x,y) = A'(x,-y)
A(-4, -3) = A'(-4,3)
B(-4, -1) = B'(-4,1)
C(-1, -1) = C'(-1,1)
D(-1, -3) = D'(-1,3)
Reflection through x-axis:
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations img_9(i)
Reflection through y-axis:
A(x,y) = A'(-x,y)
A(-4, -3) = A'(4,-3)
B(-4, -1) = B'(4,-1)
C(-1, -1) = C'(1,-1)
D(-1, -3) = D'(1,-3)
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations img_9(ii)

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
REFLECTING A FIGURE
The vertices of a triangle are J(-3, -5), K(-2, 2), and L(1, -4). Draw the figure and its reflection in
(a) the x-axis and
(b) the y-axis.

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
The vertices of a triangle are J(-3, -5), K(-2, 2), and L(1, -4).
Reflection about the x-axis:
J(-3, -5) = J'(-3,5)
K(-2, 2) = K'(-2,-2)
L(1, -4) = L'(1,4)
Big Ideas Math Answers Grade 8 Chapter 2 Transformations img_10(i)
when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Reflection about the y-axis:
J(-3, -5) = J'(3,-5)
K(-2, 2) = K'(2,2)
L(1, -4) = L'(-1,-4)
BIM Grade 8 Answers Chapter 2 Transformations img_10(ii)

Question 5.
WHICH ONE DOESN’T BELONG?
Which transformation does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 27

Answer: 3rd figure is different from other figures. Because all the other three pictures are reflections of each other except the third one. The third picture is pointed in the same direction but all the other three figures are in opposite direction.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
You design a logo using the figure shown at the left. You want both the x-axis and the y-axis to be lines of reflection. Describe how to use reflections to complete the design. Then draw the logo in the coordinate plane.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 28

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
A(-4,2), B(-2,2), C(0,0), D(-2,0)
Reflection about the x-axis:
A(-4,2) = A'(-4,-2)
B(-2,2) = B'(-2,-2)
C(0,0) = C'(0,0)
D(-2,0) = D'(-2,0)
when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Reflection about the y-axis:
A(-4,2) = A”(4,2)
B(-2,2) = B”(2,2)
C(0,0) = C”(0,0)
D(-2,0) = D”(2,0)
Now to complete the Logo again we have to take a reflection of the image figure about the y-axis. In this way, the logo will be symmetric about both axis.
A”(4,2) = A”‘(4,-2)
B”(2,2) = B”‘(2,-2)
C”(0,0) = C”‘(0,0)
D”(2,0) = D”‘(-2,0)
Bigideas Math Answer Key Grade 8 Chapter 2 img_11

Question 7.
DIG DEEPER!
You hit the golf ball along the path shown, so that its final location is a reflection in the y-axis of its starting location.
a. Does the golf ball land in the hole? Explain.
b. Your friend tries the shot from the same starting location. He bounces the ball of the wall at the point (-0.5, 7) so that its path is a reflection. Does the golf ball land in the hole?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 29

Answer:
a. Coordinates of the location of golf ball = (2,4)
Coordinates of location of hole = (-3,4)
Location of ball after reflection through y-axis = (2,4) = (-2,4)
But the location of the hole is (-3,4)
So the ball will not go into a hole and it will miss the hole by 1 unit.
Hence the ball will not go into the hole.
b. Yes, when the ball bounces at the point (-0.5,7) then it will land in the hole.

Reflections Homework & Practice 2.2

Review & Refresh

The vertices of a quadrilateral are P(-1, -1), Q(0, 4), R(3, 1), and S(1, -2). Draw the figure and its image after the translation.

Question 1.
7 units down

Answer:
Big Ideas Math 8th Grade Answer Key for Chapter 2 img_12
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = 0, b = -7
P'(-1+a,-1+b) = P'(-1+0,-1-7) = P'(-1,-8)
Q'(0+a,4+b) = Q'(0+4,4-7) = Q'(4,-3)
R'(3+a, 1+b) = R'(3+0,1-7) = R'(3,-6)
S'(1+a,-2+b) = S'(1+0,-2-7) = S'(1,-9)
Thus the coordinate of the image is P'(-1,-8), Q'(4,-3), R'(3,-6), and S'(1,-9)

Question 2.
3 units left and 2 units up

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = -3, b = 2
P'(-1+a,-1+b) = P'(-1-3,-1+2) = P'(-4,1)
Q'(0+a,4+b) = Q'(0-3,4+2) = Q'(-3,6)
R'(3+a, 1+b) = R'(3-2,1+2) = R'(0,3)
S'(1+a,-2+b) = S'(1-3,-2+2) = S'(-2,0)
Thus the coordinate of the image are P'(-4,1), Q'(-3,6), R'(0,3) and S'(-2,0)
Big Ideas Math Answers Grade 8 Ch 2 Transformations img_11

Question 3.
(x + 4, y – 1)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = 4, b = -1
P'(-1+a,-1+b) = P'(-1+4,-1-1) = P'(3,-2)
Q'(0+a,4+b) = Q'(0+4,4-1) = Q'(4,3)
R'(3+a, 1+b) = R'(3+4,1-1) = R'(7,0)
S'(1+a,-2+b) = S'(1+4,-2-1) = S'(5,-3)
Thus the coordinate of the image are P'(3,-2), Q'(4,3), R'(7,0) and S'(5,-3)
Big ideas math answers grade 8 chapter 2 transformations img_12

Question 4.
(x – 5, y – 6)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = -5, b = -6
P'(-1+a,-1+b) = P'(-1-5,-1-6) = P'(-6,-7)
Q'(0+a,4+b) = Q'(0-5,4-6) = Q'(-5,-2)
R'(3+a, 1+b) = R'(3-5,1-6) = R'(-2,-5)
S'(1+a,-2+b) = S'(1-5,-2-6) = S'(-4,-8)
Thus the coordinate of the image are P'(-6,-7), Q'(-5,-2), R'(-2,-5) and S'(-4,-8)
BIM 8th Grade Answer Key Chapter 2 Transformations img_13

Tell whether the angles are complementary, supplementary or neither.

Question 5.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 30

Answer:
108° + 82° = 190°
Thus the angle is neither supplementary nor complementary.

Question 6.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 31

Answer: Complementary

Explanation:
43° + 47° = 90°
Two angles are called complementary when their measures add to 90 degrees.

Question 7.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 32

Answer:
38° + 62° = 100°
Hence the given angle is neither supplementary nor complementary.

Question 8.
36 is 75% of what number?
A. 27
B. 48
C. 54
D. 63

Answer: B. 48

Explanation:
Let x be the unknown value.
75% of x = 36
75% × x = 36
75/100 × x = 36
3/4x × x = 36
3x = 36 × 4
3x = 144
x = 144/3
x = 48
Thus the correct answer is option B.

Concepts, Skills, &Problem Solving
DESCRIBING RELATIONSHIPS
Describe the relationship between the given point and the point A(5, 3) in terms of reflections. (See Exploration 1, p. 49.)

Answer:
We know that when a point is reflected about x-axis then y-coordinate becomes the opposite.
P(x,y) = P'(x,-y)
We know that when a point is reflected about y-axis then x-coordinate becomes opposite.
P(x,y) = P'(-x,y)
Given: A(5,3), B(5,-3)
Hence the point A is reflected about the x-axis to get point B.

IDENTIFYING A REFLECTION
Tell whether the blue figure is a reflection of the red figure.

Question 12.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 33

Answer: No

Explanation:
The blue figure is not the mirror image of the red figure. If the red figure were flipped then the right of the blue and red figure should be facing each other. So, the blue figure is not a reflection of red figure.

Question 13.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 34

Answer: Yes

Explanation:
The blue figure is the mirror image of the red figure. If the red figure were flipped it will result in the blue figure. So, the blue figure is a reflection of red figure.

Question 14.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 35

Answer: Yes

Explanation:
The blue figure is the mirror image of the red figure. If the red figure were flipped it will result in the blue figure. So, the blue figure is a reflection of red figure.

Question 15.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 36

Answer: No

Explanation:
The blue figure is not the mirror image of the red figure. If the red figure were flipped then the right of the blue and red figure should be facing each other. So, the blue figure is not a reflection of red figure.

Question 16.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 37

Answer: Yes

Explanation:
The blue figure is the mirror image of the red figure. If the red figure were flipped it will result in the blue figure. So, the blue figure is a reflection of red figure.

Question 17.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 38

Answer: No

Explanation:
The blue figure is not the mirror image of the red figure. If the red figure were flipped then the right of the blue and red figure should be facing each other. So, the blue figure is not a reflection of red figure.

REFLECTING FIGURES
Draw the figure and its reflection in the x-axis. Identify the coordinates of the image.

Question 18.
A(3, 2), B(4, 4), C(1, 3)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
A(3, 2), B(4, 4), C(1, 3)
Reflection about the x-axis:
A(3, 2) = A'(3,-2)
B(4, 4) = B'(4,-4)
C(1, 3) = C'(1,-3)
Thus the coordinate of the image are A'(3,-2), B'(4,-4), C'(1,-3)
Bigideas Math Answers Grade 8 Chapter 2 img_14

Question 19.
M(-2, 1), N(0, 3), P(2, 2)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
M(-2, 1), N(0, 3), P(2, 2)
Reflection about the x-axis:
M(-2, 1) = M'(-2,-1)
N(0, 3) = N'(0,-3)
P(2, 2) = P'(2,-2)
Thus the coordinate of the image are M'(-2,-1), N'(0,-3), P'(2,-2)
Big Ideas Math Grade 8 Ch 2 Answer Key img_15

Question 20.
H(2, -2), J(4, -1), K(6, -3), L(5, -4)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
H(2, -2), J(4, -1), K(6, -3), L(5, -4)
Reflection about the x-axis:
H(2, -2) = H'(-2,-1)
J(4, -1) = J'(4,1)
K(6, -3) = K'(6,3)
L(5, -4) = L'(5,4)
Thus the coordinate of the image are H'(-2,-1), J'(4,1), K'(6,3) and L'(5,4)
Big Ideas Math Answers Grade 8 Chapter 2 Transformations img_16

Question 21.
D(-2, -5), E(0, -1), F(2, -1), G(0, -5)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
D(-2, -5), E(0, -1), F(2, -1), G(0, -5)
Reflection about the x-axis:
D(-2, -5) = D'(-2,5)
E(0, -1) = E'(0,1)
F(2, -1) = F'(2,1)
G(0, -5) = G'(0,5)
Thus the coordinate of the image are D'(-2,5), E'(0,1), F'(2,1), G'(0,5)
Big ideas Math Answers Grade 8 Chapter 2 Transformations img_17

REFLECTING FIGURES
Draw the figure and its reflection in the y-axis. Identify the coordinates of the image.

Question 22.
Q(-4, 2), R(-2, 4), S(-1, 1)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
Q(-4, 2), R(-2, 4), S(-1, 1)
Reflection about the x-axis:
Q(-4, 2) = Q'(4,2)
R(-2, 4) = R'(2,4)
S(-1, 1)= S'(1,1)
Thus the coordinate of the image is Q'(4,2), R'(2,4), S'(1,1)
Big Ideas Math Grade 8 Chapter 2 solution Key img_18

Question 23.
T(4, -2), U(4, 2), V(6, -2)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
T(4, -2), U(4, 2), V(6, -2)
Reflection about the y-axis
T(4,-2) = T'(-4,-2)
Y(4,2) = U'(-4,2)
V(6,-2) = V'(-6,-2)
Thus the coordinates of the figure are T'(-4,-2), U'(-4,2), V'(-6,-2)
Big Ideas Math Grade 8 Chapter 2 transformations answer key img_19

Question 24.
W(2, -1), X(5, -2), Y(5, -5), Z(2, -4)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
W(2, -1), X(5, -2), Y(5, -5), Z(2, -4)
Reflection about the y-axis:
W(2,-1) = W'(-2,-1)
X(5,-2) = X'(-5,-2)
Y(5,-5) = Y'(-5,-5)
Z(2,-4) = Z'(-2,-4)
Thus the coordinates of the figure are W'(-2,-1), X'(-5,-2), Y'(-5,-5), Z'(-2,-4)
Big Ideas Math Grade 8 2nd Chapter Answer Key for Transformations img_20

Question 25.
J(2, 2), K(7, 4), L(9, -2), M(3, -1)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
J(2, 2), K(7, 4), L(9, -2), M(3, -1)
Reflection about the y-axis
J(2, 2) = J(-2,2)
K(7, 4) = K'(-7,4)
L(9, -2) = L'(-9,-2)
M(3, -1) = M'(-3,-1)
Thus the coordinates of the figure are J(-2,2), K'(-7,4), L'(-9,-2), M'(-3,-1)
BIM Grade 8 Solution Key Chapter 2 Transformations img_21

Question 26.
REASONING
Which letters look the same when reflected in the line?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 39

Answer:
The letters which will look the same after being reflected through horizontal line are
B, C, D, E, H, I, K, O, X

STRUCTURE
The coordinates of a point and its image after a reflection are given. Identify the line of reflection.

Question 27.
(2, -2) → (2, 2)

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given A(2, -2) → A'(2, 2)
Here we can see that x-coordinate of both A & A’ is the same but the y-coordinate of A’ is just the opposite of A. This means that A’ is the reflection of A about the x-axis.
Hence the point A is reflected about the x-axis to get point A’.

Question 28.
(-4, 1) → (4, 1)

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given A(-4, 1) → A'(4, 1)
Here we can see that y-coordinate of both A & A’ is the same but the x-coordinate of A’ is just the opposite of A. This means that A’ is the reflection of A about y-axis.
Hence the point A is reflected about the y-axis to get point A’.

Question 29.
(-2, -5) → (4, -5)

Answer:
Given,
A(-2, -5) → A'(4, -5)
We observe that y-coordinate of both A and A’ is same but the x-coordinate of A’ is not opposite of A. This means that A’ is the reflection of A about a line x = a.
a = (x2+x1)/2 = (4-2)/2 = 2/2 = 1
Hence the point (-2,-5) is reflected about the line x = 1 to get point (4,-5)

Question 30.
(-3, -4) → (-3, 0)

Answer:
Given,
B(-3, -4) → B'(-3, 0)
We observe that x-coordinate of both B and B’ is the same but the y-coordinate of B’ is not the opposite of B. This means that A’ is the reflection of A about a line x = a.
b = (y2+y1)/2 = (0-4)/2 = -4/2 = -2
Hence the point (-3,-4) is reflected about the line y = -2 to get point (-3,0)

TRANSFORMING FIGURES
Find the coordinates of the figure after the transformations.

Question 31.
Translate the triangle 1 unit right and 5 units down. Then reflect the image in the y-axis.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 40

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given:
R(-4,1)
S(-4,4)
T(-2,1)
a = 1 and b = -5
R(-4,1) = R'(-4+a, 1+b) = R'(-4+1, 1-5) = R'(-3, -4)
S(-4,4) = S'(-4+a, 4+b) = S'(-4+1, 4-5) = S'(-3, -1)
T(-2,1) = T'(-2+a, 1+b) = T'(-2+1, 1-5) = T'(-1, -4)
Thus the coordinates of the image are R'(-3, -4), S'(-3, -1), T'(-1, -4)

Question 32.
Reflect the trapezoid in the x-axis. Then translate the image 2 units left and 3 units up.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 41

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
Given, W(-2,-2), X(-2,1), Y(2,1), and Z(4,-2)
Now reflection about the x-axis:
W(-2,-2) = W'(-2,2)
X(-2,1) = X'(-2,-1)
Y(2,1) = Y'(2,-1)
Z(4,-2) = Z'(4,2)
Thus the coordinates of the image: W'(-2,2), X'(-2,-1), Y'(2,-1), Z'(4,2)
Now translating the above image point:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given:
W(-2,-2), X(-2,1), Y(2,1), and Z(4,-2)
a = -2 and b = 3
W(-2,2) = W”(-2+a, 2+b) = W”(-2-2, 2+3) = W”(-4, 5)
X(-2,-1) = X”(-2+a, -1+b) = X”(-2-2, -1+3) = X”(-4, 2)
Y(2,-1) = Y”(2+a, -1+b) = Y”(2-2, -1+3) = Y”(0, 2)
Z(4,2) = Z”(4+a, 2+b) = Z”(4-2, 2+3) = Z”(2, 5)
Thus the coordinates of the image: W”(-4, 5), X”(-4, 2), Y”(0, 2), Z”(2, 5)

Question 33.
REASONING
In Exercises 31 and 32, is the original figure identical to the final image? Explain.

Answer: Yes, in exercises 31 and 32 the original figure is identical to the final image. Because the type of transformation used is reflection and translation. The shape and size of the image figure do not change when there is reflection or translation. The only position of the image changes in both cases when compared to the position of the original figure.

Question 34.
CRITICAL THINKING
Hold a mirror to the left side of the photo of the vehicle.
a. What word do you see in the mirror?

Answer: The word which we will see in the mirror will be AMBULANCE. Because the word is written in mirror image form on the vehicle.

b. Why do you think it is written that way on the front of the vehicle?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 42

Answer: Ambulance

Explanation:
It is written in that way because the ambulance will be behind any vehicle then the word “AMBULANCE” will correctly appear in the Rear-view mirror of the front vehicle.

Question 35.
DIG DEEPER!
Reflect the triangle in the line y = x. How are the x- and y-coordinates of the image related to the x- and y-coordinates of the original triangle?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 43

Answer:
When a point is reflected about the line y = x then both x and y-coordinate become opposite.
A(x, y) = A'(-x, -y)
Given,
D(-1,-3)
E(-1,1)
F(-3,1)
Reflection about the line y = x
D(-1,-3) = D'(1,-3)
E(-1,1) = E'(1,-1)
F(-3,1) = F'(3,-1)
Hence the coordinates of the image: D'(1,-3), E'(1,-1), F'(3,-1)
Big Ideas Math Grade 8 Chapter 2 Solution Key img_21

Lesson 2.3 Rotations

EXPLORATION 1
Work with a partner.
a. For each figurebelow, draw the figure in the coordinate plane. Then copy the axes and the figure onto a piece of transparent paper. Turn the transparent paper and align the origin and the axes with the coordinate plane. For each pair of figures, describe the angle of rotation.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 44

  • point
  • triangle
  • line segment
  • rectangle

Big Ideas Math Answers Grade 8 Chapter 2 Transformations 45
b. When you rotate figures, what do you notice about sides, angles, and parallel lines?
c. Describe the relationship between each point below and the point A(3, 6) in terms of rotations.
d. What are the coordinates of a point P(x, y) after a rotation 90° counterclockwise about the origin? 180°? 270°?

