Common Core Math

An Area is a space or a region occupied by any shape. Square is a two-dimensional figure with four sides, it is also known as a quadrilateral. The area of a square is defined as the number of square units needed to fill a square. In other words, a square is a product of two sides, length, and breadth that are equal. In the square, all four angles are also equal. Square has four equal sides and four vertices. The diagonals of the square are equal and bisect each other at 90 degrees, angles of the square are equal to 90 degrees. If a square is cut by a plane from the center, then both halves are symmetrical.

Learn completely about the Area of a Square Definition, Formula, and Units. Get to know the Properties of a Square, Solved Examples on How to Calculate the Area of a Square, etc.

Do Read: Practice Test on Area and Perimeter of Square

Area of a Square – Definition

The area of a square is defined as the total number of unit squares in the shape of a square. A Square is a rectangle whose length and breadth are equal which means all four sides are equal.

Area of a Square Formula

The area of a square formula is used for calculating the occupied region. It is measured in square units.

Area of a square=side x side

Perimeter of a Square Formula

The perimeter of the square is equal to the sum of all its four sides. The unit of the perimeter remains the same as that of the side-length of a square.

Perimeter=side+side+side+side=4side unit

Perimeter =4 x side of the square = 4a unit

Where ‘a’ is the length of the side of the square.

Also, Read: Perimeter of a Square

Length of Diagonal of a Square

The length of the diagonals of the square is equal to s√2, where s is the side of the square. The length of the diagonals is equal to each other. By Pythagoras theorem, diagonal is the hypotenuse and the two sides of the triangle formed by the diagonal of the square, are perpendicular and base. Hypotenuse²=Base² + Perpendicular²

Hence,Diagonal²=side² + side²

Diagonal=√2side², d=s√2

where d is the length of the diagonal of a square and s is the side of the square.

Diagonal of Square

It is a line segment that connects two opposite vertices of the square. Suppose we have four vertices, thus we can have two diagonals within a square. The diagonal of a square is always greater than its sides.

The relation between Diagonal ‘d’ and Side ‘a’ of a square is, d=a√2

The relation between Diagonal ‘d’ and Area ‘A’ of a square is, d=√(2A)

The relation between Diagonal ‘d’ and Perimeter ‘P’ of a square is,d=p/2√2

Area of Square Units

Usually, the Area of the Square is measured in the Square Units since each side of the square is the same. Some of the Common Units of Measurement and their equivalent are expressed below.

• 1 m=100 cm
• 1 sq.m=10,000 sq.cm
• 1 km=1000 m
• 1 sq.km=1,000,000 sq.m

Properties of a Square

Have a glance at the Properties of Square and get an idea of them so that you might find them useful while solving related problems. They are as follows

1. The square has 4 vertices and 4 sides.
2. All four interior angles are equal to 90⁰.
3. The opposite sides of the square are parallel to each other.
4. The two diagonals of the square are equal to each other.
5. The length of diagonals is greater than the sides of the square.

Solved Problems on Area of a Square

1. Find the Area of a square of the Side 2.5 cm?

Solution:

Given, Area of a square of the side is  2.5 cm

Area of a square = length x length

Substitute the given value in the above formula,

= 2.5 x  2.5 sq. cm

= 6. 25 sq. cm

2. Find the area of a square of side 49 m?

Solution: 

Given, Area of a square of the side is 49 m

Area of a square  = length  x  length

Substitute the given value in the above formula,

= 49  x  49 sq. m.

= 2401 sq. m.

3. Let a square has a side equal to 3 cm. Find out its area, length of diagonal, and perimeter?

Solution:

Given, side of the square, s= 3 cm

Area of a square= s²

Therefore, by substituting the value of the side, we get,

3² = 9 cm²

Length of the diagonal of  square = s√2

By substituting the value, we get,

3 x 1.414 = 4.242

The perimeter of the square = 4 xs

substituting the value in the above formula, we get,

4 x 3= 12 cm.

4. Find the area of a square clipboard whose side measures 120 cm?

Solution:

Given, side of the clipboard  = 140 cm = 1.4 m

Area of the clipboard  = side x side

substituting the value in the above formula, we get,

Area of the clipboard  = 140 cm x 140 cm

= 19,600 sq . cm

= 1.96 sq . m

5. A wall that is 40 m long and 30 m wide is to be covered by square tiles. The side of each tile is  2 m. Find the number of tiles required to cover the wall?

Solution: 

Given, Length of the wall = 40 m

The breadth of the wall = 30 m

Area of the wall = Length x breadth

substituting the length and breadth values, we get

Area of the wall= 40 m x 30 m = 1200 sq. m

side of one tile = 2 m

Area of one tile = side x side

substitute the value in the above formula, we get

Area of one tile =2 m x 2 m= 4 sq. m

No. of tiles required = Area of wall/ Area of a tile

substitute the value, we get

No.of tiles required = 1200/ 4 = 300 tiles

FAQ’s on Area of a Square

 1. What is the Area of a Square Formula?

The area of a square can be calculated by using a formula side x side square units.

2. Define Perimeter and Area of a Square?

 The perimeter of a square is the sum of all the four sides of a square, whereas the Area of a square is defined as the region or the space occupied by a square in the two-dimensional space.

3. How is a Square different from a Rectangle?

A square has all its sides are equal in length whereas a rectangle has only its opposite sides are equal in length.

4. What are the real-life examples of a Square?

Some of the real-life examples of a  square are Carrom Board, Square tiles, Square shaped tables, Cubes, Chess Board, etc.

Triangle is one of the topics in geometry. The word ‘Tri’ in a triangle indicates three. Yes, Triangle is designed with three lines and is shaped as a closed curve with three lines. Each line is considered as a side of the triangle. Totally, one triangle has three sides or faces. The points, where the two sides of the triangle are intersected that particular points are called vertices and the angles are formed at the point of vertices.

Symbol of Triangle

The symbol of the triangle is a closed loop with three sides.

Here, the name of the triangle is XYZ.

Properties of the Triangle

Have a glance at the Properties of Triangle listed below and they are along the lines

• Basically, the triangle has three sides. Here, the sides of the triangle are XY, YZ, and ZX.
• The vertices of the triangle are X, Y, and Z.
• Angles of the triangle are XYZ, YXZ, and XZY.
• The sum of the three angles is equal to 180°.
• The sum of any two sides of the triangle must be greater than the third side of the triangle. That is, XY + YZ = ZX or YZ + ZX = XY or XY + XZ = YZ.
• In Triangles, we have two types of angles. They are interior angles and exterior angles. Interior angles are formed inside of the intersected point of the sides. Exterior angles are formed outside of the intersected points of the sides.

Also, Read:

• Perimeter of a Triangle
• Medians and Altitudes of a Triangle
• Area and Perimeter of the Triangle

Area of a Triangle

Area of the Triangle is equal to 1 / 2 * base * height. Here, Area is denoted by ‘A’, the base is denoted by ‘b’ and the height is denoted by ‘h’.

Area (A) = (1 / 2) b*h.

Perimeter of a Triangle

Perimeter of a triangle is equal to half of the sum of the three sides of the triangle. Perimeter is denoted by s and sides of the triangle are denoted as a, b, c.

Perimeter (s) = (a+ b + c) / 2.

• We can find out the area of the triangle by using the perimeter and sides of the triangle. That is, Area = √s(s – a) (s – b) (s – c).

Types of Triangles

We have different types of triangles. Based on the sides and angles, triangles are classified into different types. They are

1. Equilateral Triangle
2. Isosceles Triangle
3. Scalene Triangle
4. Acute angled Triangle
5. Obtuse angled Triangle
6. Right-sided angle Triangle

1. Equilateral Triangle: Three sides of the triangle, as well as three angles of the triangle, are equal and it is called as Equilateral Triangle.

Area of the Equilateral Triangle is equal to √3 / 4 *(side)^2.

2. Isosceles Triangle: In this type of Triangle, two sides of the triangle and two angles are equal.

The area of an Isosceles Triangle is equal to (base * height) / 2.

3. Scalene Triangle: A Triangle that is generated with three different sides and three different angles is called a Scalene Triangle.

Area of the Scalene Triangle is equal to (1 / 2) * base * height.

4. Acute Angled Triangle: Three internal angles of the triangle are measured less than 90° and it is called an Acute Angled Triangle.

Area of the Acute angled triangle is equal to (1 / 2) * base * height.

5. Obtuse Angled Triangle: Generally, three angles there is a triangle. If any one of the interior angles is measured as greater than 90°, then it is called as Obtuse Angled Triangle.

Area of the obtuse angle is equal to (1 / 2) * base * height.

6. Right-sided angle Triangle: The angle between two sides is equal to 90° and the sum of the remaining two angles must be equal to 90°, then it is called a Right-sided angle triangle.

Area of the Right-angled triangle is equal to (1 / 2) * base * height.

Solved Problems on Triangle

1. The Base of the triangle is 4 cm and the height of the triangle is 9 cm. find the area of the triangle?

Solution:
As per the given data, the base of the triangle (b) = 4 cm.
Height of the triangle (h) = 9 cm.
Area of the triangle (A) = (1 / 2) * base * height.
A = (1 / 2) * 4 * 9.
A = 2 * 9 =18 cm².

The area of the triangle is equal to 18 cm².

2. A triangle has an area of 60 cm² and the base of the triangle is 30 cm. Find the height of the triangle?

Solution:
As per the given information, the Area of the triangle (A) = 60 cm².
Base of the triangle (b) = 30 cm².
Area of the triangle (A) = (1 / 2) * base * height.
60 = (1 / 2) * 30 * h.
60 *2 =30 * h.
120 = 30 * h.
H = 120 / 30 = 4 cm.

Height of the triangle = 4 cm.

3. Two sides of the Isosceles triangle are 10 cm each while the third side is 15 cm. Find the area of the isosceles triangle?

Solution:
As per the given data, two sides of the Isosceles triangle = 10 cm
The third side of the triangle = 15 cm.
Area of the Isosceles triangle (A) = √s (s – a) (s – b) (s – c).
S = (a + b + c) / 2
Here, a = 10, b = 10, c =15
S = (10 + 10 + 15) / 2 = 35 / 2 = 17.5
A = √ 17.5 ( 17.5 – 10) (17.5 – 10) ( 17.5 – 15).
A = √17.5 (7.5) (7.5) (2.5) = 588.27.

The area of the Isosceles Triangle is equal to 588.27.

4. Three sides of the triangle are 12 cm, 10 cm, 10 cm. Find the perimeter of the triangle?

Solution:
As per the given information, three sides of the triangle are 12 cm, 10 cm, 10 cm.
Perimeter of the triangle = (a + b + c) / 2.
Here, a = 12 cm, b = 10 cm, c = 10 cm.
Perimeter = (12 + 10 + 10) / 2 = 32 / 2 = 16 cm.

The perimeter of the triangle = 16 cm.

5. Area of the triangle is 20 cm² and the height of the triangle is 60 cm. What is the base of the triangle?

Solution:
As per the given information, the Area of the triangle (A) = 20 cm².
Height of the triangle (h) = 60 cm.
Area of the triangle (A) = (1 / 2) * base * height.
20 = (1 / 2) * base * 60.
20 *2 = base * 60.
40 / 60 = base.
Base = 4 / 6.