2.3 Lesson

Try It
Tell whether the blue figure is a rotation of the red figure about the origin. If so, give the angle and direction of rotation.

Question 1.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 46

Answer: Yes blue figure is the rotation of red figure about the origin.

Explanation:
When we rotate the red figure 180 degrees clockwise or anti-clockwise about the origin we will get the same figure as the blue figure.

Question 2.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 47

Answer: Blue figure is not the rotation of the red figure.

Explanation:
When the red figure is rotated about the origin in any direction the distance of the center point of both the red figure and the blue figure will be the same from the origin point (0,0). The distance between the center of the object and the center of rotation always remains the same.
Thus Blue figure is not the rotation of the red figure.

Try It

The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 3.
J(-4, -4), K(-4, 2), L(-1, 0), M(-2, -3); 180° about the origin

Answer:
When a point is rotated 180 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-x, -y)
Given, J(-4, -4), K(-4, 2), L(-1, 0), M(-2, -3)
Rotation about the origin
J(-4, -4) = J'(4,4)
K(-4, 2) = K'(4,-2)
L(-1, 0) = L'(1,0)
M(-2, -3) = M'(2,3)
Hence the coordinate of the image are J'(4,4), K'(4,-2), L'(1,0), M'(2,3)

Question 4.
P(-3, 2), Q(6, 1), R(-1, -5); 90° counterclockwise about the origin

Answer:
When a point is rotated 90 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-y, x)
Given,
P(-3, 2), Q(6, 1), R(-1, -5)
Rotation about the origin
P(-3, 2) = P'(-2,-3)
Q(6, 1) = Q'(-1,6)
R(-1, -5) = R'(5,-1)
Hence the coordinate of the image is P'(-2,-3), Q'(-1,6), R'(5,-1)

Question 5.
A(5, 3), B(4, -1), C(1, -1); 90° clockwise about the origin

Answer:
When a point is rotated 270 degrees counterclockwise about the origin then both x and y-coordinates gets interchanged and the x-coordinate becomes the opposite.
A(x, y) = A'(y, -x)
Given,
A(5, 3), B(4, -1), C(1, -1)
Rotation about the origin
A(5, 3) = A'(3,-5)
B(4, -1) = B'(-1,-4)
C(1, -1) = C'(-1,-1)
Hence the coordinate of the image are A'(3,-5), B'(-1,-4), C'(-1,-1)

Try It

Question 6.
The vertices of a triangle are P(-1, 2), Q(-1, 0), and R(2, 0). Rotate the triangle 180° about the origin, and then reflect it in the x-axis. What are the coordinates of the image?

Answer:
When a point is rotated 180 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-x, -y)
Given,
P(-1, 2), Q(-1, 0), and R(2, 0)
Rotation about the origin
P(-1, 2) = P'(1,-2)
Q(-1, 0) = Q'(1,0)
R(2, 0) = R'(-2,0)
Hence the coordinate of the image is P'(1,-2), Q'(1,0), R'(-2,0)
Now reflecting above image point about x-axis:
When a point is reflected about the x-axis then the y-coordinate becomes opposite.
A(x, y) = A'(x, -y)
Given,
P'(1,-2), Q'(1,0), R'(-2,0)
Rotation about the origin
P'(1,-2) = P”(1,2)
Q'(1,0) = Q”(1,0)
R'(-2,0) = R”(-2,0)
Hence the coordinate of the image are P'(1,2), Q'(1,0), R'(-2,0)

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
IDENTIFYING A ROTATION
Tell whether the blue figure is a rotation of the red figure about point P. If so, give the angle and direction of rotation.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 48

Answer:
Yes, the blue figure is the rotation of the red figure about the origin.

Explanation:
Because when we will rotate the red figure 90 degrees anti-clockwise about the origin we will get the same figure as the blue figure.
By this, we can say that the blue figure is the result of the rotation of red figure by 90 degrees in the clock or anti-clockwise direction.

Question 8.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 49
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 50

Answer:
The statement which different from all other 3 statement is:
What are the coordinates of the image after a 270 degrees clockwise rotation about the origin?
Now coordinate of both the image are:
The rotation of an object 90 degrees clockwise is equal to the rotation of 270 degrees counterclockwise.
we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanges and x-coordinate becomes opposite.
A(x, y) = A'(y, -x)
Given,
A(2, 4)
B(4, 4)
C(4, 1)
Rotating 90 degrees clockwise about the origin
A(2,4) = A'(4,-2)
B(4, 4) = B'(4,-4)
C(4, 1) = C'(1,-4)
Hence the coordinate of the image are: A'(4,-2), B'(4,-4), C'(1,-4)
Image of statement which different from all 3 statement
The rotation of an object 270 degrees clockwise is equal to the rotation of 90 degrees counterclockwise.
we know that when a point is rotated 90 degrees counterclockwise about origin then both coordinate gets interchanges and x-coordinate becomes opposite
P(x, y) = P'(-y, x)
Given,
A(2, 4)
B(4, 4)
C(4, 1)
Rotating 90 degrees clockwise about the origin
A(2, 4) = A'(-4, 2)
B(4, 4) = B'(-4, 4)
C(4, 1) = C'(-1, 4)
Hence the coordinate of the image are: A'(4,-2), B'(-4,4), C'(-1,4)
Third statement “what are the coordinates of the image after a 270 degrees clockwise rotation about origin?” is different.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 9.
You move the red game piece to the indicated location using a rotation about the origin, followed by a translation. What are the coordinates of the vertices of the game piece after the rotation? Justify your answer.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 51

Answer:
To move the red game piece in the indicated location the game piee should be rotated 90 degrees in clockwise direction. The rotation of red game piece 90 degrees clockwise is same as the rotation of 270 degrees counterclockwise.
we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanges and x-coordinate becomes opposite.
A(x,y) = A'(y, -x)
Conner point of red game piece:
A(0,-1), B(0,0), C(1,0), D(1,1), E(-2,1), F(-2,0), G(-1,0), H(-1,-1)
Rotating 90 degrees clockwise about the origin:
A(0,-1) = A'(-1,0)
B(0,0) = B'(0,0)
C(1,0) = C'(0,-1)
D(1,1) = D'(1,-1)
E(-2,1) = E'(1,2)
F(-2,0) = F'(0,2)
G(-1,0) = G'(0,1)
H(-1,-1) = H'(-1,1)
Hence the coordinate of corner of red game piece are A'(-1,0), B'(0,0), C'(0,-1), D'(1,-1), E'(1,2), F'(0,2), G'(0,1), H'(-1,1)

Question 10.
DIG DEEPER!
Skytypingis a technique that airplanes use to write messages in the sky. The coordinate plane shows a message typed in the sky over a city, where the positive y-axis represents north. What does the message say? How can you transform the message so that it is read from north to south?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 52

Answer: HELLO

Explanation:
The message above on the coordinate plane can be transformed from north to south by rotating the image 90 degrees anticlockwise.

Rotations Homework & Practice 2.3

Review & Refresh

Tell whether the blue figure is a reflection of the red figure.

Question 1.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 53

Answer: Yes, the blue figure is the reflection of red figure.

Explanation:
Because the blue figure is the exact mirror image of the red figure. If the red figure will be flipped it will result in the blue figure. So, the blue figure is the reflection of red figure.

Question 2.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 54

Answer: No, the blue figure is not the reflection of red figure.

Explanation:
Because the blue figure is not the mirror image of the red figure. If the red figure will be flipped it will not result in the blue figure. So blue figure is not reflection of red figure.

Find the circumference of the object. Use 3.14 or \(\frac{22}{7}\) for π.

Question 3.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 55

Answer:
Given diameter of disk D = 28 cm
Circumference of the circular disk is π × D
C = π × 28
C = 22/7 × 28
C = 22 × 4
C = 88 cm
Thus the circumference is 88 cm.

Question 4.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 56

Answer:
Given the diameter of disk D = 11.4 in
Circumference of the circular disk is π × D
C = π × 11.4
C = 22/7 × 11.4
C = 3.14 × 11.4
C = 35.796 in
Thus the circumference is 35.796 in

Question 5.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 57

Answer:
Given diameter of disk r = 0.5 ft
Circumference of the circular disk is 2π × r
C = 2π × 0.5
C = 6.28 × 0.5
C = 3.14 ft
Thus the circumference is 3.14 ft

Concepts, Skills, &Problem Solving

DESCRIBING RELATIONSHIPS
Describe the relationship between the given point and the point (2, 7) in terms of rotations. (See Exploration 1, p. 55.)

Question 6.
B(7, -2)

Answer:
Given,
A(7, 2) = B(7, -2)
Here we can see that after rotation x and y coordinate are interchanged and the y-coordinate is opposite. And we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinates get interchanged and the x-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Hence the above rotation is 270 degrees counterclockwise about the origin.

Question 7.
C(-7, 2)

Answer:
Given,
A(7, 2) = C(-7, 2)
Here we can see that after rotation x and y coordinate are interchanged and the y-coordinate is opposite. And we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanged and x-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Hence the above rotation is 90 degrees counterclockwise about the origin.

Question 8.
D(-2, -7)

Answer:
Given,
A(2, 7) = C(-2, -7)
Here we can see that after rotation x and y coordinate are interchanged and the y-coordinate is opposite. And we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanged and x-coordinate becomes opposite.
P(x, y) = P'(-x, -y)
Hence the above rotation is 180 degrees counterclockwise about the origin.

IDENTIFYING A ROTATION
Tell whether the blue figure is a rotation of the red figure about the origin. If so, give the angle and direction of rotation.

Question 9.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 58

Answer: No the blue figure is not the rotation of the red figure.

Explanation:
Because if the blue triangle were the result of the rotation of the red triangle then the hypotenuse of the blue triangle should have been parallel to the x-axis. so, it is not the case of rotation.

Question 10.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 59

Answer: Yes, the blue figure is the result of the rotation of the red figure.

Explanation:
Because if the red figure is rotated 90 degrees in a counterclockwise direction it will result in a blue figure.

Question 11.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 60

Answer: Yes, the blue figure is the result of the rotation of the red figure.

Explanation:
If the red figure is rotated 180 degrees in counterclockwise or clockwise direction it will result in blue figure.

Question 12.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 61

Answer: Yes, the blue figure is the result of the rotation of the red figure.

Explanation:
If the red figure is rotated 90 degrees in a clockwise direction it will result in blue figure.

ROTATING A FIGURE
The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 13.
A(2, -2), B(4, -1), C(4, -3), D(2, -4)
90° counterclockwise about the origin

Answer:
We know that when a point is rotated 90 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given,
A(2, -2), B(4, -1), C(4, -3), D(2, -4)
Rotating 90 degrees counterclockwise about the origin
A(2, -2) = A'(2,2)
B(4, -1) = B'(1,4)
C(4, -3) = C'(3,4)
D(2, -4) = D'(4,2)
Hence the coordinates of the image are A'(2,2), B'(1,4), C'(3,4), D'(4,2)

Question 14.
F(1, 2), G(3, 5), H(3, 2) 180° about the origin

Answer:
We know that when a point is rotated 180 degrees counterclockwise or clockwise direction about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-x, -y)
Given,
F(1, 2), G(3, 5), H(3, 2)
Rotating 180 degrees about the origin
F(1, 2) = F'(-1,-2)
G(3, 5) = G'(-3,-5)
H(3, 2) = H'(-3,-2)
Hence the coordinates of the image are F'(-1,-2), G'(-3,-5), H'(-3,-2)

Question 15.
J(-4, 1), K(-2, 1), L(-4, -3)
90° clockwise about the origin

Answer:
The rotation of an object 90 degrees clockwise is equal to the rotation of 270 degrees counterclockwise.
We know that when a point is rotated 270 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Given,
J(-4, 1), K(-2, 1), L(-4, -3)
Rotating 90 degrees clockwise about the origin
J(-4, 1) = J'(1,4)
K(-2, 1) = K'(1,2)
L(-4, -3) = L'(-3,4)
Hence the coordinates of the image are J'(1,4), K'(1,2), L'(-3,4)

Question 16.
P(-3, 4), Q(-1, 4), R(-2, 1), S(-4, 1)
270° clockwise about the origin

Answer:
The rotation of an object 270 degrees clockwise is equal to the rotation of 90 degrees counterclockwise.
We know that when a point is rotated 90 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given,
P(-3, 4), Q(-1, 4), R(-2, 1), S(-4, 1)
Rotating 90 degrees clockwise about the origin
P(-3, 4) = P'(-4,-3)
Q(-1, 4) = Q'(-4,-1)
R(-2, 1) = R'(-1,-2)
S(-4, 1) = S'(-1,-4)
Hence the coordinates of the image are P'(-4,-3), Q'(-4,-1), R'(-1,-2), S'(-1,-4)

Question 17.
W(-6, -2), X(-2, -2), Y(-2, -6), Z(-5, -6)
270° counterclockwise about the origin

Answer:
We know that when a point is rotated 270 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Given,
W(-6, -2), X(-2, -2), Y(-2, -6), Z(-5, -6)
Rotating 90 degrees clockwise about the origin
W(-6, -2) = W'(-2,6)
X(-2, -2) = X'(-2,2)
Y(-2, -6) = Y'(-6,2)
Z(-5, -6) = Z'(-6,5)
Hence the coordinates of the image are W'(-2,6), X'(-2,2), Y'(-6,2), Z'(-6,5)

Question 18.
A(1, -1), B(5, -6), C(1, -6)
90° counterclockwise about the origin

Answer:
We know that when a point is rotated 90 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given,
A(1, -1), B(5, -6), C(1, -6)
Rotating 90 degrees clockwise about the origin
A(1, -1) = A'(1,1)
B(5, -6) = B'(6,5)
C(1, -6) = C'(6,1)
Hence the coordinates of the image are A'(1,1), B'(6,5), C'(6,1)

Question 19.
YOU BE THE TEACHER
The vertices of a triangle are A(4, 4), B(1, -2), and C(-3, 0). Your friend finds the coordinates of the image after a rotation 90° clockwise about the origin. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 62

Answer:
We know that when a point is rotated 270 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Given,
A(4, 4), B(1, -2), and C(-3, 0).
Rotating 90 degrees clockwise about the origin
A(4, 4) = A'(4,-4)
B(1, -2) = B'(-2,-1)
C(-3,0) = C'(0,3)
Hence the coordinates of the image are A'(4,-4), B'(-2,-1), C'(0,3)
By this I can say that my friend is not correct.

Question 20.
PROBLEM SOLVING
A game show contestant spins the prize wheel shown. The arrow remains in a fixed position while the wheel rotates. The wheel stops spinning, resulting in an image that is a rotation 270° clockwise about the center of the wheel. What is the result?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 62.1

Answer: Free spin

Explanation:
The arrow is located at 90 degrees in the counterclockwise direction of free spin. So when the wheel is rotated 270 degrees in a clockwise direction the arrow will be on the free spin column.

PATTERN
A figure has rotational symmetry if a rotation of 180° or less produces an image that fits exactly on the original figure. Determine whether the figure has rotational symmetry. Explain your reasoning.

Question 21.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 63

Answer: Yes the given figure has rotational symmetry.

Explanation:
The given figure in the problem is rotated 120 degrees in any direction clockwise or counterclockwise then it will produce the same identical image. Since 120 degrees is less than 180 degrees so it will have rotational symmetry.

Question 22.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 64

Answer: No the given figure does not have rotational symmetry.

Explanation:
The given figure in the problem will produce the same identical image only when it is rotated 360 degrees. Since 360 degrees is greater than 180 degrees so it will not have rotational symmetry.

Question 23.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 65

Answer: Yes the given figure has rotational symmetry.

Explanation:
The given figure in the problem will produce the same identical image only when it is rotated 180 degrees. Since the maximum angle for rotational symmetry is 180 degrees so it will have rotational symmetry.

USING MORE THAN ONE TRANSFORMATION
The vertices of a figure are given. Find the coordinates of the image after the transformations given.

Question 24.
R(-7, -5), S(-1, -2), T(-1, -5)
Rotate 90° counterclockwise about the origin. Then translate 3 units left and 8 units up.

Answer:
We know that when a point is rotated 90 degrees counterclockwise about origin then both coordinates gets interchanges and y-coordinate becomes opposite.
P(x,y) = P'(-y,x)
Given, R(-7, -5), S(-1, -2), T(-1, -5)
Rotating 90 degrees counterclockwise about the origin
R(-7,-5) = R'(5,-7)
S(-1, -2) = S'(2,-1)
T(-1, -5) = T'(5,-1)
The coordinate of the image are R'(5,-7), S'(2,-1), T'(5,-1)
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given, R'(5,-7), S'(2,-1), T'(5,-1) and a = -3, b = 8
R'(5+a, -7+b) = R”(5-3, -7+8) = R”(2,1)
S'(2+a, -1+b) = R”(2-3, -1+8) = S”(-1,7)
R'(5+a, -1+b) = R”(5-3, -1+8) = T”(2,7)
The coordinate of the image are R”(2,1), S”(-1,7), T”(2,7)

Question 25.
J(-4, 4), K(-3, 4), L(-1, 1), M(-4, 1) Reflect in the x-axis, and then rotate 180° about the origin.

Answer:
We know that when a point is reflected about x-axis then y-coordinate becomes opposite.
A(x, y) = A'(x, -y)
Given J(-4, 4), K(-3, 4), L(-1, 1), M(-4, 1)
Reflection about the x-axis:
J(-4, 4) = J'(-4,-4)
K(-3, 4) = K'(-3,-4)
L(-1, 1) = L'(-1,-1)
M(-4, 1) = M'(-4,-1)
The coordinate of the image are J'(-4,-4), K'(-3,-4), L'(-1,-1), M'(-4,-1)
Now rotating the above image 180 degrees about the origin.
We know that when a point is reflected about x-axis then y-coordinate becomes opposite.
A(x, y) = A'(-x, -y)
J'(-4,-4), K'(-3,-4), L'(-1,-1), M'(-4,-1)
Rotating 180 degrees about the origin:
J'(-4,-4) = J”(4,4)
K'(-3,-4) = K”(3,4)
L'(-1,-1) = L”(1,1)
M'(-4,-1) = M”(4,1)
The coordinate of the image are J”(4,4), K”(3,4), L”(1,1), M”(4,1)

CRITICAL THINKING
Describe two different sequences of transformations in which the blue figure is the image of the red figure.

Question 26.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 66

Answer:
Two different ways of translating a red figure in to blue figure:
1. First rotate the red figure 90 degrees in the counterclockwise direction and then translate that image 5 units towards the left to get the blue figure.
2. First rotate the red figure 90 degrees in a clockwise direction and then translate that image 1 unit towards the Right and 5 units Up to get the blue figure.

Question 27.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 67

Answer:
Two different ways of translating a red figure in to blue figure:
1. First rotate the red figure 90 degrees in the counterclockwise direction and then translate that image 1 unit towards the left and 1 Down to get the blue figure.
2. First rotate the image in the x-axis and then translate that image 4 units towards the left and 2 units Up to get the blue figure.

Question 28.
REASONING
A trapezoid has vertices A(-6, -2), B(-3, -2), C(-1, -4), and D(-6, -4).
a. Rotate the trapezoid 180° about the origin. What are the coordinates of the image?

Answer:
A(x, y) = A'(-x, -y)
Given,
A(-6, -2), B(-3, -2), C(-1, -4), and D(-6, -4).
Rotating 180 degrees about the origin:
A(-6, -2) = A'(6,2)
B(-3, -2) = B'(3,2)
C(-1, -4) = C'(1,4)
D(-6, -4) = D'(6,4)
The coordinates of the image of trapezoid vertices are A'(6,2), B'(3,2), C'(1,4), D'(6,4)

b. Describe a way to obtain the same image without using rotations.

Answer:
In the above question, we can see that the coordinates of all the vertices of the trapezoid are negative and all the coordinates of the image vertices are positive. So there is another way to get the vertices of the image.
First, reflect the trapezoid in the x-axis and then in the y-axis or first reflect the trapezoid in the y-axis and then in the x-axis.

ROTATING A FIGURE
The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 29.
D(2, 1), E(2, -2), F(-1, 4)
90° counterclockwise about vertex D

Answer:
P(x, y) = P'(-(y – b) + a, (x -a) + b)
Given,
D(2, 1), E(2, -2), F(-1, 4)
(a, b) = (2, 1)
Rotation about the point D(2,1)
D(2, 1) = D'(2,1)
E(2, -2) = E'(-(-2-1) + 2, (2 – 2) + 1) = E'(5, 1)
F(-1, 4) = F'(-(4 – 1) + 2, (-1 – 2) + 1) = F'(-1, -2)
Hence the coordinate of the image: D'(2,1), E'(5, 1), F'(-1, -2)

Question 30.
L(-4, -3), M(-1, -1), N(2, -2)
180° about vertex M

Answer:
When a point is rotated 180 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite and ‘b’ and ‘a’ are subtracted from x and y coordinate respectively.
P(x, y) = P'(-(x – a) + b, -(y – b) + a)
Given,
L(-4, -3), M(-1, -1), N(2, -2)
Rotation about the point M(-1, -1):
L(-4, -3) = L'(-(-4 + 1) – 1, -(-3 + 1) – 1) = L'(2, 1)
M(-1, -1) = M'(-1, -1)
N(2, -2) = N'(-(2 + 1) – 1, -(-2 + 1) – 1) = N'(-4, 0)
Hence the coordinate of the image are L'(2, 1), M'(-1, -1), N'(-4, 0)

Question 31.
W(-5, 0), X(-1, 4), Y(3, -1), Z(0, -4)
270° counterclockwise about vertex W

Answer:
When a point is rotated 270 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite and ‘b’ and ‘a’ are subtracted from x and y coordinate respectively.
P(x, y) = P'(-(x – a) + b, -(y – b) + a)
Given,
W(-5, 0), X(-1, 4), Y(3, -1), Z(0, -4)
Rotation about the point W(-5, 0):
W(-5, 0) = W'(-5, 0)
X(-1, 4) = X'((4 – 0) – 5, -(-1 + 5) + 0) = X'(-1, -4)
Y(3, -1) = Y'((-1 – 0) – 5, -(3 + 5) + 0) = Y'(-6, -8)
Z(0, -4) = Z'((-4 – 0) – 5, -(0 + 5) + 0) = Z'(-9, -5)
Hence the coordinate of the image are W'(-5, 0), X'(-1, -4), Y'(-6, -8), Z'(-9, -5)

Question 32.
D(-3, -4), E(-5, 2), F(1, -1), G(3, -7)
270° clockwise about vertex E.

Answer:
When a point is rotated 90 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite and ‘b’ and ‘a’ are subtracted from x and y coordinate respectively.
P(x, y) = P'(-(x – a) + b, -(y – b) + a)
Given,
D(-3, -4), E(-5, 2), F(1, -1), G(3, -7)
Rotation about the point E(-5, 2):
D(-3, -4) = D'(-(-4 – 2) – 5, -(-3 + 5) + 2) = D'(1, 4)
E(-5, 2) = E'(-5, 2)
F(1, -1) = F'(-(-1 – 2) – 5, (1 + 5) + 2) = F'(-2, 8)
G(3, -7) = G'(-(-7 – 2) – 5, (3 + 5) + 2) = G'(4, 10)
Hence the coordinate of the image are D'(1, 4), E'(-5, 2), F'(-2, 8), G'(4, 10)

Question 33.
LOGIC
You want to find the treasure located on the map at Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 68. You are located at Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 69. The following transformations will lead you to the treasure, but they are not in the correct order. Find the correct order. Use each transformation exactly once.

  • Rotate 180° about the origin.
  • Reflect in the y-axis.
  • Rotate 90° counterclockwise about the origin.
  • Translate 1 unit right and 1 unit up.

Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 68.1

Answer:
The correct order of transformation to get the treasure are:

  • Rotate 180° about the origin.
  • Rotate 90° counterclockwise about the origin.
  • Reflect in the y-axis.
  • Translate 1 unit right and 1 unit up.

Question 34.
DIG DEEPER!
You rotate a triangle 90° counterclockwise about the origin. Then you translate its image 1 unit left and 2 units down. The vertices of the final image are (-5, 0), (-2, 2), and (-2, -1). What are the vertices of the original triangle?

Answer:
When a point is rotated 90 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite.
Let the three vertices of the triangle be: (x1, y1), (x2, y2), (x3, y3)
P(x, y) = P'(-y, x)
Rotating 90 degrees counterclockwise about the origin:
A(x1, y1) = A'(-y1, x1)
B(x2, y2) = B'(-y2, x2)
C(x3, y3) = C'(-y3, x3)
Now translating the image of the vertex for the final image
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x, y) = A'(x + a, y + b)
Given,
A'(-y1, x1), B'(-y2, x2), C'(-y3, x3) and a = -1, b = -2
A'(-y1 + a, x1 + b) = A”(-y1 – 1, x1 – 2)
B'(-y2 + a, x2 + b) = B”(-y2 – 1, x2 – 2)
C'(-y3 + a, x3 + b) = C”(-y3 – 1, x3 – 2)
The given coordinate of vertex point of final image are: (-5, 0), (-2, 2) and (-2, -1)
Now comparing the coordinate of the final image
(-y1 – 1, x1 – 2) = (-5, 0) so y1 = 4 and x1 = 2
(-y2 – 1, x2 – 2) = (-2, 2) so y2 = 1 and x2 = 4
(-y3 – 1, x3 – 2) = (-2, 1) so y3 = 1 and x3 = 1
Hence the vertices of original triangle are (2, 4), (4, 1) and (1, 1)

Lesson 2.4 Congruent Figures

EXPLORATION 1

Work with a partner.
a. For each pair of figures whose vertices are given below, draw the figures in a coordinate plane. Then copy one of the figures onto a piece of transparent paper. Use transformations to try to obtain one of the figures from the other figure.

  • A(-5, 1), B(-5, -4), C(-2, -4) and D(1, 4), E(1, -1), F(-2, -1)
  • G(1, 2), H(2, -6), J(5, 0) and L(-1, -2), M(-2, 6), N(-5, 0)
  • P(0, 0), Q(2, 2), R(4, -2) and X(0, 0), Y(3, 3), Z(6, -3)
  • A(0, 4), B(3, 8), C(6, 4), D(3, 0) and
    F(-4, -3), G(-8, 0), H(-4, 3), J(0, 0)
  • P(-2, 1), Q(-1, -2), R(1, -2), S(1, 1) and
    W(7, 1), X(5, -2), Y(3, -2), Z(3, 1)

Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 69.1
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 69.2
b. Which pairs of figures in part(a) are identical? Explain your reasoning.
c. FigureA and FigureB are identical. Do you think there must be a sequence of transformations that obtains Figure A from Figure B? Explain your reasoning.

2.4 Lesson

Try It

Question 1.
A triangle has vertices X(0, 4), Y(4, 4), and Z(4, 2). Is △XYZ congruent to any of the triangles in Example 1? Explain.

Answer:
Big Ideas Math Grade 8 Chapter 2 transformation key img_22
After plotting the triangle XYZ on the coordinate plane we can say that the triangle XYZ is congruent to triangle PQR among all the given triangle in the figure. In fact, if triangle PQR is rotated 90 degrees clockwise of 270 degrees counterclockwise it will result in the triangle XYZ.

Try It

Question 2.
Describing a different Sequence of rigid motions between the figures.

Answer:
Different sequence of rigid motion to get the blue figure from the red figure are:
1. First rotate the red figure 90 degrees clockwise and the origin.
2. Then translate the image 4 units Right and 1 unit Up.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
IDENTIFYING CONGRUENT FIGURES
Use the coordinate plane shown.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 70
a. Identify any congruent figures.

Answer: a. After seeing the figure we can say that triangle ABCD is congruent to triangle JKLM.

b. A rectangle has vertices W(4, 1), X(4, 2), Y(1, 2), and Z(-1, -1). Is Rectangle WXYZ congruent to any of the rectangles in the coordinate plane? Explain.

Answer:
Big ideas math Grade 8 ch 2 solution key img_23
Rectangle WXYZ is not congruent to any of the rectangles in the given figure because rectangle WXYZ is square of 3 units sides and the other rectangle in the figure does not have all the sides of 3 units.

RIGID MOTIONS
The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 4.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 71

Answer:
The sequence of the rigid motions from the red-figure to the blue figure:
1. First we will rotate the red figure 180 degrees clockwise or anticlockwise about the origin because the given red figure in the 4th quadrant and the blue figure is in the 2nd quadrant.
2. Then we will translate the image 1 unit left because one vertex (-1,-4) of red figure is on the negative side of the x-axis.

Question 5.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 72

Answer:
The sequence of the rigid motions from red figure to blue figure:
1. First we will rotate the red figure 90 degrees clockwise about the origin because the given red figure in the 1st quadrant.
2. Then we will translate the image 3 units right and 1 unit down.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
In the coordinate plane at the left, each grid line represents 50 feet. Each figure represents a pasture.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 73
a. Are the figures congruent? Use rigid motions to justify your answer.

Answer: No the blue figure and red figure are not congruent.

Explanation:
By reflecting the red figure about the y-axis and translating the image 4 units Up we will not the same blue figure. So both figure are not congruent to each other.

b. How many feet of fencing do you need to enclose each pasture?

Answer:
Given the length of each grid line = 50 feet
Total feet of fencing = 50 × total number of grid line along the boundary
For red figure fencing: 50 × 12 = 600 feet
For blue figure fencing: 50 × 12 = 600 feet

Question 7.
A home decorator uses a computer to design a floor tile. How can the decorator transform the tile as shown?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 74

Answer:
First, rotate the given tiles about 90 degrees in the clockwise direction and then take the mirror image about the vertical axis.

Congruent Figures Homework & Practice 2.4

Review & Refresh

The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 1.
A(1, 3), B(2, 5), C(3, 5), D(2, 3)
90° counterclockwise about the origin

Answer:
When a point is rotated 90 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-y, x)
Given,
A(1, 3), B(2, 5), C(3, 5), D(2, 3)
Rotating 90 degrees counterclockwise about the origin:
A(1, 3) = A'(-3, 1)
B(2, 5) = B'(-5, 2)
C(3, 5) = C'(-5, 3)
D(2, 3) = D'(-3, 2)
Hence the coordinate of the image are A'(-3, 1), B'(-5, 2), C'(-5, 3), D'(-3, 2)

Question 2.
F(-2, 1), G(-1, 3), H(3, 1)
180° about the origin

Answer:
When a point is rotated 180 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-x, -y)
Given,
F(-2, 1), G(-1, 3), H(3, 1)
Rotating 90 degrees counterclockwise about the origin:
F(-2, 1) = F'(2,-1)
G(-1, 3) = G'(1,-3)
H(3, 1) = H'(-3,-1)
Hence the coordinate of the image are F'(2,-1), G'(1,-3), H'(-3,-1)

Factor the expression using the greatest common factor.

Question 3.
4n – 32

Answer:
4n – 32
Take 4 as a common factor.
4(n – 8)
Thus the greatest common factor is 4(n – 8)

Question 4.
3w + 66

Answer:
3w + 66
Take 3 as a common factor.
3(w + 22)
Thus the greatest common factor is 3(w + 22)

Question 5.
2y – 18

Answer:
2y – 18
Take 2 as a common factor.
2(y – 9)
Thus the greatest common factor is 2(y – 9).

Concepts, Skills, & Problem Solving
TRANSFORMING FIGURES
The vertices of a pair of figures are given. Determine whether the figures are identical. (See Exploration 1, p. 63.)

Question 6.
G(0, 0), H(3, 2), J(1, -2) and L(-1, 0), M(2, 2), N(0, -3)

Answer:
Big Ideas Math 8th Grade Solution Key Chapter 2 img_24
After plotting the triangles GHJ and LMN we can say that the triangle LMN are bigger compared to the other triangle. Thus both the triangles are not identical.

Question 7.
A(-2, -1), B(-2, 2), C(-1, 1), D(-1, -2) and F(-2, 0), G(-1, 1), H(2, 1), J(1, 0)

Answer:
Answer Key for BIM Grade 8 Chapter 2 tranformations img_25
By seeing both the quadrilaterals ABCD and FGHJ we can say that they are identical.

IDENTIFYING CONGRUENT FIGURES
Identify any congruent figures in the coordinate plane.

Question 8.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 75

Answer:
On observing the diagram in the given figure we can see that the shape and size of pentagon ABCDE and pentagon FKJHG are the same. The length of each side of both the pentagon is the same. Thus they are congruent.

Question 9.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 76

Answer:
By seeing the above figure we can say that the shape and size of parallelogram EFGH and parallelogram BCDA are the same. The length of each side of both the parallelogram are same. Parallelogram BCDA can be obtained by rotating parallelogram EFGH 90 degrees clockwise and translating its image. Hence the parallelogram, EFGH is congruent to BCDA.

DESCRIBING A SEQUENCE OF RIGID MOTIONS
The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 10.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 77

Answer:
The sequence of rigid motions between the red and blue figures are:
1. First we will rotate the red figure 90 degrees clockwise about the origin because the given red figure is in the 2nd quadrant and the blue figure is in the 1st quadrant.
2. Then we will translate the image 1 unit left and 1 unit Down because one vertex of the red figure is at (-1, 1)

Question 11.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 78

Answer:
The sequence of rigid motions between the red and blue figures are:
1. First we will rotate the red figure 180 degrees clockwise or anticlockwise about the origin because the given red figure is in 4th quadrant and the blue figure is in the 2nd quadrant.
2. Then we will translate the image 1 unit Right and 1 unit Down because one vertex of red figure is at (2, -2)

Question 12.
YOU BE THE TEACHER
Your friend describes a sequence of rigid motions between the figures. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 79

Answer:
When a point is reflected about x-axis then the y-coordinate becomes opposite.
A(x, y) = A'(x, -y)
Coordinates of red figure are A(1, -1), B(3, -1), C(4, -3), D(2, -3)
Reflection about the x-axis:
A(1, -1) = A'(1, 1)
B(3, -1) = B'(3, 1)
C(4, -3) = C'(4, 3)
D(2, -3) = D'(2, 3)
Now translating the above image point 5 unit left.
Coordinate of the vertex of blue figure are: A”(-4, 1), B”(3,1), C”(-1, 3), D”(-3,3)
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given:
A(1, 1), B(3, 1), C(4, 3), D(2, 3) and a = -5, b = 0
A”(1+a, 1+b) = A”(1-5, 1+0) = A”(-4, 1)
B”(3+a, 1+b) = B”(3-5, 1+0) = B”(-2,1)
C”(4+a,3+b) = C”(4-5, 3+0) = C”(-1, 3)
D”(2+a,3+b) = D”(2-5, 3+0) = D”(-3,3)
Hence the coordinate of image are A”(-4, 1), B”(-2,1), C”(-1, 3), D”(-3,3)
Since the coordinate of the vertex of the blue is the same in both ways.
We can say that the blue figure is obtained by the rigid motion of the red figure.

NAMING CORRESPONDING PARTS
The figures are congruent. Name the corresponding angles and the corresponding sides.

Question 13.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 80

Answer:
Corresponding sides of the congruent figure are
AD = EH
AB = EF
BC = FG
CD = GH
Corresponding angles of the congruent figure are
∠A = ∠E
∠B = ∠F
∠C = ∠G
∠D = ∠H

Question 14.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 81

Answer:
Corresponding sides of the congruent figure are
PQ = WV
QR = VZ
RS = ZY
ST = YX
TP = XW
Corresponding angles of the congruent figure are
∠P = ∠W
∠Q = ∠V
∠R = ∠Z
∠S = ∠Y
∠T = ∠X

Question 15.
MODELING REAL LIFE
You use a computer program to transform an emoji. How can you transform the emoji as shown?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 82

Answer:
First, take the reflection of that emoji about vertical line and then rotate that image 90 degrees clockwise to get that given emoji.

Question 16.
CRITICAL THINKING
Two figures are congruent. Are the areas of the two figures the same? the perimeters? Explain your reasoning.

Answer:

  • The size of both figures should be the same.
  • The shape of both the figures should be the same.
  • All the corresponding angles should be the same.
  • Both the area and perimeter of two congruent figures are the same.

Question 17.
DIG DEEPER!
The houses are identical.
a. What is the length of side LM?

Answer:
Length of LM = length of CD
length of CD = 32 feet
So, the length of LM is 32 feet

b.Which angle of JKLMN corresponds to ∠D?
Answer:
∠D = ∠M
Thus ∠M corresponds to ∠D

c. Side AB is congruent to side AE. What is the length of side AB? What is the perimeter of ABCDE?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 83.

Answer:
AE = JN
The length of JN is 20 ft
So, the length of AE = 20 ft
Perimeter of ABCDE = AB + BC + CD + DE + EA
Perimeter of ABCDE = 20 + 12 + 32 + 12 + 20 = 96 feet
Thus the Perimeter of ABCDE  is 96 feet

Question 18.
REASONING
Two constellations are represented by the figures in the coordinate plane shown. Are the figures congruent? Justify your answer.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 84

Answer:
The above figure can be tranformed into below figure by rotating the figure 180 degrees clockwise or counterclockwise about the origin and translating the image 8 units Right and 8 units Up to get the above figure.