The base of the triangle is equal to 2 / 3.

Frequently Asked Questions on Triangle

1. What is Triangle?

Triangle is made up of three lines and it is a closed curve.

2. What are the properties of the Triangle?

Properties of the triangle are

• Triangle has three sides, three angles, and three vertices.
• The Sum of the three angles must be equal to 180°.
• The Sum of any two angles must be greater than the third angle.

3. What are the types of Triangles?

Based on the angles and sides triangles are divided into 6 types. They are

Classification of triangles based on the sides.

1. Scalene Triangle
2. Isosceles Triangle
3. Equilateral Triangle

Classification of triangles based on the angle

1. Acute Angle Triangle
2. Obtuse Angle Triangle
3. Right Angle Triangle

4. What are the basic formulas for triangles?

The basic formulas of triangles are Area and Perimeter Formulas. They are
Area of the Triangle (A) = (1 / 2) * base * height.
Or
Area (A) = √s(s – a) (s – b) (s – c).
Perimeter of the Triangle (s) = (a + b + c) / 2.

5. Difference between the Isosceles triangle and scalene triangle?

Isosceles Triangle: Two sides of the triangle are equal and the third side is different, then it is called as Isosceles Triangle.
Scalene Triangle: If the three sides of the triangle values are different, then it is called a scalene triangle.

Probability is a possibility of outcome for a nonoccurrence event. That means, when we are not sure about the outcome or result of an event, at that moment we can apply the probability method to the event. So that, we will know the chances of outcome of an event. For example, if we are trying to flip a coin and we don’t know the result or outcome of a coin that means, either it may be heads or tails. In such a case, we can use the probability method. If you want to know What is the Probability for Rolling Two Dices

Get to know the Probability When Two Dice are Rolled, Solved Examples on How to Calculate the Two Dice Rolling Probability, etc.  Also, find the Possible Outcomes Whe Two Dice are Rolled by checking out the Probability Table.

Two Dice Rolling Probability

In order to determine the probability of a dice roll we need to know two things namely

• Size of the Sample Space or Set of Possible Outcomes
• How often an Event Occurs

If you throw a single die the sample space is equal to values on the die i.e. (1, 2, 3, 4, 5, 6). Since th die is fair each number in the set occurs only once. To obtain the probability of rolling any number on the die we divide the event frequency by the size of sample space.

In the same way, When Two Dice are Rolled calculating the Probability becomes difficult. Here, Rolling One Die is independent of the other. One roll has no effect on the other and while dealing with the independent events we use the multiplication rule.

Also, Read:

• Probability of Tossing Three Coins
• Coin Toss Probability

Possibilities of Outcomes When Two Dice are Rolled

As said above, when we throw two dice, there is the possibility to get 36 outcomes. Have a look at the possibilities below.
Probability of an event = Number of favorable outcomes/ Total number of outcomes
Two dice are thrown at a time. Here, one die is x and another one is y.
X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}

Probability Table for Rolling Two Dice 

The Possible Outcomes When Two Dice are Rolled is given below. Total Possible Outcomes is equal to the Product of sample space of the first die(6) and the sample space of the second die(6) that is 36.

Possibility of outcomes are {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Total number of outcomes are 36.

Two Dice Probability Examples

1. Two dice are rolled. Find the Probability of the sum of scores is an even number?

Solution:
Two dices are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.
Add the scores of the two dices. That is
{(1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)}.
So, the number of possibilities of the sum of even numbers is 18.
The probability of an event = number of favorable outcomes/ total number of outcomes.

Probability of sum of an even numbers = 18 / 36 = 1 / 2.

2. Two dice are rolled. Find the Probability of the sum of scores is an odd number?

Solution:
Two dices are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of two dices are 36.
Add the scores of the two dices. That is
{(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)}.
So, the number of possibilities of the sum of odd numbers is 18.
The probability of an event = number of favorable outcomes/ total number of outcomes.

Probability of sum of an odd numbers = 18 / 36 = 1 / 2.

3. Two dice are rolled. Find the probability of the sum of 2, 4, and 12?

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.
(i) Number of favorable outcomes of the sum of 2 is (1,1).
So, only 1 outcome.
The total number of outcomes = 36.
So, the probability of an event = number of favorable outcomes/ total number of outcomes.
Probability of sum of 2 = 1/36.
(ii) Number of favorable outcomes of sum of 4 are {(1,3), (2,2), (3,1)}.
So, the number of favorable outcomes is 3.
The total number of outcomes = 36.
So, the probability of an event = number of favorable outcomes/ total number of outcomes.
Probability of sum of 4 = 3/36 = 1/12.
(iii) Number of favorable outcomes of the sum of 12 are {(6,6)}.
So, a number of favorable outcomes is 1.
The total number of outcomes = 36.
So, the probability of an event = number of favorable outcomes/ total number of outcomes.
Probability of sum of 12 = 1/36.

4. Two dice are rolled. P is the event that the sum of the numbers shown on the two dice is 5, and Q is the event that at least one of the dice shows up a 3?
Are the two events (i) mutually exclusive, (ii) exhaustive? Give arguments in support of your answer.

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.
P is sum of 5. That is, P = {(1,4), (2,3), (3,2), (4,1)}.
Q is at least one of the dice shows up 3. That is, Q = {(1,3), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,3), (5,3), (6,3)}
(i) Mutually Exclusive
P ∩ Q = {(2, 3), (3, 2)} ≠ ∅.
Hence, P and Q are not mutually exclusive.
(ii) Exhaustive
P∪ Q ≠ S.

Therefore, P and Q are not exhaustive events.

5. Two dice are thrown simultaneously. Find the probability of (i) doublet (ii) product of 6 (iii) Divisible by 4 (iv) total of at least 10 (v) sum of 8?

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of two dices is 36.
(i) Doublets
Possibilities of doublets are {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} = 6.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of an event = 6 / 36 = 1 / 6.
(ii) Product of 6.
Possibilities of product of 6 is {(1,6), (2,3), (3,2), (6,1)} = 4.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of an event = 4 / 36 = 1 / 9.
(iii) Divisible by 4.
Possibilities of Divisible by 4 is {(1,4), (2,2), (2,6), (3,1), (3,5), (4,1), (4,4), (5,3), (6,2), (6,6)} = 10.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of an event = 10 / 36 = 5 / 18.
(iv) Total of at least 10.
Possibilities of Total of at least 10 are {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)} = 6.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of an event = 6 / 36 = 1 / 6.
(v) Sum of 8.
Possibilities of sum of 8 are {(2,6), (3,5), (4,4), (5,3), (6,2)} = 5.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of an event = 5 / 36.

6. Two dice are thrown. Find the probability of
(i) multiple of 4.
(ii) multiple of 5.
(iii) prime number as the sum.
(iv) product as 2.
(v) sum as < = 5.
(vi) getting a multiple of 4 on one die and multiple of 2 on another die.

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.
(i) Multiple of 4.
Possibilities of multiple of 4 are {(1,3), (2,2), (2,6), (3,1), (3,5), (4,4), (5,3), (6, 2), (6,6)} = 9.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of Multiple of 4 = 9 / 36 = 1 / 4.
(ii) Multiple of 5.
Possibilities of multiple of 5 are {(1,4), (2,3), (3,2), (4,1), (4,6), (5,5), (6,4)} = 7.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of Multiple of 5 = 7 / 36.
(iii) Prime number as sum
Prime numbers are 1,2,3,5,7,11,13…..
Possibilities of prime number as sum = {(1,1), (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5)} = 15.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of Prime number as sum = 15 / 36 = 5 / 12.
(iv) Product as 2
Possibilities of product as 2 are {(1,2), (2,1)} = 2.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of Product as 2 = 2 / 36 = 1 / 18.
(v) sum as < = 5.
Possibilities of sum as < = 5 are {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)} = 10.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of sum as <=5 = 10 / 36 = 5 / 18.
(vi) getting a multiple of 4 on one die and multiple of 2 on another die.
Possibilities of getting a multiple of 4 on one die and multiple of 2 on another die are {(4, 2), (4, 4), (4,6), (2,4), (6,4)} = 5.
Probability of an event = number of favorable outcomes / total number of outcomes.

Probability of getting a multiple of 4 on one die and multiple of 2 on another die = 5 / 36.

7. Two dice are thrown. Find out the (i) the odds in favor of getting the sum 4, and (ii) the odds against getting the sum 3.

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.

(i) The odds in favor of getting the sum 4.
Possibilities of getting the sum 4 = {(1,3), (2,2), (3,1)} = 3.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of getting the sum 4 = 3 / 36 = 1 / 12.
Odds in favor of getting the sum 4 = probability of getting the sum 4 / (1 – probability of getting the sum 4).
Odds in favor of getting the sum 4 = (1 / 12) / (1 – (1 / 12).
= (1 / 12) / (11 / 12).
=1 / 11.

Finally, the odds I favor of getting the sum of 4 is equal to 1 / 11.

(ii) the odds against getting the sum 3.
Possibilities of getting the sum 3 = {(1,2), (2,1)} = 2.
Probability of an event = number of favorable outcomes / total number of outcomes.
Probability of getting the sum 3 = 2 / 36 = 1 / 18.
Odds in favor of getting the sum 3 = probability of getting the sum 3 / (1 – the probability of getting the sum 3).
Odds in favor of getting the sum 3 = (1 / 18) / (1 – (1 / 18).
= (1 / 18) / (17 / 18).
=1 / 17.

Finally, the odds I favor of getting the sum 3 is equal to 1 / 17.

8. Two dice are thrown. Find the probability that the numbers on the two dices are different?

Solution:
Two dice are rolled at a time.
That is, X = {1,2,3,4,5,6} and Y = {1,2,3,4,5,6}.
The total number of outcomes of the two dices is 36.
Possibilities of the numbers on the two dices are different = {(1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5)} = 30.
Probability of an event = number of favorable outcomes / total number of outcomes.

Probability of the numbers on the two dices are different = 30 / 36 = 5 / 6.

A triangle is a polygon having 3 sides and three vertices. The sum of interior angles of a triangle is 180 degrees. Depending on the side length triangles are divided into three types they are equilateral triangle, isosceles triangle, and scalene triangle. And based on the angle measurement, triangles are again classified into three various types they are right, acute, oblique triangles. Both median, altitude is the lines in the triangle. Here, we will learn more about the Medians and Altitudes of a Triangle.

Also, Read

• Area and Perimeter of the Triangle
• The triangle on the Same Base and between Same Parallels Theorem
• Perimeter of a Triangle

Median of a Triangle – Definition

Median in a triangle is nothing but the straight line that joins one vertex and midpoint of the side that is opposite to the vertex. A triangle can have a maximum of three medians and the point of intersection of three medians is called the center of the triangle.

In △ ABC, AD is the median that divides a side BC into two equal parts. So, BD = CD.

Properties of a Triangle Median

• All triangles have 3 medians, each one from the triangle vertex. Here AD, BE, CF are the 3 medians of the triangle ABC.
• The three medians meet at a single point.
• The point where 3 medians meet is called the centroid of the triangle. Here O is the centroid of △ ABC.
• Each median of a triangle divides the triangle into two smaller triangles having the same area.
• So, 3 medians divide a triangle into 6 smaller triangles of equal area.