Lesson 2.5 Dilations

Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 85

EXPLORATION 1

Work with a partner. Use geometry software.
a. Draw a polygon in the coordinate plane. Then dilate the polygon with respect to the origin. Describe the scale factor of the image.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 86
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 86.1
b. Compare the image and the original polygon in part(a). What do you notice about the sides? the angles?
c. Describe the relationship between each point below and the point A(x, y) in terms of dilations.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 87
d. What are the coordinates of a point P(x, y) after a dilation with respect to the origin by a scale factor of k?

2.5 Lesson

Try It

Tell whether the blue figure is a dilation of the red figure.

Question 1.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 88

Answer:
No the blue figure is not the dilation of the red figure.

Explanation:
Blue and red figure has same size and same shape but the blue figure is reflection about vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

Question 2.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 89

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding verrtices meeting at a point. This means that blue figure is dilation of the red figure.

Try It

Question 3.
WHAT IF?
Triangle ABC is dilated by a scale factor of 2. What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: A (1, 3), B (2, 3), C (2, 1) and scale factor = 2
Dilating the figure by scale factor of 2
A (1, 3) = A'(1 . 2, 3 . 2) = A'(2, 6)
B (2, 3) = B'(2 . 2, 3 . 2) = B'(4, 6)
C (2, 1) = C'(2 . 2, 1 . 2) =  C'(4, 2)
Hence the coordinates of the image are A'(2, 6), B'(4, 6),  C'(4, 2)

Try It

Question 4.
WHAT IF?
Rectangle WXYZ is dilated by a scale factor of \(\frac{1}{4}\). What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the rectangle: W(-4, -6), X(-4, 8), Y(4, 8), Z(4, -6)
scale factor = 0.25
W(-4, -6) = W'(-4 × 0.25, -6 × 0.25) = W'(-1, -1.5)
X(-4, 8) = X'(-4 × 0.25, 8 × 0.25) = X'(-1, 2)
Y(4, 8) = Y'(4 × 0.25, 8 × 0.25) = Y'(1, 2)
Z(4, -6) = Z'(4 × 0.25, -6 × 0.25) = Z'(1, -1.5)
Hence the coordinates of the image are W'(-1, -1.5), X'(-1, 2), Y'(1, 2), Z'(1, -1.5)

Try It

Question 5.
WHAT IF?
Trapezoid ABCDis dilated using a scale factor of 3, and then rotated 180° about the origin. What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
a is the scaling factor
Given points of trapezoid: A(-2, -1), B(-1,1), C(0,1), D(0,-1) scale factor = 3
Dilating the figure by scale factor of 3
A(-2, 1) = A'(-2 . 3, -1 . 3) = A'(-6, -3)
B(-1, 1) = B'(-1 . 3, 1 . 3) = B'(-3, 3)
C(0, 1) = C'(0 . 3, 1 . 3) = C'(0, 3)
D(0, -1) = D'(0 . 3, -1 . 3) = D'(0, -3)
Thus the coodrinate of the image are A'(-6, -3), B'(-3, 3), C'(0, 3), D'(0, -3)
when a point is rotated 180 degrees about the origin then both x and y coordinate becomes opposite.
P(x, y) = P'(-x, -y)
Image points: A'(-6, -3), B'(-3, 3), C'(0, 3), D'(0, -3)
Rotating 180 degrees about the origin:
A'(-6, -3) = A”(6, 3)
B'(-3, 3) = B”(3, -3)
C'(0, 3) = C”(0, -3)
D'(0, -3) = D”(0, 3)
Thus the coodrinate of the image are A”(6, 3), B”(3, -3), C”(0, -3), D”(0, 3)

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

IDENTIFYING A DILATION
Tell whether the blue figure is a dilation of the red figure.

Question 6.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 90

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is dilation of the red figure.

Question 7.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 91

Answer: No the blue figure is not the dilation of the red figure.

Explanation:
Blue and red figure has same size and same shape but the blue figure is reflection about vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

Question 8.
DILATING A FIGURE
The vertices of a rectangle are J(4, 8), K(12, 8), L(12, 4), and M(4, 4). Draw the image after a dilation with a scale factor of \(\frac{1}{4}\). Identify the type of dilation.

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
a is scale factor
Given, vertices of a rectangle are J(4, 8), K(12, 8), L(12, 4), and M(4, 4), scale factor = 0.25
J(4, 8) = J'(4 × 0.25, 8 × 0.25) = J'(1, 2)
K(12, 8) = K'(12 × 0.25, 8 × 0.25) = K'(3, 2)
L(12, 4) = L'(12 × 0.25, 4 × 0.25) = L'(3, 1)
M(4, 4) = M'(4 × 0.25, 4 × 0.25) = M'(1, 1)
Hence the coordinates of the image are J'(1, 2), K'(3, 2), L'(3, 1), M'(1, 1)
Big Ideas Math Answers Grade 8 Chapter 2 Transformations img_26

Question 9.
VOCABULARY
How is a dilation different from other transformations?

Answer:
The difference between dilation and other transformations are

  • In the case of dilate the size of the figure after the dilation either decrease or increase but the shape of the figure before and after dilation remains same. Also after the dilation the corresponding angles will be congruent.
  • In case of other transformations such as Rotation, translation, reflection the shape and size of figure before and after transformation remains the same.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 10.
A photograph is dilated to fit in a frame, so that its area after the dilation is 9 times greater than the area of the original photograph. What is the scale factor of the dilation? Explain.

Answer: The scale factor of length and breadth will be 3.

Explanation:
Given,
The area after the dilation is 9 times greater than the area of the original photograph.
Area = length × breadth
p = 3 × 3
Hence the scale factor of length and breadth will be 3.

Question 11.
DIG DEEPER!
The location of a water treatment plant is mapped using a coordinate plane, where each unit represents 1 foot. The plant has vertices (0, 0), (0, 180), (240, 180), and (240, 0). You dilate the figure with a scale factor of \(\frac{1}{3}\). What are the coordinates of the image? What do you need to change so that the image accurately represents the location of the plant? Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 92

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
a is the scaling factor
Location of water treatment plant: A(0, 0), B(0, 180), C(240, 180), D(240, 0)
scale factor = \(\frac{1}{3}\)
Dilating the figure by scale factor of \(\frac{1}{3}\)
A(0, 0) = A'(0 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = A'(0, 0)
B(0, 180) = B'(0 × \(\frac{1}{3}\), 180 × \(\frac{1}{3}\)) = B'(0, 60)
C(240, 180) = C'(240 × \(\frac{1}{3}\), 180 × \(\frac{1}{3}\)) = C'(80, 60)
D(240, 0) = D'(240 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = D'(80, 0)
Hence the coordinates of the image are A'(0, 0), B'(0, 60), C'(80, 60), D'(80, 0)

Dilations Homework & Practice 2.5

Review & Refresh

The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 1.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 93

Answer:
Sequence of rigid motion between the red and blue figure are
1. First rotate the blue figure 90 degrees in counterclockwise direction about the orgin because blue figure in 1st quadrant and red figure is in 3rd quadrant.
2. Then translate the image 1 unit left and 4 units down.

Question 2.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 94

Answer:
Sequence of rigid motion between the red and blue figure are
1. First reflect the blue fiure about x-axis. The image after the reflection will lies in the 3rd quadrant with same orientation.
2. Then translate the image 5 units Right.

Tell whether the ratios form a proportion.

Question 3.
3 : 5 and 15 : 20

Answer:
When two ratios are equal then it is called as proportion.
Given,
Given 3 : 5 and 15 : 20
3/5 and 15/20
3/5 and 3/4
Since the above two ratio are not equal hence they are not proportion

Question 4.
2 to 3 and 12 to 18

Answer:
When two ratios are equal then it is called as proportion.
Given,
2 to 3 and 12 to 18
2/3 and 12/18
2/3 and 2/3
Since the above two ratio are equal hence they are proportion.

Question 5.
7 : 28 and 12 : 48

Answer:
When two ratios are equal then it is called as proportion.
Given,
7 : 28 and 12 : 48
7/28 and 12/48
1/4 and 1/4
Since the above two ratio are equal hence they are proportion.

Concepts, Skills, &Problem Solving

DESCRIBING RELATIONSHIPS
Describe the relationship between the given point and the point A(8, 12) in terms of dilations. (See Exploration 1, p. 69.)

Question 6.
B(16, 24)

Answer:
Given a point and its image: A(8,12), B(16, 24)
Here we can see that both x-coordinate and y-coordinate of image point have increased to double.
This means that in this case, the image figure has become larger by the scale factor of 2 with respect to the origin.
Hence the dilation scale factor is 2.

Question 7.
C(2, 3)

Answer:
Given a point and its image: A(8,12), C(2, 3)
Here we can see that both x-coordinate and y-coordinate of image point has decreased to one-fourth.
This means that in this case, the image figure has become smaller by the scale factor of 0.25 with respect to the origin.
Thus the dilation scale factor is 1/4.

Question 8.
D(6, 9)

Answer:
Given a point and its image: A(8,12), D(6, 9)
Here we can see that both x-coordinate and y-coordinate of image point has decreased to three-fourth.
This means that in this case the image figure has become smaller by the scale factor of 0.75 with respect to the origin.
Thus the dilation scale factor is 3/4.

IDENTIFYING A DILATION
Tell whether the blue figure is a dilation of the red figure.

Question 9.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 95

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of the red figure are congruent to the angles of the red figure. Also there will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is a dilation of the red figure.

Question 10.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 96

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of the red figure are congruent to the angles of the red figure. Also there will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is a dilation of the red figure.

Question 11.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 97

Answer: No the blue figure is not the dilation of the red figure.

Explanation:
The Blue and red figure has same size and same shape but the blue figure is a reflection of vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

Question 12.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 98

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is dilation of the red figure.

Question 13.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 99

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is dilation of the red figure.

Question 14.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 100

Answer: No the blue figure is not the dilation of the red figure.

Explanation:
Blue and red figure has same size and same shape but the blue figure is reflection about vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

DILATING A FIGURE
The vertices of a figure are given. Draw the figure and its image after a dilation with the given scale factor. Identify the type of dilation.

Question 15.
A(1, 1), B(1, 4), C(3, 1); k = 4

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: A(1, 1), B(1, 4), C(3, 1) and scale factor = 4
Dilating the figure by scale factor by 4
A(1, 1) = A'(1 × 4, 1 × 4) = A'(4, 4)
B(1, 4) = B'(1 × 4, 4 × 4) = B'(4, 16)
C(3, 1) = C'(3 × 4, 1 × 4) = C'(12, 4)
Hence the coordinate of the image are A'(4, 4), B'(4, 16), C'(12, 4)
Big Ideas Math 8th Grade Solution Key Chapter 2 Transformations img_27

Question 16.
D(0, 2), E(6, 2), F(6, 4); k = 0.5

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: D(0, 2), E(6, 2), F(6, 4) and scale factor = 0.5
Dilating the figure by scale factor by 0.5
D(0, 2) = D'(0 × 0.25, 2 × 0.25) = D'(0, 1)
E(6, 2) = E'(6 × 0.25, 2 × 0.25) = E'(3, 1)
F(6, 4) = F'(6 × 0.25, 4 × 0.25) = F'(3, 2)
Hence the coordinate of the image are D'(0, 1), E'(3, 1), F'(3, 2)
BIM 8th grade solution key for chapter 2 transformations img_28

Question 17.
G(-2, -2), H(-2, 6), J(2, 6); k = 0.25

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: G(-2, -2), H(-2, 6), J(2, 6) and scale factor = 0.25
G(-2, -2) = G'(-2 × 0.25, -2 × 0.25) = G'(-0.5, -0.5)
H(-2, 6) = H'(-2 × 0.25,6 × 0.25) = H'(-0.5, 1.5)
J(2, 6) = G'(2 × 0.25, 6 × 0.25) = J'(0.5, 1.5)
Hence the coordinate of the image are G'(-0.5, -0.5), H'(-0.5, 1.5), J'(0.5, 1.5)
Big Ideas Math Book Answers Grade 8 Chapter 2 Transformations img_29

Question 18.
M(2, 3), N(5, 3), P(5, 1); k = 3

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: M(2, 3), N(5, 3), P(5, 1) and scale factor = 3
M(2, 3) = M'(2 × 3, 3 × 3) = M'(6, 9)
N(5, 3) = N'(5 × 3, 3 × 3) = N'(15, 9)
P(5, 1) = P'(5 × 3, 1 × 3) = P'(15, 3)
Hence the coordinate of the image are M'(6, 9), N'(15, 9), P'(15, 3)
Big Ideas Math Grade 8 Chapter 2 Solution Key img_30

Question 19.
Q(-3, 0), R(-3, 6), T(4, 6), U(4, 0); k = \(\frac{1}{3}\)

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
Q(-3, 0), R(-3, 6), T(4, 6), U(4, 0) and scale factor = \(\frac{1}{3}\)
Q(-3, 0) = Q'(-3 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = Q'(-1, 0)
R(-3, 6) = R'(-3 × \(\frac{1}{3}\), 6 × \(\frac{1}{3}\)) = R'(-1, 2)
T(4, 6) = T'(4 × \(\frac{1}{3}\), 6 × \(\frac{1}{3}\)) = T'(4/3, 2)
U(4, 0) = U'(4 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = U'(4/3, 0)
Hence the coordinate of the image are Q'(-1, 0), R'(-1, 2), T'(4/3, 2), U'(4/3, 0)
Big Ideas Math Grade 8 ch 2 transformations answer key img_31

Question 20.
V(-2, -2), W(-2, 3), X(5, 3), Y(5, -2); k = 5

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
V(-2, -2), W(-2, 3), X(5, 3), Y(5, -2), scaling factor = 5
Dilating the figure by scale factor of 5
V(-2, -2) = V'(-2 × 5, -2 × 5) = V'(-10, -10)
W(-2, 3) = W'(-2 × 5, 3 × 5) = W'(-10, 15)
X(5, 3) = X'(5 × 5, 3 × 5) = X'(25, 15)
Y(5, -2) = Y'(5 × 5, -2 × 5) = Y'(25, -10)
Hence the coordinate of the image are V'(-10, -10),W'(-10, 15), X'(25, 15), Y'(25, -10)
Big Ideas Math Grade 8 ch 2 answer key img_32

Question 21.
YOU BE THE TEACHER
Your friend finds the coordinates of the image of △ABC after a dilation with a scale factor of 2. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 101

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
The points of triangle: A(2, 5), B(2, 0), C(4, 0)
scale factor = 2
Dilating the figure by scale factor of 2
A(2, 5) = A'(2 × 2, 5 × 2) = A'(4, 10)
B(2, 0) = B'(2 × 2, 0 × 2) = B'(4, 0)
C(4, 0) = C'(4 × 2, 0 × 2) = C'(8, 0)
Hence the coordinate of the image are A'(4, 10), B'(4, 0), C'(8, 0)
By this, we can say that my friend is correct.

FINDING A SCALE FACTOR
The blue figure is a dilation of the red figure. Identify the type of dilation and find the scale factor.

Question 22.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 102

Answer:
Scale factor = side length of image/side length of original figure
Scale factor = A’B’/AB = 6/3 = 2
Scale factor = 2
Hence, type of dilation is enlargement with scale factor of 2.

Question 23.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 103

Answer:
Scale factor = side length of image/side length of original figure
Scale factor = X’Y’/XY= 2/8 = 1/4
Scale factor = 1/4
Hence, the type of dilation is reduction with the scale factor of 1/4

Question 24.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 104

Answer:
Scale factor = side length of image/side length of the original figure
Scale factor = J’K’/JK = 15/10 = 3/2
Scale factor = 3/2
Hence, the type of dilation is reduction with the scale factor of 3/2

Question 25.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 105

Answer:
Scale factor = side length of image/side length of original figure
Scale factor = Q’R’/QR = 4/8 = 1/2
Scale factor = 1/2
Hence, type of dilation is reduction with scale factor of 1/2

USING MORE THAN ONE TRANSFORMATION
The vertices of a figure are given. Find the coordinates of the image after the transformations given.

Question 26.
A(-5, 3), B(-2, 3), C(-2, 1), D(-5, 1)
Reflect in the y-axis. Then dilate using a scale factor of 2.

Answer:
We know that when a point is reflected about y-axis then is x-coordinate becomes opposite.
A(-5, 3), B(-2, 3), C(-2, 1), D(-5, 1)
A(x, y) = A'(-x, y)
A(-5, 3) = A'(5, 3)
B(-2, 3) = B'(2, 3)
C(-2, 1) = C'(2, 1)
D(-5, 1) = D'(5, 1)
Coordinate of the image are A'(5, 3), B'(2, 3), C'(2, 1), D'(5, 1)
A(-5, 3), B(-2, 3), C(-2, 1), D(-5, 1)
Reflect in the y-axis. Then dilate using a scale factor of 2
A'(5, 3) = A”(5 × 2, 3 × 2) = A”(10, 6)
B'(2, 3) = B”(2 × 2, 3 × 2) = B”(4, 6)
C'(2, 1) = C”(2 × 2, 1 × 2) = C”(4, 2)
D'(5, 1) = D”(5 × 2, 1 × 2) = D”(10, 2)
Coordinate of the image are A”(10, 6), B”(4, 6), C”(4, 2), D”(10, 2)
Big ideas math grade 8 chapter 2 solution key img_33

Question 27.
F(-9, -9), G(-3, -6), H(-3, -9).
Dilate using a scale factor of \(\frac{2}{3}\). Then translate 6 units up.

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
F(-9, -9), G(-3, -6), H(-3, -9) and scale factor of \(\frac{2}{3}\)
Dilating the figure by scale factor of \(\frac{2}{3}\)
F(-9, -9) = F'(-9 × 2/3, -9 × 2/3) = F'(-6, -6)
G(-3, -6) = G'(-3 × 2/3, -6 × 2/3) = G'(-2, -4)
H(-3, -9) = H'(-3 × 2/3, -9 × 2/3) = H'(-2, -6)
Coordinate of the image are F'(-6, -6), G'(-2, -4), H'(-2, -6)
Now translating above image 6 units up
F'(-6, -6), G'(-2, -4), H'(-2, -6) and a = 0, b = 6
F”(-6 + a, -6 + b) = F”(-6 + 0, -6 + 6) = F”(-6, 0)
G”(-2 + a, -4 + b) = G”(-2 + 0, -4 + 6) = G”(-2, 2)
H”(-2 + a, -6 + b) = H”(-2 + 0, -6 + 6) = H”(-2, 0)
Coordinate of the image are F”(-6, 0), G”(-2, 2), H”(-2, 0)
BIM Grade 8 Answers Chapter 2 img_34

Question 28.
J(1, 1), K(3, 4), L(5, 1)
Rotate 90° clockwise about the origin. Then dilate using a scale factor of 3.