Altitude of a Triangle – Definition

The altitude is a straight line that starts from the triangle vertex and stretches till the opposite side of the vertex making a right angle with the side of the triangle.

Properties of Altitude of a Triangle

• Each triangle has 3 altitudes. Here AD, BE, CF are the altitudes of the triangle ABC.
• An altitude is also called the shortest distance from the vertex to the opposite side of a triangle.
• Three altitudes always meet at a single point.
• The point of intersection of three altitudes is called the ortho-center of the triangle. Here O is called the ortho-center of triangle ABC.
• The altitude of a triangle may lie inside or outside the triangle.

Median and Altitude of an Isosceles Triangle

Isosceles Triangle is a type of triangle that has two sides or angles of equal measurement. The median and altitude of an isosceles triangle have some particular features. They are along the lines.

• The Median, angle bisector is the same in an isosceles triangle when the altitude is drawn from the vertex to the base.
• Altitude, median, angle bisector interchange in case of an isosceles triangle.
• The Median, and altitude of the isosceles triangle are the same.

Medians and Altitudes of Triangles Examples

Example 1:

The given angles of a triangle ABC are in the ratio of 1 : 2 : 3. Evaluate all the angles of △ ABC.

Solution:

Let the first angle of triangle A is x.

So, ∠B = 2x, ∠C = 3x

We know that sum of all angles in a triangle is 180°

x + 2x + 3x = 180°

6x = 180

x = \(\frac { 180 }{ 6 } \)

x = 30°

So, ∠A = 30°

∠B = 2 x 30° = 60°

∠C = 3 x 30° = 90°

Therefore, the angles in a scalene triangle are different.

Example 2:

Construct ΔABC whose sides are AB = 4 cm, BC = 6 cm and AC = 5 cm and locate its orthocentre.

Solution:

Draw ΔABC using the given measurements.

Construct altitudes from any two vertices (A and C) to their opposite sides (BC and AB respectively).

The point of intersection of the altitudes O is the orthocentre of the given ΔABC.

Example 3:

Construct the centroid of ΔABC whose sides are AB = 6 cm, BC = 7 cm, and AC = 5 cm.

Solution:

Draw ΔABC using the given measurements.

Construct the perpendicular bisectors of any two sides (AC and BC) to find the mid points D and E of AC and BC respectively.

Draw the medians AE and BD and let them meet at G.

The point G is the centroid of the given ΔABC.

FAQs on Medians and Altitudes of a Triangle

1. How many medians can a triangle have?

Median is a line segment that connects a vertex to the mid point of the opposite side. Every triangle has exactly 3 medians each from one vertex.

2. What is the difference between the median and altitude of a triangle?

The altitude is a perpendicular bisector that falls on any side of the triangle and the median meets the side of a triangle at the midpoint. For an isosceles triangle, the altitude drawn to the base of a triangle is called the median, median drawn to the triangle base is called the altitude.

3. What is an ortho-center in a triangle?

The point where all three altitudes intersect is called the ortho-center of the triangle. The ortho-center may lie either inside or outside of a triangle.

The construction of angles is an important part of geometry as this knowledge is extended for the construction of other geometric figures. Constructing angles of unknown and unknown measures can be possible with geometric tools like compass, ruler, protractor. Here we will learn about the Construction of Angles by Using Compass in the following sections. You will find the Construction of Angles using Compass Examples with Solutions explained step by step.

Construction of Angles by Using Compass – Introduction

An angle is defined as the figure formed by two rays meeting at a common endpoint. The representation of angle is ∠. Construction of angles by using compass means you need to make an angle between two straight lines just with the compass. Take one straight line and draw an arc on the line with the compass with any radius from two ends of the line. Get the simple steps in the below-mentioned sections of this page.

Steps to Construct Angle of 60° with Compass

Check out the detailed steps to construct an angle of 60 degrees by using a compass.

• Draw a ray AB.
• By taking either A or B as a center and any suitable radius draw an arc above the ray and cutting at a point P.
• Now, with P as the center and having the same radius, draw another arc that meets the previous arc at C.
• Join AC or BC and produce it to D.
• Then ∠ADB = 60°.

More Related Articles

• Related Angles
• Pairs of Angles
• Types of Angles

Construction of Angles by Compass Examples

Example 1:

Construct an acute angle of 30° by using the compass?

Solution:

Take any ray OA

With O as a center and any radius draw an arc on OA at point P and draw another arc with P as the center, same radius.

The point of intersection of two arcs is Q.

∠AOR = 60°

• With center O and any convenient radius draw an arc cutting OA and OB at P and Q respectively.
• With centre P and radius more than \(\frac { 1 }{ 2 } \)(PQ), draw an arc in the interior of ∠AOB.
• With center Q and the same radius, as in step III, draw another arc intersecting the arc in step III at R.
• Join OR and product it to any point C.
• The angle ∠AOC is the angle of measure 30º.

Example 2:

Construct an obtuse angle of 120° by using the compass.

Solution:

• Draw a ray OA.
• By taking O as the center and any convenient radius, draw an arc cutting OA at P.
• By taking P as a center and same radius draw an arc, cutting the first arc at Q.
• With Q as the center and the same radius, draw an arc cutting the arc drawn in step II at R.
• Join OR and produce it to any point C.
• ∠AOC measures 120º.

Example 3:

Construct an angle of 90° by using the compass.