Answer:
The rotation of an object 90 degrees clockwise is equal to the rotation of 270 degrees counterclockwise.
When a point is rotated 270 degrees counterclockwise about the origin then both x and y-coordinates gets interchanged and the x-coordinate becomes the opposite.
A(x, y) = A'(y, -x)
J(1, 1), K(3, 4), L(5, 1)
Rotate 90° clockwise about the origin.
J(1, 1) = J'(1, -1)
K(3, 4) = K'(4, -3)
L(5, 1) = L'(1, -5)
Coordinate of the image are J'(1, -1), K'(4, -3), L'(1, -5)
Now dilate using a scale factor of 3.
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
J'(1, -1) = J”(1 . 3, -1 . 3) = J”(3, -3)
K'(4, -3) = K”(4 . 3, -3 . 3) = K”(12, -9)
L'(1, -5) = L”(1 . 3, -5 . 3) = L”(3, -15)
Coordinate of the image are J”(3, -3), K”(12, -9), L”(3, -15)
BIM 8th Grade Answers Ch 2 transformations img_35

Question 29.
LOGIC
You can use a flashlight and a shadow puppet (your hands) to project shadows on the wall.
a. Identify the type of dilation.p

Answer: The type of dilation is an enlargement

b. What does the flashlight represent?

Answer: Flashlight represents center of dilation because all the line connecting shadow and hand meet at the flashlight.
c. The length of the ears on the shadow puppet is 3 inches. The length of the ears on the shadow is 4 inches. What is the scale factor?

Answer: Scale factor = length of ears on shadow/length of ears on puppet
Scale factor = 4/3
d. Describe what happens as the shadow puppet moves closer to the flashlight. How does this affect the scale factor?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 106

Answer:
As the flashlight will move closer the shadow will become larger. Also, the scale factor will increase.

Question 30.
REASONING
A triangle is dilated using a scale factor of 3. The image is then dilated using a scale factor of \(\frac{1}{2}\). What scale factor can you use to dilate the original triangle to obtain the final image? Explain.

Answer:
Given the first scale factor of triangle S1 = 3
Given second scale factor of triangle S2 = 1/2
We know that the final scale factor S = S1 × S2
Final Scale factor S = 3 × 1/2 = 3/2
Hence, the scale factor of the final image will be the multiplication of the first and second dilation scale factor and the final scale factor will be 3/2.

CRITICAL THINKING
The coordinate notation shows how the coordinates of a figure are related to the coordinates of its image after transformations. What are the transformations? Are the figure and its image congruent? Explain.

Question 31.
(x, y) → (2x + 4, 2y – 3)

Answer:
Given, (x, y) → (2x + 4, 2y – 3)
We can see that both x-coordinate and y-coordinate has been multiplied by 2 this means that the point has been dilated by the scale factor of 2.
Also, 4 has been added to x-coordinate while 3 is added to y-coordinate which means that obtained after the dilation has been translated 4 unit Right and 3 units Down.
The final image will not be congruent because after the dilation the size of the image either increases or decreases that depend on the type of dilation.

Question 32.
(x, y) → (-x – 1, y – 2)

Answer:
Given, (x, y) → (-x – 1, y – 2)
We can see that 1 has been subtracted from x-coordinate while 2 is subtracted from y-coordinate which means that will image has translated 1 unit left and 2 units down. And also x-coordinate is opposite which means the image has been reflected about the y-axis.
Hence, transforms translation of 1 unit left and 2 units down followed by reflection about y-axis.

Question 33.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 107

Answer:
Given, (x, y) → (1/3x, -1/3y)
We can see that both x-coordinate and y-coordinate has been multiplied by 2 this means that the point has been dilated by the scale factor of 1/3. Also, y-coordinate is opposite which means that image obtained after the dilation has been reflected about the x-axis.
Thus transforms are dilation with the scale factor of 1/3 followed by reflection about the x-axis.

STRUCTURE
The blue figure is a transformation of the red figure. Use coordinate notation to describe the transformation. Explain your reasoning.

Question 34.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 108

Answer:
Coordinates of original figure A(1, 1) B(1, 2), C(2, 1)
Coordinates of red figure A'(2, 3) B'(2, 6), C'(4, 3)
Scale factor of x-coordinate = x-coordinate of image/x-coordinate of image = 2/1 = 2
Scale factor of y-coordinate = y-coordinate of image/y-coordinate of image = 3/1 = 3
Thus to transfer the red-figure into the blue figure x-coordinate of all the points has been multiplied by 2 and the y-coordinate of all the points has been multiplied by 3.

Question 35.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 109

Answer:
Coordinates of original figure A(4, 4) B(4, 8), C(8, 8), D(8, 4)
Coordinates of red figure A'(1, 2) B'(1, 4), C'(2, 4), D'(2. 2)
Scale factor of x-coordinate = x-coordinate of image/x-coordinate of image = 1/4 = 0.25
Scale factor of y-coordinate = y-coordinate of image/y-coordinate of image = 2/4 = 1/2 = 0.5
Thus to transfer the red figure into the blue figure x-coordinate of all the points has been multiplied by 0.25 and the y-coordinate of all the points has been multiplied by 0.50

Question 36.
NUMBER SENSE
You dilate a figure using a scale factor of 2, and then translate it 3 units right. Your friend translates the same figure 3 units right and then dilates it using a scale factor of 2. Are the images congruent? Explain.

Answer:
Blue the final image in both the case will be of the same shape and size.
Yes, the image in both cases will be the same.

Question 37.
PROBLEM SOLVING
The vertices of a trapezoid are A(-2, 3), B(2, 3), C(5, -2), and D(-2, -2). Dilate the trapezoid with respect to vertex A using a scale factor of 2. What are the coordinates of the image? Explain the method you used.

Answer:
When the points of a given figure are dilated about a point we simply multiply the distance of each side by the given scale factor. The coordinate of one point remains the same about which dilation occurs.
The vertices of a trapezoid are A(-2, 3), B(2, 3), C(5, -2), and D(-2, -2).
Scale factor = 2
So, here the coordinate of point A(-2, 3) will remains the same but all the other coordinates of points B’, C’, D’ will change according to the distance between each side of the trapezoid.
Big Ideas Math Key Grade 8 Chapter 2 transformations img_36
Image of the figure after dilating by a scale factor of 2
A(-2, 3) = A'(-2, 3)
B(2, 3) = B'(6, 3)
C(5, -2) = C'(12, -7)
D(-2, -2) = D'(-2, -7)
Thus the coordinate of the image are A'(-2, 3), B'(6, 3), C'(12, -7), D'(-2, -7)

Question 38.
DIG DEEPER!
A figure is dilated using a scale factor of -1. How can you obtain the image without using a dilation? Explain your reasoning.

Answer:
When a figure is dilated using a scale factor of -1 then both the x and y-coordinate of the image will become opposite.
Example:
A'(x . -1, y . -1) = A'(-x, -y)
But there are two ways to get the same image:
1. By rotating the figure 180 degrees clockwise or anticlockwise
A(x, y) rotating 180 degrees about the origin = A'(-x, -y)
2. By rotating the figure about the x-axis and y axis
A(x, y) reflecting about the origin = A'(-x, -y)

Lesson 2.6 Similar Figures

EXPLORATION 1

Work with a partner. Use geometry software.
a. For each pair of figures whose vertices are given below, draw the figures in a coordinate plane. Use dilations and rigid motions to try to obtain one of the figures from the other figure.

  • A(-3, 6), B(0, -3), C(3, 6) and G(-1, 2), H(0, 1), J(1, 2)
  • D(0, 0), E(3, 0), F(3, 3) and L(0, 0), M(0, 6), N(-6, 6)
  • P(1, 0), Q(4, 2), R(7, 0) and X(-1, 0), Y(-4, 6), Z(-7, 0)
  • A(-3, 2), B(-1, 2), C(-1, -1), D(-3, -1) and F(6, 4), G(2, 4), H(2, -2), J(6, -2)
  • P(-2, 2), Q(-1, -1), R(1, -1), S(2, 2) and W(2, 8), X(3, 3), Y(7, 3), Z(8, 8)

Big Ideas Math Answers 8th Grade Chapter 2 Transformations 110
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 111
b. Is a scale drawing represented by any of the pairs of figures in part(a)? Explain your reasoning.
c. Figure A is a scale drawing of Figure B. Do you think there must be a sequence of transformations that obtains Figure A from Figure B? Explain your reasoning.

2.6 Lesson

Try It

Question 1.
A triangle has vertices D(0, 4), E(5, 4), and F(5, 0). Is △DEF similar to △ABC and △JKL in Example 1? Explain.

Answer:
Given coordinate of the triangle ABC: A(0, 3), B(3, 3), C(3, 0)
Given coordinate of the triangle DEF: D(0, 4), E(5, 4), F(5, 0)
Given coordinate of the triangle JKL: J(0, 6), K(6, 6), L(6, 0)
Here we can see that there is no fixed relation between the coordinate between triangle ABC and DEF or triangle ABC and JKL. So no triangle is dilation with the triangle ABC.
Hence, triangle ABC is not similar △DEF and △JKL.

Try It

Question 2.
Can you reflect the red figure first, and then perform the dilation to obtain the blue figure? Explain.

Answer:
Because the final image will not depend on the order of transformation. When we will first reflect red figure then the image will be of the same size and after the dilation of the image obtained after reflection we will get the same image so we can use any two method but the final image will be the same

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
IDENTIFYING SIMILAR FIGURES
In the coordinate plane at the left, determine whether Rectangle ABCD is similar to Rectangle EFGH. Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 112

Answer: No rectangle ABCD is not similar to rectangle EFGH

Explanation:
Because the orientation of rectangle ABCD is not the same as the rectangle EFGH. Also, rectangle ABCD is not dilated with rectangle EFGH. So there is no similarity transformation between rectangle ABCD and rectangle EFGH.

Question 4.
SIMILARITY TRANSFORMATION
The red triangle is similar to the blue triangle. Describe a similarity transformation between the figures.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 113

Answer:
Coordinate of left vertex of the red triangle: A(5, 5)
Coordinate of the same vertex after dilation: A'(10, 10)
Now coordinate left vertex of the blue triangle: A”(0, 2)
So, the value of a = 0 – 10 = -10 and b = 2 – 10 = -8
It is given that the red triangle is similar to the blue triangle so the steps of transformation:
First, dilate the figure by the scale factor of 2 and then translate the image 10 units left and 8 units down.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
A medical supplier sells gauze in large and small rectangular sheets. A large sheet has a length of 9 inches and an area of 45 square inches. A small sheet has a length of 4 inches and a width of 3 inches. Are the sheets similar? Justify your answer.

Answer:
Condition for the rectangular sheets to be similar is that all the corresponding sides of bigger and smaller rectangular sheets should be in proportional.
Area of larger rectangular sheets a = 45 sq in
length of larger rectangular sheets l1 = 9 in
Width = a/l = 45/9 = 5 in
length of smaller rectangular sheets l2 = 4 in
width of smaller rectangular sheets b2 = 3 in
condition for similarity l1/l2 = b1/b2
9/4 ≠ 5/3
These sheets are not similar

Question 6.
The sail on a souvenir boat is similar in shape to the sail on a sailboat. The sail on the sailboat is in the shape of a right triangle with a base of 9 feet and a height of 24 feet. The height of the souvenir’s sail is 3 inches. What is the base of the souvenir’s sail?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 114

Answer:
Given,
The base of sail on sailboat b1 = 9 ft
height of sail on sailboat h1 = 24 ft
Given height of sail on souvenir boat: h2 = 3in = 0.25 ft
h1/h2 = b1/b2
24/0.25 = 9/h2
h2 = (9 × 0.25)/24 = 0.9375 ft = 1.125 in
Thus the height of sail of a souvenir boat is 1.125 in

Question 7.
DIG DEEPER!
A coordinate plane is used to represent a cheerleading formation. The vertices of the formation are A(4, 4), B(0, 8), C(4, 4), and D(0, 6). A choreographer creates a new formation similar to the original formation. Three vertices of the new formation are J(-2, -2), K(0, -4), and L(2, -2). What is the location of the fourth vertex? Explain.

Answer:
The vertices of the formation are A(4, 4), B(0, 8), C(4, 4), and D(0, 6). A choreographer creates a new formation similar to the original formation.
We observe the image point carefully that both the x and y coordinate of the image point is just half of the original point and each y-coordinate is opposite.
A(4, 4) = J(-2, -2)
B(0, 8) = K(0, -4)
C(4, 4) = L(2, -2)
D(0, 6) = M(x, y)
This means that the point A, B, C are dilated by using a scale factor of 0.5 and the image obtained from the dilation is reflected about the x-axis.
So, the image point is D(0, 6) = M(0, -3)
Big Ideas Math Answers Grade 8 Chapter 2 Transformations img_37

Similar Figures Homework & Practice 2.6

Review & Refresh

Tell whether the blue figure is a dilation of the red figure.

Question 1.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 115

Answer: No

Explanation:
Because the shape and size of both red and blue figure are the same which is not the property of dilation. The blue figure is the result of the reflection of the red figure 180 degrees in the clockwise or counterclockwise direction.

Question 2.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 116

Answer: Yes

Explanation:
When we see both red and blue figures closely we observe that all the angles of the red figure are congruent to the blue figure. Also, there will be the lines connecting corresponding vertices meeting at a point. This means that the blue figure is a dilation of red figure.

Question 3.
You solve the equation S = lw + 2wh for w. Which equation is correct?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 117

Answer: Option C

Explanation:
S = lw + 2wh
lw + 2wh = S
Taking w as a common factor
w(l + 2h) = s
w = s/(l + 2h)
Thus the correct answer is option C.

Concepts, Skills, &Problem Solving
TRANSFORMING FIGURES
The vertices of a pair of figures are given. Determine whether a scale drawing is represented by the pair of figures. (See Exploration 1, p. 77.)

Question 4.
A(-8, -2), B(-4, 2), C(-4, -2) and G(2, -1), H(4, -1), J(2, -3)

Answer: Yes

Explanation:
BIM Grade 8 Answer Key Chapter 2 Transformations img_38
After plotting both triangles we see that the original figure is exactly double of the image. Each side of the original triangle is double the length of the image triangle. So given vertices pair represent a scale drawing.
Scale factor = 2

Question 5.
A(0, 3), B(3, 4), C(5, 3), D(3, 2) and F(-4, 4), G(-1, 5), H(5, 3), J(3, 2)

Answer: No

Explanation:
BIM Grade 8 Answers Ch 2 img_39
After plotting both given figures we can see that there is no transformation relation between the original figure and the image figure. So, the given vertices pair does not represent a scale drawing.

IDENTIFYING SIMILAR FIGURES
Determine whether the figures are similar. Explain your reasoning.(See Exploration 1, p. 77.)

Question 6.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 118

Answer: Rectangle ABCD is similar to rectangle EFGH.

Explanation:
Because when you see both the given figure we can see that all the corresponding angles of rectangle ABCD and rectangle EFGH are equal. And also the corresponding sides of both rectangle are in proportional.
∠A = ∠E, ∠B = ∠F, ∠C = ∠G, ∠D = ∠H
AB/EF = BC/FG = GH/CD = DA/HE = 1/2
Hence rectangle ABCD is similar to rectangle EFGH.

Question 7.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 119

Answer: Both the triangle are not similar

Explanation:
Because when you see both the given figure we can see that all the corresponding angles of triangle ABC and triangle JKL are equal. And also the corresponding sides of both triangles are not proportional.
AB/JK ≠ KL/BC ≠ CA/LJ
Hence triangle ABC is not similar to triangle JKL

IDENTIFYING SIMILAR FIGURES
Draw the figures with the given vertices in a coordinate plane. Which figures are similar? Explain your reasoning.

Question 8.
Rectangle A: (0, 0), (4, 0), (4, 2), (0, 2)
Rectangle B: (0, 0), (6, 0), (6, 3), (0, 3)
Rectangle C: (0, 0), (4, 0), (4, 2), (0, 2)

Answer: Rectangle A and B are similar

Explanation:
Rectangle A: (0, 0), (4, 0), (4, 2), (0, 2)
Big Ideas Math Grade 8 Chapter 2 Transformations img_39(i)
Rectangle B: (0, 0), (6, 0), (6, 3), (0, 3)
Big Ideas Math Grade 8 Chapter 2 Transformations img_39(ii)
Rectangle C: (0, 0), (4, 0), (4, 2), (0, 2)
Big Ideas Math Grade 8 Chapter 2 Transformations img_39(iii)
By seeing the above figure we can say that rectangle A and rectangle B are similar and rectangle A and Rectangle C are congruent.

Question 9.
FigureA: (4, 2), (2, 2), (2, 0), (4, 0)
Figure B: (1, 4), (4, 4), (4, 1), (1, 1)
Figure C: (2, 1), (5, 1), (5, 3), (2, 3)

Answer: Rectangle A and B are similar

Explanation:
FigureA: (4, 2), (2, 2), (2, 0), (4, 0)
Big Ideas Math Grade 8 Chapter 2 Transformations img_40(i)
Figure B: (1, 4), (4, 4), (4, 1), (1, 1)
Big Ideas Math Grade 8 Chapter 2 Transformations img_40(ii)
Figure C: (2, 1), (5, 1), (5, 3), (2, 3)
Big Ideas Math Grade 8 Chapter 2 Transformations img_40(iii)
Rectangles A and B are similar because in rectangle A and B all the corresponding angles are equal and also all the corresponding sides are equal.

DESCRIBING A SIMILARITY TRANSFORMATION
The red figure is similar to the blue figure. Describe a similarity transformation between the figures.

Question 10.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 120

Answer:
1. First rotate the red figure 90 degrees anticlockwise because the red figure is in the first quadrant and the blue figure is in the second quadrant.
2. Then dilate the image obtained after the rotation by the scale factor of 2 because the blue figure is double the size of the red figure.
Scale factor = side of the image/side of the original figure = 4/2 = 2

Question 11.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 121

Answer:
First, dilate the red figure by the scale figure of 3 because the blue figure is triple the size of red figure.
scale factor = side of image/side of original figure = 6/2 = 3

Question 12.
MODELING REAL LIFE
A barrier in the shape of a rectangle is used to retain oil spills. On a blueprint, a similar barrier is 9 inches long and 2 inches wide. The width of the actual barrier is 1.2 miles. What is the length of the actual barrier?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 122

Answer:
Given,
Width of the actual barrier = 1.2 miles
Width of the barrier in the blueprint = 2 inches
2 inch dimension of blueprint = 1.2 miles of original
So, 1 inch dimension of blueprint = 1.2/2 = 0.6 miles.
Since the length of the barrier in the blueprint = 9 inches,
Thus the length of the actual barrier = 9(0.6) = 5.4 miles.

Question 13.
LOGIC
Are the following figures always, sometimes, or never similar? Explain.
a. two triangles
b. two squares
c. two rectangles

Answer:
a. Two triangles sometimes two triangles are similar when all the corresponding angles are equal and all the corresponding sides lengths are in proportion.
b. Two squares always two square are similar only when all the sides are proportional and all the angles are equal.
c. Two rectangles are similar when all the corresponding angles are equal but the lengths of the corresponding sides are not always in proportion.

Question 14.
CRITICAL THINKING
Can you draw two quadrilaterals each having two 130° angles and two 50° angles that are not similar? Justify your answer.

Answer:
Quadrilateral 1: 50°, 50°, 130°, 130° (trapezoid)
Quadrilateral 2: 50°, 130°, 50°, 130° (parallelogram)
Big Ideas Math Grade 8 ch 2 transformations answer key img_41

Question 15.
REASONING
The sign is rectangular.
a. You increase each side length by 20%. Is the new sign similar to the original? Explain your reasoning.

Answer:
Given,
You increase each side length by 20%
Scale factor = 1 + percentage increase/100 = 1 + 20/100 = 1.2

b. You increase each side length by 6 inches. Is the new sign similar to the original? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 123

Answer:
No, because when length and width are of a different size then adding 6 inches on each side will not increase the figure in a fixed proportion. So the corresponding length will be not proportional. This means that both the figure will be not in dilation, hence not similar.

Question 16.
DIG DEEPER!
A person standing 20 feet from a streetlight casts a shadow as shown. How many times taller is the streetlight than the person? Assume the triangles are similar.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 124

Answer:
Length of shadow l1 = 10 ft
height of man b1 = 6 ft
Total length of bigger triangle l2 = 20 + 10 = 30 ft
l1/l2 = b1/b2
10/30 = 6/b2
b2 = 180/10
b2 = 18 ft
The ratio of the height of streetlight and man is: 18/6 = 3
Hence, the streetlight is 3 times taller than that person.

Question 17.
GEOMETRY
Use a ruler to draw two different isosceles triangles similar to the one shown. Measure the heights of each triangle.
a. Are the ratios of the corresponding heights equivalent to the ratios of the corresponding side lengths?

Answer:
b1/b2 = h1/h2
6/3 = 4/2 = 2
Hence the ratio of corresponding heights is equivalent to the ratio of corresponding side lengths.

b. Do you think this is true for all similar triangles? Explain.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 125

Answer:
Yes, this will be true for all the similar triangles because the heights of the two similar triangles are multiplied by the same amount as the sides.

Question 18.
CRITICAL THINKING
Given △ABC ∼ △DEF and △DEF ∼ △JKL, is △ABC ∼ △JKL? Justify your answer.

Answer:
When △ABC is similar to △DEF and △DEF is similar to △JKL, then △ABC is similar to △JKL.

Lesson 2.7 Perimeters and Areas of Similar Figures

EXPLORATION 1

Work with a partner. Draw a rectangle in the coordinate plane.
a. Dilate your rectangle using each indicated scale factor k. Then complete the table for the perimeter P of each rectangle. Describe the pattern.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 126
b. Compare the ratios of the perimeters to the ratios of the corresponding side lengths. What do you notice?
c. Repeat part(a) to complete the table for the area A of each rectangle. Describe the pattern.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 127
d. Compare the ratios of the areas to the ratios of the corresponding side lengths. What do you notice?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 128
e. The rectangles shown are similar. You know the perimeter and the area of the red rectangle and a pair of corresponding side lengths. How can you find the perimeter of the blue rectangle? the area of the blue rectangle?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 500

2.7 Lesson

Try It

Question 1.
The height of Figure A is 9 feet. The height of a similar Figure B is 15 feet. What is the value of the ratio of the perimeter of A to the perimeter of B?

Answer: The ratio of the perimeter of A to B is 3/5

Explanation:
We know that when two figures are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = Height of figure A/Height of figure B
Perimeter of figure A/Perimeter of figure B = 9/15 = 3/5
Thus the ratio of the perimeter of A to B is 3/5

Try It

Question 2.
The base of Triangle P is 8 meters. The base of a similar Triangle Q is 7 meters. What is the value of the ratio of the area of P to the area of Q?

Answer:
We know that when two figures are similar then the value of the ratio of their area is equal to the square of the value of the ratio of their corresponding side lengths.
The base of Triangle P is 8 meters. The base of a similar Triangle Q is 7 meters.
b1 = 8 m
b2 = 7 m
Area of triangle P/Area of triangle Q = base of triangle P/base of triangle Q

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

COMPARING PERIMETERS OF SIMILAR FIGURES
Find the value of the ratio (red to blue) of the perimeters of the similar figures.

Question 3.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 130

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
l1 = 9
l2 = 7
Perimeter of red figure/Perimeter of blue figure = Side length of red figure/Side length of blue figure
Perimeter of red figure/Perimeter of blue figure = 9/7
Thus the ratio of the perimeter of red to blue figure is 9/7

Question 4.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 131

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
b1 = 8
b2 = 10
Perimeter of red figure/Perimeter of blue figure = base length of red figure/base length of blue figure
Perimeter of red figure/Perimeter of blue figure = 8/10 = 4/5
Thus the ratio of the perimeter of red to blue triangle is 4/5

COMPARING AREAS OF SIMILAR FIGURES
Find the value of the ratio (red to blue) of the areas of the similar figures.

Question 5.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 132

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = (Side length of figure A/Side length of figure B)²
l1 = 12
l2 = 8
Perimeter of red figure/Perimeter of blue figure = side length of red figure/side length of blue figure
Perimeter of red figure/Perimeter of blue figure = (12/8)² = (3/2)² = 9/4
Thus the ratio of the perimeter of red to blue figure is 9/4

Question 6.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 133

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Area of figure A/Area of figure B = (Side length of A/Side length of B)²
l1 = 12
l2 = 8
Area of red figure/Area of blue figure = side length of red figure/side length of blue figure
Area of red figure/Area of blue figure = (4/5)² = 16/225
Thus the ratio of the perimeter of red to blue triangle is 16/225

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
Two similar triangular regions are prepared for development.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 134
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 136
a. It costs $6 per foot to install fencing. How much does it cost to surround the forest with a fence?

Answer:
Given,
It costs $6 per foot to install fencing.
Perimeter of grassland/perimeter of forest = Height of grassland/Height of forest
h1 = 60 yards
The perimeter of grassland = 240 yards
Height of forest h2 = 45 yards
Perimeter of grassland/perimeter of forest = 60/45
240/ perimeter of forest = 60/45
the perimeter of forest = 180 yards
Convert from yards to feet
180 yards = 540 feet
Thus the cost of fencing forest = 6 × 540 = $3,240

b. The cost to prepare 1 square yard of grassland is $15 and the cost to prepare 1 square yard of forest is $25. Which region costs more to prepare? Justify your answer.

Answer:
Perimeter of grassland/perimeter of forest = (Height of grassland/Height of forest)²
Height of grassland h1 = 60 yard
Height of forest h2 = 45 yards
Area of grassland = 2400 yd²
Cost to prepare 1 sq yd of grassland = $15
Cost to prepare 1 sq yd of forest = $25
Area of forest = (2400 × 9)/16
Thus the area of forest is 1350 yd²
Cost to prepare grassland = $15 × 2400 = $36,000
Cost to prepare of forest = $25 × 1350 = $33,750
Thus the grassland will cost more to prepare.

Question 8.
DIG DEEPER!
You buy a new television with a screen similar in shape to your old television screen, but with an area four times greater. The size of a television screen is often described using the distance between opposite corners of the screen. Your old television has a 30-inch screen. What is the size of your new television screen? Explain.

Answer:
Area of ΔABC/Area of ΔDEF = (Side length of AB/Side length of DE)²
Let the area of the screen of old television be x
Let the area of the screen of new television be 4x
l1 = 30 in
Area of the screen of new television/Area of the screen of new television= (distance of the screen of new television/distance of the screen of old television)²
4x/x = (distance of the screen of new television/30)²
distance of the screen of new television = 30 × 2 = 60 inch
Hence the distance of the screen of the new television is 60 inches.

Perimeters and Areas of Similar Figures Homework & Practice 2.7

Review & Refresh

The red figure is similar to the blue figure. Describe a similarity transformation between the figures.

Question 1.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 501

Answer:
First, dilate the red figure using the scale factor of 3 because the side lengths of the blue figure are 3 times the side length of the red figure.
Scale factor = 6/2 = 3
Now reflect the image obtained after a dilation about the y-axis because both red and blue triangle is facing each other.

Question 2.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 137

Answer:
First, dilate the red figure using the scale factor of 0.5 because the side lengths of the blue figure are 3 times the side length of the red figure.
Scale factor = 2/4 = 0.5
Then rotate the image obtained after dilation in direction 90 degrees clockwise about the origin.

Find the area of the figure.

Question 3.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 138

Answer:
We know that the formula for the area of trapezoid = Base × height
h = 16 cm
b = 9 cm
Area of figure = 16 × 9 = 144 sq. cm
Hence the area of the given figure is 144 sq. cm

Question 4.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 139

Answer:
h = 5 in
b = 3 in
We know that,
A = 1/2 × b × h
A = 1/2 × 5 × 3
A = 7.5 sq. cm
Thus the area of the given figure is 7.5 sq. cm

Question 5.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 140

Answer:
h = 5 km
b1 = 6 km
b2 = 8 km
We know that,
Area of trapezoid = 1/2 × h × (b1 + b2)
A = 1/2 × 5 × 14 = 35 sq. km
Hence the area of the trapezoid is 35 sq. km

Concepts, Skills, &Problem Solving
COMPARING SIMILAR FIGURES
Dilate the figure using the indicated scale factor k. What is the value of the ratio (new to original) of the perimeters? the areas? (See Exploration 1, p. 83.)

Question 6.
a triangle with vertices (0, 0), (0, 2), and (2, 0); k = 3

Answer:
When the points of a given figure are dilated we simply multiply each x-coordinate and y-coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
where a is the scale factor
Given a triangle with vertices (0, 0), (0, 2), and (2, 0); k = 3
A(0, 0) = A'(0 . 3, 0 . 3) = A'(0, 0)
B(0, 2) = B'(0 . 3, 2 . 3) = B'(0, 6)
C(2, 0) = C'(2 . 3, 0 . 3) = C'(6, 0)
The coordinates of the image are A'(0, 0), B'(0, 6), C'(6, 0)
AB = √(2 – 0)² – (0 – 0)² = 2
A’B’ = √(6 – 0)² – (0 – 0)² = 6
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of new triangle/Perimeter of the original triangle = Distance of A’B’/Distance of AB = 6/2 = 3
Area of new triangle/Area of the original triangle = (Distance of A’B’/Distance of AB)² = (6/2)² = 3² = 9

Question 7.
a square with vertices (0, 0), (0, 4), (4, 4), and (4, 0); k = 0.5

Answer:
When the points of a given figure are dilated we simply multiply each x-coordinate and y-coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
where a is the scale factor
a square with vertices (0, 0), (0, 4), (4, 4), and (4, 0); k = 0.5
A(0, 0) = A'(0 . 0.5, 0 . 0.5) = A'(0, 0)
B(0, 4) = B'(0 . 0.5, 4 . 0.5) = B'(0, 2)
C(4, 4) = C'(4 . 0.5, 4 . 0.5) = C'(2, 2)
D(4, 0) = D'(4 . 0.5, 0 . 0.5) = D'(2, 0)
Coordinates of the image are A'(0, 0), B'(0, 2), C'(2, 2), D'(2, 0)
AB = √(4 – 0)² + (0 – 0)² = 4
A’B’ = √(2 – 0)² + (0 – 0)² = 2
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of new square/Perimeter of the original square = Distance of A’B’/Distance of AB = 2/4 = 1/2
Area of new square /Area of the original square = (Distance of A’B’/Distance of AB)² = (2/4)² = 1/4

PERIMETERS AND AREAS OF SIMILAR FIGURES
Find the values of the ratios (red to blue) of the perimeters and areas of the similar figures.

Question 8.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 141

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure= 11/6
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (11/6)² = 121/36

Question 9.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 142

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure= 5/8
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (5/8)² = 25/64

Question 10.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 143

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure= 4/7
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (4/7)² = 16/49

Question 11.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 144

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure = 14/9
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (14/9)² = 196/81

USING SIMILAR FIGURES
The figures are similar. Find x.

Question 12.
The ratio of the perimeters is 7 : 10.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 145

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of the figure B = Distance of figure A/Distance of figure B
7/10 = x/12
x = 84/10
x = 8.4
Thus the value of x is 8.4

Question 13.
The ratio of the perimeters is 8 : 5.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 146

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of the figure B = Distance of figure A/Distance of figure B
8/5 = x/16
x = 25.6
Thus the value of x is 25.6

Question 14.
COMPARING AREAS
The playing surfaces of two foosball tables are similar. The ratio of the corresponding side lengths is 10:7. What is the ratio of the areas?

Answer:
Area of figure A /Area of figure B = (Distance of figure A/Distance of figure B)²
Area of figure A /Area of figure B = (10/7)²
Area of figure A /Area of figure B = 100/49
Hence, the ratio of their areas is 100/49

Question 15.
CRITICAL THINKING
The ratio of the side length of Square A to the side length of Square B is 4:9. The side length of Square A is 12 yards. What is the perimeter of Square B?

Answer:
Given,
The ratio of the side length of Square A to the side length of Square B is 4/9.
The side length of Square A is 12 yards.
side length of Square A/side length of Square B = 4/9
12 /side length of Square B = 4/9
side length of Square B = 27 yards
We know that,
The perimeter of the square is = 4s
The perimeter of the square B = 4 × 27 = 108 yards

Question 16.
MODELING REAL LIFE
The cost of the piece of fabric shown is $1.31. What would you expect to pay for a similar piece of fabric that is 18 inches by 42 inches?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 147

Answer:
Given,
l = 21 in
w = 9 in
Area of the rectangle = l × w
A = 21 × 9 = 189 sq. in
The cost of the piece of fabric shown is $1.31
The cost of 1 sq. in of fabric = 1.31/189
l = 18 in
b = 42 in
Area of new fabric = 18 × 42 = 756 sq. in
Given the cost of new fabric = 1.31/189 × 756 = $5.24
Hence the cost of the new fabric is $5.24

Question 17.
PROBLEM SOLVING
A scale model of a merry-go-round and the actual merry-go-round are similar.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 148
a. How many times greater is the base area of the actual merry-go-round than the base area of the scale model? Explain.

Answer:
Radius of model merry go round = 6 in
Radius of actual merry go round = 10 ft = 120 in
Area of base of actual merry/Area of base of model merry = (Radius of actual merry/Radius of model merry)²
Area of base of actual merry/Area of base of model merry = (120/6)² = 400

b. What is the base area of the actual merry-go-round in square feet?

Answer:
The radius of model merry go round = 6 in
Radius of actual merry go round = 10 ft = 120 in
Area of base of actual merry = 450 sq. in
Area of base of actual merry/Area of base of model merry = (Radius of actual merry/Radius of model merry)²
Area of base of actual merry/450 = (120/6)² = 400
Area of base of actual merry = 400 × 450 = 180000 sq. in = 1250 ft²

Question 18.
STRUCTURE
The circumference of Circle K is π. The circumference of Circle L is 4π. What is the value of the ratio of their circumferences? of their radii? of their areas?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 149

Answer:
Given,
The circumference of Circle K is π
The circumference of Circle L is 4π.
circumference of Circle = 2πr
2πr = π
The radius of circle K r1 = 1/2
2πr = 4π
The radius of circle K r2 = 2
The ratio of their circumference = π/4π = 1/4
The ratio of radius of both circle = 1/4
The ratio of their area = π(r1)²/π(r2)² = 1/16

Question 19.
GEOMETRY
A triangle with an area of 10 square meters has a base of 4 meters. A similar triangle has an area of 90 square meters. What is the height of the larger triangle?

Answer:
Given,
A triangle with an area of 10 square meters has a base of 4 meters.
A similar triangle has an area of 90 square meters.
Area of the triangle = bh/2
h = 2a/b
h = (2 × 10)/4
h = 5 meters
Area of larger triangle/Area of smaller triangle = (height of larger triangle/height of smaller triangle)²
90/10 = (height of larger triangle/5)²
3 = (height of larger triangle/5)
Thus the height of larger triangle = 3 × 5 = 15 meters

Question 20.
PROBLEM SOLVING
You need two bottles of fertilizer to treat the flower garden shown. How many bottles do you need to treat a similar garden with a perimeter of 105 feet?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 150

Answer:
The sides of the above figure are 4ft, 15 ft, 18 ft, 5 ft
Perimeter = 4ft + 15 ft + 18 ft + 5 ft = 42 ft
Number of bottle of fertilizer used in the above garden = 2
Number of bottle of fertilizer used in 1 ft = 2/42
Fertilizer used for 105 ft = 2/42 × 105 = 5
Thus 5 bottles will be used for 105 feet

Question 21.
REPEATED REASONING
Three square mirrors are used for a light reflection experiment. The ratio of the side length of Mirror A to the side length of Mirror B is 5 : 6. The ratio of the area of Mirror B to the area of Mirror C is 16 : 25. The perimeter of Mirror C is 280 centimeters. What is the area of Mirror A? Justify your answer.