Solution:

• Draw a ray OA.
• With O as a center and any radius, draw an arc on OA at point P.
• By taking P as the center, the same radius, draw another arc cutting the first arc at point Q.
• With the point, Q as a center and the same radius draw an arc cutting the arc drawn in step II at R.
• By taking point Q as the center and same radius, draw an arc.
• With R as a center and the same radius, draw an arc, cutting the arc drawn in step V at B.
• Draw OB and extend it to C.
• ∠AOC measures an angle is 90º.

Example 4:

Construct an angle of 45° by using the compass.

Solution:

First of all, draw an angle of 90º.

Draw a bisector for 90º to measure 45º.

Check problems on trigonometric identities along with the solutions. Find the step by step procedure to know the trigonometric identities problems. Refer to all the solutions present in the below sections to prepare for the exam. Scroll down the page to get the Trigonometric Identities Word Problems and study material. Know the various formulae involved in solving trigonometric identities below. Assess your knowledge level taking the help of the Practice Problems on Trigonometric Identities available.

Also, Read:

• Worksheet on Trigonometric Identities
• Trigonometrical Ratios Table

Problem 1:

The angle of inclination from a point on the ground 30 feet away to the top of Lakeland’s Armington Clocktower is 60°. Find the height of the clock tower which is nearest to the foot?

Solution:

As given in the question,

The angle of inclination (θ) = 60°

The height from the ground(a) = 30 feet

To find the height of the Clocktower to the nearest foot, we use the formula

tan θ = h/a

tan 60° = h / 30

h = 30 tan 60°

h = 9.60 ≅ 10 feet

Therefore, the height of the clocktower to the nearest foot is 10 feet.

Hence, the final solution is 10 feet.

Problem 2:

Mary wants to determine the California redwood tree height, there are two sightings available from the ground which one is 200 feet directly behind the other. If the angles of inclination (Θ) are 45° and 30° respectively, how tall is the tree to the nearest foot?

Solution:

As given in the question,

Length at which trees are slighting = 200

The angle of inclinations = 45° and 30°

To find the inclination on the first tree, we apply the Pythagorean theorem,

tan 45° = h/x

h = x tan 45° is the (1) equation

tan 30° = h/(200+x)

h = (200 + x) tan 30 is the (2) equation

From both the equations,

x tan 45° = (200 + x) tan 30°

x tan 45° = 200 tan 30° + x tan 30°

x tan 45° – x tan 30° = 200 tan 30

Divide the equation with tan 45° – tan 30°

x (tan 45° – tan 30°) / tan 45° – tan 30° = 200 tan 30° / tan 45° – tan 30°

x = 273.21 feet

h = 273.21 tan 45

h = 273. 21 feet

h ≅ 273 feet

Therefore, the height of the tree to the nearest foot = 273 feet

Thus, the final solution = 273 feet

Problem 3:

A tree that is standing vertically on the level ground casts the 120 foot long shadow. The angle of elevation from the end of the shadow of the top of the tree is 21.4°. Find the height of the tree to the nearest foot?

Solution:

As given in the question,

Length of the foot-long shadow = 120

The inclination of the tree = 21.4°

To find the height of the tree to the nearest foot, we apply the Pythagorean theorem

tan θ = 0/a

tan 214° = h/120

h = 120 tan 214°

h = 47.03

h ≅ 47 feet

The height of the tree to the nearest foot = 47 feet

Thus, the final solution is 47 feet

Problem 4:

The broadcast tower which is for the radio station WSAZ (“Carl” and “Jeff”‘s home of algebra) has 2 enormous flashing red lights on it. Of the 2 enormous flashing lights, one is at the very top and the other one is few feet below the top. From that point to the base of the tower it is 5000 feet away from level ground, the top light angle of elevation is 7.970° and the light angle of elevation of the second light is 7.125°. Find the distance between the nearest foot and the lights.

Solution:

As given in the question,

Height of the tower = 5000 feet

The angle of elevation of top light = 7.970

The angle of elevation of second light = 7.125°

To find the distance between the nearest foot and the lights, we have to use the Pythagorean theorem

tan θ = h/5000

h = 5000 tan 7.97°

tan β = h-x/5000

h-x = 5000 tan (7.125)°

x = h – 5000 tan (7.125)°

x = 5000 tan 7.97° – 5000 tan 7.125°

x = 5000 (tan 7.97° – tan 7.125°)

x = 75.04 feet

x ≅ 75  feet

Therefore, 75 feet is the distance between the nearest foot and the lights

Thus, the final solution is 75 feet.

Problem 5:

Find the solution of tan (θ) = sin (θ) sec (θ)?

Solution:

sinθ/ cosθ = sinθ secθ

sinθ. (1/cosθ) = sinθ secθ

sinθ secθ = sinθ secθ

tanθ = sinθ secθ

= sinθ(1/cosθ)

= sinθ/cosθ

= tanθ

∴ Hence it is proved

Problem 6:

Prove that (sec(θ)-tan(θ))(sec(θ)+tan(θ))=1

Solution:

sec²(θ)-tan²(θ)=1

1/cos²(θ)-sin²(θ)/cos(θ)=1

(1-sin²(θ))/cos²(θ)=1

cos²(θ)/cos²(θ)=1

(1-sin²(θ))/cos²(θ)

1/cos²(θ)-sin²(θ)/cos²(θ)=sec²(θ)-tan²(θ)

(sec(θ)-tan(θ))(sec(θ)+tan(θ))

Therefore, (sec(θ)-tan(θ))(sec(θ)+tan(θ)) = 1

∴Hence, it is proved

Problem 7:

Prove that sec(θ)/(1-tan(θ))=1/(cos(θ)-sin(θ))

Solution:

1/((cos(θ)-sin(θ)).1/cos(θ).1/cos(θ)

(1/cos(θ))/((cos(θ)-sin(θ))-(cos(θ))=sec(θ)/(1-tan(θ))

Therefore, sec(θ)/(1-tan(θ))=1/(cos(θ)-sin(θ))

∴Hence, it is proved

Problem 8 :

An aeroplane over the Pacific sights an atoll at a 20° angle of depression. If the plane is 435 ma above water, how many kilometres is it from a point 435m directly above the centre of a troll?