Answer:
Given,
Three square mirrors are used for a light reflection experiment.
The ratio of the side length of Mirror A to the side length of Mirror B is 5 : 6 = 5/6
The ratio of the area of Mirror B to the area of Mirror C is 16 : 25 = 16/25
The perimeter of Mirror C is 280 centimeters
Side length of mirror C = Perimeter/4 = 280/4 = 70 cm
(Side length of mirror B/Side length of mirror C)² = Area of mirror A/Area of mirror B
(Side length of mirror B/Side length of mirror C)² = 16/25
(Side length of mirror B/Side length of mirror C) = 4/5
The side length of mirror B = 4/5 × Side length of mirror C
Side length of mirror B = 4/5 × 70 = 56 cm
A = s × s
A = 56 cm × 56 cm = 3136 sq. cm
Area of mirror A/Area of mirror B = (Side length of mirror A/Side length of mirror B)²
Area of mirror A/3136 = (5/6)²
Area of mirror A = 25/36 × 3136
Area of mirror A = 2177.7 sq. cm

Transformations Connecting Concepts

2 Connecting Concepts

Using the Problem-Solving Plan

Question 1.
A scale drawing of a helipad uses a scale of 1 ft : 20 ft. The scale drawing has an area of 6.25 square feet. What is the area of the actual helipad?
Understand the problem.
You know the scale of the drawing and the area of the helipad in the drawing. You are asked to find the area of the actual helipad.
Make a plan.
A scale drawing is similar to the actual object. So, use the scale 1 ft : 20 ft and the ratio 6.25 ft2 : A ft2 to write and solve a proportion that represents the area A of the actual helipad.
Solve and Check.
Use the plan to solve the problem. Then check your solution.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 150.1

Answer: 125 sq. ft

Question 2.
The locations of three cargo ships are shown in the coordinate plane. Each ship travels at the same speed in the same direction. After 1 hour, the x- and y-coordinates of Ship A increase 80%. Use a translation to describe the change in the locations of the ships. Then find the new coordinates of each ship.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 151
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 152

Question 3.
All circles are similar. A circle with a radius of 2 inches is dilated, resulting in a circle with a circumference of 22π inches. What is the scale factor? Justify your answer.

Answer:
Given,
A circle with a radius of 2 inches is dilated, resulting in a circle with a circumference of 22π inches.
C = 2π . r
22π = 2π . 2
p = 2π . 2
Thus the scale factor is 2.

Performance Task

Master Puppeteer

At the beginning of this chapter, you watched a STEAM Video called “Shadow Puppets.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 153

Transformations Connecting Concepts

2 Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 154

Graphic Organizers
You can use a Summary Triangle to explain a concept. Here is an example of Summary Triangle for translating a figure.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 155

Choose and complete a graphic organizer to help you study the concept.

  1. reflecting a figure
  2. rotating a figure
  3. congruent figures
  4. dilating a figure
  5. similar figures
  6. perimeters of similar figures
  7. areas of similar figures

Big Ideas Math Answers 8th Grade Chapter 2 Transformations 156

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 157

2.1 Translations (pp. 43–48)

Tell whether the blue figure is a translation of the red figure.

Question 1.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 158

Answer: No

Explanation:
The answer is no because in the case of translation the size of the figure does not change, only the position of the figure changes. But here the size of the blue figure is larger as compared to the red figure so this is not the translation. Here the blue figure is the result of the dilation of red figure.

Question 2.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 159

Answer: Yes

Explanation:
Yes, because in the case of translation the size of the figure does not change, only the position of the figure changes. Here the size of both blue figure and the red figure is the same but there is only a change in the position of the red figure to get blue figure. Here the blue figure is the result of the translation of the red figure.

Question 3.
The vertices of a quadrilateral are W(1, 2), X(1, 4), Y(4, 4), and Z(4, 2). Draw the figure and its image after a translation 3 units left and 2 units down.

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: W(1, 2), X(1, 4), Y(4, 4), and Z(4, 2) and a = -3, b = -2
W'(1+a, 2+b) = W'(1-3, 2-2) = W'(-2,0)
X'(1+a, 4+b) = X'(1-3, 4-2) = X'(-2,2)
Y'(4+a, 4+b) = Y'(4-3, 4-2) = Y'(1, 2)
Z'(4+a, 2+b) = C'(4-3, 2-2) = Z'(1,0)
Hence the coordinate of image are W'(-2,0), X'(-2,2), Y'(1, 2), Z'(1,0)
big ideas math answers grade 8 chapter 2 img_41

Question 4.
The vertices of a triangle are A(-1, -2), B(-2, 2), and C(-3, 0). Draw the figure and its image after a translation 5 units right and 1 unit up.

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(-1, -2), B(-2, 2), and C(-3, 0) and a = 5, b = 1
A'(-1+a, -2+b) = A'(-1-5, -2+1) = A'(4,-1)
B'(-2+a, 2+b) = B'(-2+5, 2+1) = B'(3,3)
C'(-3+a, 0+b) = C'(-3+5, 0+1) = C'(2, 1)
Hence the coordinate of image are A'(4,-1), B'(3,3), C'(2, 1)
BIM Grade 8 Chapter 2 Answer Key img_42

Question 5.
Your locker number is 20 and your friend’s locker number is 33. Describe the location of your friend’s locker relative to the location of your locker.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 159.1

Answer:
The location of my friend’s locker is first 1 locker Down and then 3 locker Right.

Question 6.
Translate the triangle 4 units left and 1 unit down. What are the coordinates of the image?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 160

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(3, 5), B(6, 3), and C(4, 1) and a = -4, b = -1
A'(3+a, 5+b) = A'(3-4, 5-1) = A'(-1,4)
B'(6+a, 3+b) = B'(6-4, 3-1) = B'(2,2)
C'(4+a, 1+b) = C'(4-4, 1-1) = C'(0, 0)
Hence the coordinate of image are A'(-1,4), B'(2,2), C'(0, 0)
BIM 8th Grade Answer Key Chapter 2 img_42

Question 7.
Describe a translation of the airplane from point A to point B.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 161

Answer:
First, move the aeroplane 6 units right from point A and then 4 units down.

2.2 Reflections (pp. 49 – 54)

Tell whether the blue figure is a reflection of the red figure.

Question 8.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 162

Answer:
No, because in the above figure the plane of reflection is inclined at 45 degrees with the horizontal line. So, the reflected figure will be perpendicular to the original figure. But in the given figure both are facing each other. This means that the blue figure is not the reflection of red figure.

Question 9.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 163

Answer:
The answer is no because the blue figure is not the mirror image of the red figure. The side of the red figure is not facing the side of the blue figure this means that the blue figure is not the reflection of red figure.

Question 10.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 163.1

Answer:
The answer is yes because the blue figure is the mirror image of the red figure. The side of the red figure is facing the side of the blue figure this means that the blue figure is the reflection of red figure.

Draw the figure and its reflection in (a) the x-axis and (b) the y-axis. Identify the coordinates of the image.

Question 11.
A(2, 0), B(1, 5), C(4, 3)

Answer:
A(x, y) = A'(x, -y)
Given: A(2, 0), B(1, 5), C(4, 3)
Reflection about x-axis:
A(2, 0) = A'(2, 0)
B(1, 5) = B'(1, -5)
C(4, 3) = C'(4, -3)
BIM Grade 8 Answers Chapter 2 img_43
Reflection about y-axis:
A(x, y) = A'(-x, y)
A(2, 0) = A'(-2, 0)
B(1, 5) = B'(-1, 5)
C(4, 3) = C'(-4, 3)
BIM Grade 8 Answers Chapter 2 img_44

Question 12.
D(-5, -5), E(-5, 0), F(-2, -2), G(-2, -5)

Answer:
Given, D(-5, -5), E(-5, 0), F(-2, -2), G(-2, -5)
Reflection about x-axis:
A(x, y) = A'(x, -y)
D(-5, -5) = D'(-5, 5)
E(-5, 0) = E'(-5, 0)
F(-2, -2) = F'(-2, 2)
G(-2, -5) = G'(-2, 5)
BIM Grade 8 Answers Chapter 2 img_45
Reflection about y-axis:
A(x, y) = A'(-x, y)
D(-5, -5) = D(5, -5)
E(-5, 0) = E'(5, 0)
F(-2, -2) = F'(2, -2)
G(-2, -5) = G'(2, -5)
BIM Grade 8 Answers Chapter 2 img_46

Question 13.
The vertices of a rectangle are E(-1, 1), F(-1, 3), G(-5, 3), and H(-5, 1). Find the coordinates of the figure after reflecting in the x-axis, and then translating 3 units right.

Answer:
We know that when a point is reflected about x-axis then y-coordinate becomes the opposite.
A(x, y) = A'(x, -y)
The vertices of a rectangle are E(-1, 1), F(-1, 3), G(-5, 3), and H(-5, 1).
Reflection about x-axis:
E(-1, 1) = E'(-1, -1)
F(-1, 3) = F'(-1, -3)
G(-5, 3) = G'(-5, 3)
H(-5, 1) = H'(-5, 1)
Thus the coordinates of the image are E'(-1, -1), F'(-1, -3), G'(-5, 3), H'(-5, 1)
Now translating the image 3 units Right.
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: E'(-1, -1), F'(-1, -3), G'(-5, 3), H'(-5, 1) a = 3, b = 0
E”(-1 + a, -1 + b) = E”(-1 + 3, -1 + 0) = E”(2, -1)
F”(-1 + a, -3 + b) = F”(-1 + 3, -3 + 0) = F”(2, -3)
G”(-5 + a, 3 + b) = G”(-5 + 3, 3 + 0) = G”(-2, 3)
H”(-5 + a, 1 + b) = H”(-5 + 3, 1 + 0) = H”(-2, 1)
Thus the coordinates of the image are E”(2, -1), F”(2, -3), G”(-2, 3), H”(-2, 1)

The coordinates of a point and its image after a reflection are given. Identify the line of reflection.

Question 14.
(-1, -3) → (1, -3)

Answer:
Given,
(-1, -3) → (1, -3)
We can see that the y-coordinate of both points and its image are the same but the x-coordinate of the image is the opposite of its points.
Hence, Y-axis is the line of reflection.

Question 15.
(2, 1) → (2, -1)

Answer:
Given,
(2, 1) → (2, -1)
We can see that the x-coordinate of both points and its image are the same but the y-coordinate of the image is the opposite of its points.
Hence, X-axis is the line of reflection.

Question 16.
You perform an experiment involving angles of refraction with a laser pen. You point a laser pen from point L at a mirror along the red path and the image is a reflection in the y-axis.
a. Does the light reach a cat at point C? Explain.

Answer:
Yes, the light will reach at point C.
Because the coordinate of point L is (4, 3) and the coordinate of point C is (-4, 3) and it is given problems that laser is reflected about the y-axis.
So when point L(4, 3) is reflected about the y-axis its x-coordinates become opposite and y-coordinates remain the same.
So when point L(4, 3) is reflected about the y-axis its image will be point C(-4, -3)

b. You bounce the light off the mirror so its path is a reflection. What line of reflection is needed for the light to reach the cat?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 164

Answer: The line of reflection will be y-axis.

2.3 Rotations (pp. 55–62)

Tell whether the blue figure is a rotation of the red figure about the origin. If so, give the angle and the direction of rotation.

Question 17.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 165

Answer:
The answer is no because the blue figure is the mirror image of the red figure. The blue figure is the result of the reflection of red figure about the y-axis. Also, both red and blue figure are facing each other with the y-axis in the center of both which remains that it is not the case of rotation.

Question 18.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 166

Answer:
The answer is yes, because the red figure is in the 1st quadrant and the blue figure in third quadrant. Also, both blue and red figure are facing each other in opposite directions which blue figure is the result of the rotation of red figure.
When red figure is rotated 180 degrees counterclockwise it will result in a blue figure.

The vertices of a triangle are A(-4, 2), B(-2, 2), and (-3, 4). Rotate the triangle as described. Find the coordinates of the image.

Question 19.
180° about the origin

Answer:
We know that when a point is rotated 180 degrees about origin then both x coordinate and y coordinate becomes opposite.
A(x, y) = A'(-x, -y)
Given points: A(-4, 2), B(-2, 2), and C(-3, 4)
Rotated 180 degrees about origin:
A(-4, 2) = A'(4, -2)
B(-2, 2) = B'(2, -2)
C(-3, 4) = C'(3, -4)
The coordinate of the image are A'(4, -2), B'(2, -2), C'(3, -4)

Question 20.
270° clockwise about the origin

Answer:
We know that when a point is rotated 90 degrees counterclockwise about origin then both x coordinate and y coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given points: A(-4, 2), B(-2, 2), and C(-3, 4)
Rotating 270 degrees clockwise about the origin:
A(-4, 2) = A'(-2, -4)
B(-2, 2) = B'(-2, -2)
C(-3, 4) = C'(-4, -3)
The coordinate of the image are A'(-2, -4), B'(-2, -2), C'(-4, -3)

Question 21.
A bicycle wheel is represented in a coordinate plane with the center of the wheel at the origin. Reflectors are placed on the bicycle wheel at points (7, 4) and (-5, -6). After a bike ride, the reflectors have rotated 90° counterclockwise about the origin. What are the locations of the reflectors at the end of the bike ride?

Answer:
We know that when a point is rotated 90 degrees counterclockwise about origin then both x coordinate and y coordinate becomes opposite.
P(x, y) = P'(-y, x)
Reflectors are placed on the bicycle wheel at points (7, 4) and (-5, -6)
A(7, 4) = A'(-4, 7)
B(-5, -6) = B'(6, -5)
Hence the new coordinate of the reflector are A'(-4, 7), B'(6, -5)

2.4 Congruent Figures (pp. 63–68)

Identify any congruent figures in the coordinate plane.

Question 22.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 167

Answer:
AB = ED
BC = DC
CA = CE
∠A = ∠E
∠B = ∠D
∠C = ∠C
When we see both triangles ABC and EDC closely we observe that both the triangles are the mirror images of each other with the y-axis as the line of reflection. So all the corresponding sides are equal and also all the corresponding angles, this means that both the triangle are congruent.
Hence ΔABC is congruent to ΔEDC
GH = JK
HF = KI
FG = IJ
∠G = ∠J
∠H = ∠K
∠F = ∠I
So we can see that all the corresponding sides are equal and also all the corresponding angles. this means that both the triangles are congruent.
Hence ΔGHF is congruent to ΔJKI

Question 23.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 168

Answer:
When we observe square ABCD and square EFGH we can see that
AB = EF
BC = FG
CD = GH
DA = HE
∠A = ∠E
∠B = ∠F
∠C = ∠G
∠D = ∠H
We can see that all the corresponding sides are equal and also all the corresponding angles, this means that both are congruent.
RS = IJ
ST = JK
TU = KL
UR = LI
∠R = ∠I
∠S = ∠J
∠T = ∠K
∠L = ∠U
Hence rectangle RSTU is congruent to rectangle IJKL

The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 24.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 169

Answer:
First, rotate the blue figure 90° clockwise because the blue figure is in a vertical position but the red figure is in the horizontal position.
Translate the image 5 units right because the first image formed after the rotation will in the second quadrant but the red figure in the first quadrant.

Question 25.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 170

Answer:
First, reflect the blue figure about the y-axis because both red and blue figure is facing each other and they are the mirror image of each other.
Then translate the image 2 units up because the first image formed after reflection will be at the same distance from the x-axis but the red figure is touchung the x-axis.

Question 26.
The figures are congruent. Name the corresponding angles and the corresponding sides.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 171

Answer:
Corresponding sides are
AB = KL
BC = LM
CA = MK
Corresponding angles
∠A = ∠K
∠B = ∠L
∠C = ∠M

Question 27.
Trapezoids EFGH and QRST are congruent.
a. What is the length of side QR ?

Answer:
Length of side:
QR = EF = 3 feet

b. Which angle in QRST corresponds to ∠H?
Answer:
The angle that corresponds to ∠H is ∠T

c. What is the perimeter of QRST ?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 172

Answer:
Perimeter of QSRT = QR + RS + ST + TQ
= EF + FG + GH + HE
= 3 + 5 + 4 + 8
= 20 ft

2.5 Dilations (pp. 69–76)

Tell whether the blue figure is a dilation of the red figure.

Question 28.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 173

Answer:
The answer is no because dilation the size of the image either increases or decreases that depend on the type of dilation. We can see that both the red and blue figure are of the same size which means that the blue figure is not dilation of the red figure.

Question 29.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 174

Answer:
The answer is yes because both red and blue figures are of the same shape and all the corresponding angles are equal but the blue figure is larger than the red figure. And in dilation, the size of the image is either increases or decreases but the shape always remains the same. So blue figure is the result of dilation of red figure.

The vertices of a figure are given. Draw the figure and its image after a dilation with the given scale factor. Identify the type of dilation.

Question 30.
P(-3, -2), Q(-3, 0), R(0, 0); k = 4

Answer:
A(x, y) = A'(x . a, y . a)
Where a is the scaling factor
Given,
P(-3, -2), Q(-3, 0), R(0, 0); k = 4
P(-3, -2) = P'(-3 . 4, -2 . 4) = P'(-12, -8)
Q(-3, 0) = Q'(-3 . 4, 0 . 4) = Q'(-12, 0)
R(0, 0) = R'(0 . 4, 0 . 4) = R'(0, 0)
Thus the coordinates of the image: P'(-12, -8), Q'(-12, 0), R'(0, 0)
Bigideas Math Answer Key for Grade 8 Chapter 2 img_45

Question 31.
B(3, 3), C(3, 6), D(6, 6), E(6, 3); k = \(\frac{1}{3}\)

Answer:
A(x, y) = A'(x . a, y . a)
Where a is the scaling factor
Given,
B(3, 3), C(3, 6), D(6, 6), E(6, 3); k = \(\frac{1}{3}\)
B(3, 3) = B'(3 . \(\frac{1}{3}\), 3 . \(\frac{1}{3}\)) = B'(1, 1)
C(3, 6) = C'(3 . \(\frac{1}{3}\), 6 . \(\frac{1}{3}\)) = C'(1, 2)
D(6, 6) = D'(6 . \(\frac{1}{3}\), 6 . \(\frac{1}{3}\)) = D'(2, 2)
E(6, 3) = E'(6 . \(\frac{1}{3}\), 3 . \(\frac{1}{3}\)) = E'(2, 1)
Thus the coordinates of the image: B'(1, 1), C'(1, 2), D'(2, 2), E'(2, 1)
BIm Grade 8 Chapter 2 Answers img_46

Question 32.
The blue figure is a dilation of the red figure. Identify the type of dilation and find the scale factor.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 175

Answer:
AB = 1 unit
A’B’ = 2 units
scale factor = size of image figure/size of actual figure
scale factor = 2/1 = 2
We can see from the above figure that the size of the image figure are larger as compared to the size of the original image so it is the Enlargement dilation.

Question 33.
The vertices of a rectangle are Q(-6, 2), R(6, 2), S(6, -4), and T(-6, -4). Dilate the rectangle with respect to the origin using a scale factor of \(\frac{3}{2}\). Then translate it 5 units right and 1 unit down. What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the rectangle: Q(-6, 2), R(6, 2), S(6, -4), and T(-6, -4), scale factor = \(\frac{3}{2}\)
Q(-6, 2) = Q'(-6 . \(\frac{3}{2}\), 2 . \(\frac{3}{2}\)) = Q'(-9, 3)
R(6, 2) = R'(6 . \(\frac{3}{2}\), 2 . \(\frac{3}{2}\)) = R'(9, 3)
S(6, -4) = S'(6 . \(\frac{3}{2}\), -4 . \(\frac{3}{2}\)) = S'(9, 6)
T(-6, -4) = T'(-6 . \(\frac{3}{2}\), -4 . \(\frac{3}{2}\)) = T'(-9, -6)
Thus the coordinates of the image: Q'(-9, 3), R'(9, 3), S'(9, 6), T'(-9, -6)
Q'(-9, 3), R'(9, 3), S'(9, 6), T'(-9, -6) a = 5, b = -1
Q”(-9 + a, 3 + b) = Q”(-9 + 5, 3 – 1) = Q”(-4, 2)
R”(9 + a, 3 + b) = R”(9 + 5, 3 – 1) = R”(14, 2)
S”(9 + a, -6 + b) = S”(9 + 5, -6 – 1) = S”(14, -7)
T”(-9 + a, -6 + b) = T”(-9 + 5, -6 – 1) = T”(-4, -7)
Thus the coordinates of the image: Q”(-4, 2), R”(14, 2), S”(14, -7), T”(-4, -7)

2.6 Similar Figures

Question 34.
Determine whether the two figures are similar. Explain your reasoning.

Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 175.1

Answer:
No, the above two figures are not similar.

Question 35.
Draw figures with the given vertices in a coordinate plane. Which figures are similar? Explain your reasoning.
Triangle A: (-4, 4), (-2, 4), (-2, 0)
Triangle B: (-2, 2), (-1, 2), (-1, 0)
Triangle C: (6, 6), (3, 6), (3, 0)

Answer:
Triangle A: (-4, 4), (-2, 4), (-2, 0)
BIM Grade 8 Solution Key Ch 2 img_48
Triangle B: (-2, 2), (-1, 2), (-1, 0)
Bigideas Math Answers Grade 8 Ch 2 img_49
Triangle C: (6, 6), (3, 6), (3, 0)
BIM Answers for Grade 8 Chapter 2 img_50

The figures are similar. Find x.

Question 36.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 177

Answer:
Ratio of sides of larger triangle = Ratio of sides of smaller triangle
20/14 = x/7
x = 10
Thus the value of x is 7 inches.

Question 37.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 178

Answer:
Ratio of sides of larger parallelogram= Ratio of sides of smaller parallelogram
x/6 = 6/4
x = 9
Thus the value of x is 9 cm

2.7 Perimeters and Areas of Similar Figures (pp. 83-88)

Find the values of the ratios (red to blue) of the perimeters and areas of the similar figures.

Question 38.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 179

Answer:
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
Perimeter of red figure/Perimeter of blue figure = 6/8 = 3/4
Hence the ratio of perimeter of red triangle to blue is 3/4
Area of figure A/Area of figure B = (Side length of A/Side length of B)²
Area of red figure/Area of blue figure = (6/8)² = 9/16
Thus the ratio of the perimeter of the red to the blue triangle is 9/16

Question 39.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 180

Answer:
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
Perimeter of red figure/Perimeter of blue figure = 28/16 = 7/4
Hence the ratio of the perimeter of red rectangle to blue is 7/4
Area of figure A/Area of figure B = (Side length of A/Side length of B)²
Area of red figure/Area of blue figure = (28/16)² = 49/16
Thus the ratio of the perimeter of the red to the blue rectangle is 49/16

The figures are similar. Find x.

Question 40.
The ratio of the perimeters is 5 :7.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 181

Answer:
5/7 = 12/x
x =(12 × 7)/5
x = 16.8 cm
Thus the value of x is 16.8 cm

Question 41.
The ratio of the perimeters is 6 : 5.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 182

Answer:
6/5 = x/6
x = 36/5
x = 7.2
Thus the value of x is 7.2

Question 42.
Two photos are similar. The ratio of the corresponding side lengths is 3 : 4. What is the ratio of the areas?

Answer:
Area of photo A/Area of photo B = (Side length of photo A/Side length of photo B)²
Area of photo A/Area of photo B = (3/4)² = 9/16
Thus the ratio of the area of two photos is 9/16

Question 43.
The ratio of side lengths of Square A to Square B is 2 : 3. The perimeter of Square A is 16 inches. What is the area of Square B?

Answer:
Given,
The ratio of side lengths of Square A to Square B is 2 : 3.
The perimeter of Square A is 16 inches.
Perimeter of Square A/Perimeter of Square B = Side length of Square A/Side length of Square B
The perimeter of Square B = (16 × 3)/2
The perimeter of Square B = 24 inches
Side length of square B = Perimeter/4 = 24/4 = 6 inch
Area of square B = s × s = 6 × 6 = 36 sq. in

Question 44.
The TV screen is similar to the computer screen. What is the area of the TV screen?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 183

Answer:
Area of computer screen = 108 sq. in
The side length of the computer screen = 12 in
The side length of the TV screen = 20 in
Area of TV screen = (25 × 108)/9
Area of TV screen = 300 sq. in
Hence the area of the TV screen is 300 sq. in

Transformations Practice Test

2 Practice Test

Triangles ABC and DEF are congruent.

Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 183.1

Question 1.
Which angle of DEF corresponds to ∠C ?

Answer:
Corresponding sides
Side AB = Side DE
Side BC = Side EF
Side CA = Side ED
Corresponding angles
∠A = ∠D
∠B = ∠E
∠C = ∠F
Thus the angle correspond to ∠C is ∠F

Question 2.
What is the perimeter of DEF ?

Answer:
Corresponding sides
Side AB = Side DE
Side BC = Side EF
Side CA = Side ED
Corresponding angles
∠A = ∠D
∠B = ∠E
∠C = ∠F
Perimeter of DEF = DE + EF + FD
= AB + BC + CA
= 5 + 4 + 6
= 15 cm
Thus the perimeter of the ΔDEF is 15 cm

Tell whether the blue figure is a translation, reflection, rotation, or dilation of the red figure.

Question 3.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 184

Answer: The scale factor of a dilation is greater than 1 because the shape of the blue figure is larger than the red figure.

Question 4.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 185

Answer:
The blue figure is the reflection of the red figure. Because the blue figure is the mirror image of red figure. Also, the shape and size of both red and blue figures are the same.

Question 5.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 186

Answer:
The blue figure is the translation of the red figure. Because the shape and size of both red and blue figures are the same. And both the figure is not facing to each other. This means that the blue figure is the result of the translation of the red figure.

Question 6.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 187

Answer:
The blue figure is the result of the rotation of the red figure. Because the shape and size of both the red and blue figure are the same but the figure is horizontal and the blue figure is vertical. This means that the blue figure is the result of the rotation of the term figure.

The vertices of a triangle are A(2, 5), B(1, 2), and C(3, 1). Find the coordinates of the image after the transformations given.

Question 7.
Reflect in the y-axis.

Answer:
A(x, y) = A'(-x, y)
A(2, 5), B(1, 2), C(3, 1)
Reflection about the y-axis:
A(2, 5) = A'(-2, 5)
B(1, 2) = B'(-1, 2)
C(3, 1) = C'(-3, 1)
Thus the coordinates of the image: A'(-2, 5), B'(-1, 2), C'(-3, 1)

Question 8.
Rotate 90° clockwise about the origin.

Answer:
A(x, y) = A'(x, -y)
Given, A(2, 5), B(1, 2), C(3, 1)
A(2, 5) = A'(2, -5)
B(1, 2) = B'(1, -2)
C(3, 1) = C'(3, -1)
Thus the coordinates of the image: A'(2, -5), B'(1, -2), C'(3, -1)

Question 9.
Reflect in the x-axis, and then rotate 90° counterclockwise about the origin.

Answer:
A(x, y) = A'(x, -y)
Given, A(2, 5), B(1, 2), C(3, 1)
A(2, 5) = A'(2, -5)
B(1, 2) = B'(1, -2)
C(3, 1) = C'(3, -1)
Thus the coordinates of the image: A'(2, -5), B'(1, -2), C'(3, -1)
A'(2, -5) = A”(2, 5)
B'(1, -2) = B”(1, 2)
C'(3, -1) = C”(3, 1)
Thus the coordinates of the image: A”(2, 5), B”(1, 2), C”(3, 1)

Question 10.
Dilate with respect to the origin using a scale factor of 2. Then translate 2 units left and 1 unit up.

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: A(2, 5), B(1, 2), C(3, 1) and scale factor = 2
Dilating the figure by scale factor of 2
A (2, 5) = A'(2 . 2, 5 . 2) = A'(4, 10)
B (1, 2) = B'(1 . 2, 2 . 2) = B'(2, 4)
C (3, 1) = C'(3 . 2, 1 . 2) =  C'(6, 2)
Hence the coordinates of the image are A'(4, 10), B'(2, 4),  C'(6, 2)
Now translating image 2 unit left and 1 unit up.
Given: A'(4, 10), B'(2, 4),  C'(6, 2) a = -2, b = 1
A”(4 + a, 10 + b) = A”(4 – 2, 10 + 1) = A”(2, 11)
B”(2 + a, 4 + b) = B”(2 – 2, 4 + 1) = B”(0, 5)
C”(6 + a, 2 + b) = C”(6 – 2, 2 + 1) = C”(4, 3)
Hence the coordinates of the image are A”(2, 11), B”(0, 5), C”(4, 3)

Question 11.
In a coordinate plane, draw Rectangle A: (-4, 4), (0, 4), (0, 2), (-4, 2); Rectangle B: (-2, 2), (0, 2), (0, 1), (-2, 1); and Rectangle C:(-6, 6), (0, 6), (0, 3), (-6, 3). Which figures are similar? Explain your reasoning.

Answer:
Rectangle A: (-4, 4), (0, 4), (0, 2), (-4, 2)
BIM Grade 8 Chapter 2 Solutions img_51
Rectangle B: (-2, 2), (0, 2), (0, 1), (-2, 1)
BIM Grade 8 Chapter 2 Solutions img_52
Rectangle C:(-6, 6), (0, 6), (0, 3), (-6, 3)
BIM Grade 8 Chapter 2 Solutions img_53

Question 12.
Translate a point (x, y) 3 units left and 5 units up. Then translate the image 5 units right and 2 units up. What are the coordinates of the point after the translations?

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given: A(x,y) and a = -3, b = 5
A'(x+a, y+b) = A'(x – 3, y + 5)
Now translating image 5 units right and 2 units up.
Image after first translation: A'(x – 3, y + 5) and a = 5, b = 2
A”(x – 3 + a, y + 5 + b) = A”(x – 3 + 5, y + 5 + 2) = A”(x + 2, y + 7)
Thus the final image will be A”(x + 2, y + 7)

Question 13.
The two figures are similar.
(a) Find the value of x.

Answer:
Ratio of sides of red figure = Ratio of sides of blue figure
x/14 = 10/8
x = (10 × 14)/8
x = 17.5

(b) Find the values of the ratios (red to blue) of the perimeters and of the areas.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 188

Answer:
Perimeter of red figure/Perimeter of blue figure = Side length of red figure/Side length of blue figure
Perimeter of red figure/Perimeter of blue figure = 14/8 = 7/4
Thus the ratio of the perimeter of red to blue figure is 7/4
Area of red figure/Area of blue figure = (side length of red figure/side length of blue figure)²
Area of red figure/Area of blue figure = (14/8)² = 49/16
Thus the ratio of the perimeter of red to blue triangle is 49/16

Question 14.
A wide-screen television measures 36 inches by 54 inches. A movie theater screen measures 42 feet by 63 feet. Are the screens similar? Explain.

Answer:
Given,
A wide-screen television measures 36/54 = 2/3
A movie theater screen measures 42/63 = 2/3
We can see that the ratio of corresponding sides of the television screen is equal to the ratio of corresponding sides of the movie theatre. So television screens and movie theatres are similar.

Question 15.
You want to use the rectangular piece of fabric shown to make a pair of curtains for your window. Name the types of congruent shapes you can make with one straight cut. Draw an example of each type.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 189

Answer:
The types of congruent shapes that can be made with one straight cut
2 right triangles
2 rectangles
2 right trapezoid

Transformations Cumulative Practice

Cumulative Practice

Question 1.
A clockwise rotation of 90° is equivalent to a counterclockwise rotation of how many degrees?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 190

Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 191

Answer:
90° of clockwise rotation = (360 – 90)° of counterclockwise rotation
= 270° of counterclockwise rotation

Question 2.
The formula K = C + 273.15 converts temperatures from degrees Celsius C to Kelvin K. Which of the following formulas is not correct?
A. K – C = 273.
B. C = K – 273.15
C. C – K = -273.15
D. C = K + 273.15

Answer: C = K + 273.15

Question 3.
You want to solve the equation -3(x + 2) = 12x. What should you do first?
F. Subtract 2 from each side.
G. Add 3 to each side.
H. Multiply each side by -3.
I. Divide each side by -3.

Answer: I. Divide each side by -3.

Explanation:
-3(x + 2) = 12x
x + 2 = -4x
x = -4x – 2
x + 4x = -2
5x = -2
x = –\(\frac{2}{5}\)
Thus the correct answer is option I.

Question 4.
Which value of x makes the equation \(\frac{3}{4} x\) = 12 true?
A. 9
B. 11\(\frac{1}{4}\)
C. 16
D. 48

Answer: C. 16

Explanation:
\(\frac{3}{4} x\) = 12
3x = 12 × 4
3x = 48
x = \(\frac{48}{3}\)
x = 16
Thus the correct answer is option C.

Question 5.
A triangle is graphed in the coordinate plane. What are the coordinates of the image after a translation 3 units right and 2 units down?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 191.1
F. A'(1, 4), B'(1, 1), C'(3, 1)
G. A'(1, 2), B'(1, -1), C'(3, -1)
H. A'(-2, 2), B'(-2, -1), C'(0, -1)
I. A'(0, 1), B'(0, -2), C'(2, -2)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
A(-2, 4), B(-2, 1), C(0, 1) and a = 3, b = -2
A'(-2+a, 4+b) = A'(-2 + 3, 4 – 2) = A'(1, 2)
B'(-2+a, 1+b) = B'(-2 + 3, 1 – 2) = B'(1,-1)
C'(0+a, 1+b) = C'(0 + 3, 1 – 2) = C'(3, -1)
Coordinate of the image are: A'(1, 2), B'(1,-1), C'(3, -1)
Thus the correct answer is option G.

Question 6.
Your friend solved the equation in the box shown. What should your friend do to correct the error that he made?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 192
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 193

Answer:
–\(\frac{x}{3}\) + \(\frac{2}{5}\) = –\(\frac{7}{15}\)
–\(\frac{x}{3}\) = –\(\frac{13}{15}\)
x = 2\(\frac{3}{5}\)
Thus the correct answer is option C.

Question 7.
Your teacher dilates the rectangle using a scale factor of \(\frac{1}{2}\).
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 194
What is the area of the dilated rectangle in square inches?

Answer:
l = 10 in
b = 6 in
scale factor = \(\frac{1}{2}\)
New length after dilation = 10 × \(\frac{1}{2}\) = 5
New breadth after dilation = 6 × \(\frac{1}{2}\) = 3
Area of rectangle = l × b
A = 5 × 3 = 15 sq. in
The area of the dilated rectangle will be 5 in²

Question 8.
Your cousin earns $9.25 an hour at work. Last week she earned $222.00 How many hours did she work last week?
F. \(\frac{1}{24}\)
G. 22 hours
H. 24 hours
I. 212.75 hours

Answer: H. 24 hours

Explanation:
Given,
Your cousin earns $9.25 an hour at work.
Last week she earned $222.00
Total no. of working hour = total earning of week/rate of one hour
= \(\frac{222}{9.25}\)
= 24 hours
Thus the correct answer is option H.

Question 9.
Triangle EFG is a dilation of Triangle HIJ. Which proportion is not true for Triangle EFG and Triangle HIJ ?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 195
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 196

Answer: \(\frac{EG}{HI}\) = \(\frac{FG}{IJ}\)
The correct answer is option B.

Question 10.
The red figure is congruent to the blue figure. Which of the following is a sequence of rigid motions between the figures?
F. Reflect the red triangle in the x-axis, and then translate it 3 units left.
G. Reflect the red triangle in the x-axis, and then translate it 3 units right.
H. Reflect the red triangle in the y-axis, and then translate it 3 units left.
I. Rotate the red triangle 90° clockwise about the origin.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 197

Answer:
1. First red triangle is reflected about the x-axis because both red and blue triangles are the mirror image of each other and also the red triangle is in the 1st quadrant and the blue triangle is in 4th quadrant.
2. Then translate the image 3 unit left because the base of both red and blue triangles is not opposite to each other.
Thus the correct answer is option F.

Question 11.
Several transformations are used to create the pattern.
Part A
Describe the transformation of Triangle GLM to Triangle DGH

Answer:
Both ΔGLM and Δ DGH are of the same shape and size but their position are different so the transformation will be translated.

Part B
Describe the transformation of Triangle ALQ to Triangle GLM.

Answer:
The size of the triangle ALQ is four times the size of triangle GLM and the shape of both triangles is the same so the transformation will be dilation.

Part C
Triangle DFN is a dilation of Triangle GHM. Find the scale factor.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 198
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 198.1

Answer:
The size of the triangle DFN is double the size of triangle GHM. So the scale factor of dilation will be 2.

Question 12.
A rectangle is graphed in the coordinate plane.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 199
What are the coordinates of the image after a reflection in the y-axis?
A. J'(4, 1), K'(4, 3), L'( 1, 3), M'(-1, 1)
B. J'(-4, 1), K(-4, -3), L'(1, -3), M'(1, 1)
C. J'(1, 4), K'(3, 4), L'(3, -1), M'(1, -1)
D. J'(-4, 1), K'(-4, 3), L'(1, 3), M'(1, 1)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x, y) = A'(-x, y)
J(-4, 1), K(-4, 3), L(1, 3), M(1, 1)
Reflection about the y-axis:
J(-4, 1) = J'(4, 1)
K(-4, 3) = K'(4, 3)
L(1, 3) = L'(-1, 3)
M(1, 1) = M'(-1, 1)
Coordinate of image are: J'(4, 1), K'(4, 3), L'(-1, 3), M'(-1, 1)
Thus the correct answer is option A.

Conclusion:

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