Solution:

As given in the question,

The angle of depression = 20°

Height of the plane above water = 435ma

Height above the centre of a troll = 435m

To find the kilometers, we use the pythegorean theorem

tan = θ/A

tan 20° = 435/x

x = 435/tan20°

x = 1.195 km

Therefore, 1.195 kilometres is it from a point 435m directly above the centre of a troll

Thus, the final solution is 1.195 kilometres

Problem 9:

The force F (in pounds)on the back of a person when he or she bends over an acute angle θ (in degrees) is given by F = 0.2W sin(θ + 70)/sin12° where w is the weight in pounds of the person

a) Simplify the formula or F.

b) Find the force on the back of a person where an angle of 30° weight is 50 pounds if he bends an angle of 30°

c) How many pounds should a person weigh for his book to endure a force of 400 lbs if he bends 40°?

Solution: 

a. F = 0.2W sin (θ + 90)/sin 12°

= 0.2W [sinθ cos 90 + cosθ sin90]/sin 12°

= 0.2W [θ(sinθ) + (cosθ) (1)]/sin 12°

= 0.2W [0 + cosθ]/sin 12°

F = 0.2W(cosθ)/sin 12°

The value of F is 0.2W(cosθ)/sin 12°

b. W = 50, θ = 30°, F=?

F = 0.2W cosθ/sin 12°

F = 0.2 (50) (cos30°)/sin 12°

F = 41.45

The force on the back of a person wherein the angle of 30e weight is 50 pounds if he bends an angle of 30° is 41.45

c. F = 400, θ = 40°, W = ?

400 = 0.2(w)(cos 40°)/sin 12°

400(sin 12°)/0.2 cos 40 = 0.2 (w) (cos 40°)/0.2 cos40°

542.82 = w

Therefore, the weigh for his book to endure a force of 400 lbs if he bends 40° is 542.82 pounds

Problem 10:

An observer standing on the top of vertical cliff pots a house in the adjacent valley at an angle of depression of 12°. The cliff is 60m tall. How far is the house from the base of the cliff?

Solution:

As given in the question,

The angle of depression = 12°

Height of the cliff = 60m

To find, the distance of the house from the base of the cliff, we apply the Pythagorean theorem

tan 12° = 60/x

x = 60/tan 12°

x = 282m

282m is the distance of the house from the base of the cliff

Hence, the final solution is 282m

Problem 11:

Building A and B are across the street from each other which is 35m apart. From the point on the roof of building A, the angle of elevation of the top of building B is 24°, the angle of depression of the base of building B is 34° How tall is each building?

Solution:

As given in the question,

The angle of elevation of the top of the building = 24°

The angle of depression of the base of the building = 34°

The distance of both buildings = 35m

To find the height of each building, we apply the Pythagoras theorem,

tan 24° = c/35

c = 15.6

tan 56° = 35/a

a = 23.6m

b = a+c

b = 39.2m

A is 23.6m tall

B is 39.2m tall

Therefore, the height of building A is 23.6m tall

The height of building B is 39.2m tall

Thus the final solution is 23.6m, 39.2m

Problem 12:

In Johannesburg in June, the daily low temperature is usually around 3°C, and the daily temperature is around 18°. The temperature is typically halfway between the daily high and daily low at 10 am and 10 pm. and the highest temperatures are in the afternoon. Find out the trigonometric function which models the temperature T in Johannesburg t hours after midnight?

Solution:

As per the question,

To determine the trigonometric model, the temperature ‘T’ which is Celsius degree and the temperature in axis and then right over here is time in hours. To think about the range of temperatures, the daily temperature is around 3-degree celsius and the highest is 18°. The midpoint between 18 and 3 is 10.5 (21 divided by 2)

Let F(t) is the temperature t hours after 10 am

F(t) = 7.5sin(2Π/24 t) + 10.5

T (t) = 7.5sin (Π/12(t-10)) + 10.5

T(10) = F(0) where T(10) is temperature at 10 pm

F(0) is the temperature at 10 am

Therefore, T(10) = F(0) is the trigonometric function that models the temperature T in Johannesburg t hours after midnight

Problem 13:

A ladder is 6 meters long and reaches the wall at a point of 5m from the ground. What is the angle which the ladder will make with the wall?

Solution:

Let θ be the inclination of the ladder which it makes with the wall

As given in the question,

Length of the ladder = 6m

The distance at which the ladder touches the wall = 5m

The angle at which the ladder will make with the wall

cos θ = 5/6

θ = cos¯¹ (5/6)

θ = 33.56°

Problem 14:

The acceleration of the piston is given by a = 5.0(sinωt + cos2ωt). At what positive values of ωt less than 2Π does a = 0?

Solution:

Let ωt be the crank angle(product of time and angular velocity) in piston acceleration, a be the accelaration of a piston.

a = 5.0(sinωt + cos2ωt)

0 = 5.0(sinωt + cos2ωt)

0 = sinωt + cos2ωt

0 = sinωt + 1 – 2sin²ωt

2sin²ωt – sinωt – 1 = 0

(2sinωt + 1) (sinωt – 1) = 0

2sinωt + 1 = 0

2sinωt = -1

sinωt = -1/2

ωt = sin¯¹(-1/2)

ωt = (7π/6, 11π/6)

sinωt – 1 = 0

sinωt = 1

ωt = sin¯¹(1)

ωt = π/2

{π/2, 7π/6, 11π/6}

Problem 15:

A lighthouse at sea level is 34 mi from a boat. It is known that the top of the lighthouse is 42.5mi from the boat. Find the angle of depression from the top of the lighthouse.

Solution:

Let θ be the angle of inclination from the top of the lighthouse to the boat

Let x be the horizontal distance from the base of the lighthouse to the boat, and

r be the distance from the top of the lighthouse to the boat.

As given in the question,

The length of the lighthouse at sea level = 34 mi

The distance of the top lighthouse from the boat = 42.5 mi

To find the angle of depression from the top of the lighthouse, we apply the Pythagorean theorem

cosθ = 34/42.5

θ = cos¯¹(34/42.5)

θ = 36.87°

Skip counting means counting by a number that is not 1. Skip Counting by 4’s is an essential skill to learn when making the jump from counting to basic addition. The result of skip counting by 4s will be the multiples of 4. You can also get the result by using the addition operation. Get to know more about the concept in the following sections.

Skip Counting by 4 Chart

This skip counting by fours chart will help us to write the number to complete the series which involves skip counting by 4’s up to 25 times. Skip counting is a method of counting numbers by adding a number every time to the previous number. When we skip count natural numbers by 4 then we add 4 at each step. The highlighted numbers in the chart are skip counting by 4’s numbers. It helps in counting the number quickly and it has a huge application in the multiplication of tables.

Forward and Backward Skip Counting

Forward skip counting means we count the natural number in forward direction of a number. Here, we skip count for positive numbers. Skip counting has a huge application in the real life. if we have to count eggs that are in the 40s of a number, then we use the skip counting by fours method. Counting one egg will take more time.

Backward skip counting means counting the numbers towards negative numbers. Here, you need to subtract 4 from the result.

Also, check out

• Skip Counting by 3S

Skip Counting by 4 Examples with Answers

Example 1:

Complete the fill in the blanks using the skip counting by fours chart.

(i) 4 x 19 = ______

(ii) 4 x 23 = ______

(iii) 4 x 12 = ______

Solution:

(i) 4 x 19 = (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4)

= 76

4 x 19 means add 4 19 times.

(ii) 4 x 23 = (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4)

= 92

4 x 23 means add 4 23 times.

(iii) 4 x 12 = (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4)

= 48

4 x 12 means add 4 12 times.

Example 2:

Complete the skip counting series by 4s:

(i) ____, ____, ____, ____, 36, ____, ____, 48.

(ii) 24, 28, ____, ____, ____, ____, ____, 52.

(iii) 8, 12, 16, ____, ____, ____, ____, ____.

Solution:

(i) Subtract and add 4 from the known numbers to know the before and after numbers.

So, 36 – 4 = 32

32 – 4 = 28

28 – 4 = 24

24 – 4 = 20

36 + 4 = 40

40 + 4 = 44

Therefore, the sequence is 20, 24, 28, 32, 36, 40, 44.

(ii) Add 4 from the known numbers to know the before and after numbers.

28 + 4 = 32

32 + 4 = 36

36 + 4 = 40

40 + 4 = 44

44 + 4 = 48

48 + 4 = 52

Therefore, the sequence is 24, 28, 32, 36, 40, 44, 48, 52.

(iii) Add 4 from the known numbers to know the before and after numbers.

8 + 4 = 12

12 + 4 = 16

16 + 4 = 20

20 + 4 = 24

24 + 4 = 28

28 + 4 = 32

32 + 4 = 36

Therefore, the sequence is 8, 12, 16, 20, 24, 28, 32, 36.

Example 3:

Complete the fill in the blanks using the skip counting by fours chart.

(i) 4 x 43 = _____

(ii) 4 x 25 = _____

(iii) 4 x 67 = _____

Solution:

(i) 4 x 43 = (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4)

= 172

4 x 43 means add 4 23 times.

(ii) 4 x 25 = (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4)

= 100

4 x 25 means add 4 25 times.

(iii) 4 x 67 = (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4)

= 268

4 x 67 means add 4 67 times.

Example 4:

Complete the skip counting series by 4s:

(i) ____, ____, ____, ____, 88, 92, ____, 100.

(ii) ____, ____, 16, 20, ____, ____, ____, 36.

(iii) 72, 76, ____, ____, ____, ____, 96, ____.

Solution:

(i)

Subtract and add 4 from the known numbers to know the before and after numbers.

88 – 4 = 84

84 – 4 = 80

80 – 4 = 76

76 – 4 = 72

88 + 4 = 92

92 + 4 = 96

96 + 4 = 100

Therefore, the sequence is 72, 76, 80, 84, 88, 92, 96, 100.

(ii)

Subtract and add 4 from the known numbers to know the before and after numbers.

16 – 4 = 12

12 – 4 = 8

16 + 4 = 20

20 + 4 = 24

24 + 4 = 28

28 + 4 = 32

32 + 4 = 36

Therefore, the sequence is 8, 12, 16, 20, 24, 28, 32, 36.

(iii)

Subtract and add 4 from the known numbers to know the before and after numbers.

72 + 4 = 76

76 + 4 = 80

80 + 4 = 84

84 + 4 = 88

88 + 4 = 92

92 + 4 = 96

96 + 4 = 100

Therefore, the sequence is 72, 76, 80, 84, 88, 92, 96, 100.

FAQs on Skip Counting by 4S

1. What is skip counting?

Skip counting is a method of counting forward by any number apart from 1. If we skip count by 4 means we are adding 4 at each step to get another number.

2. How do we skip count?

Skip counting is a method of counting numbers by skipping them with a certain number.

3. How to skip count by 4?

The skip ccounting by 4’s means we have to skip 4 numbers in forward count and jump to next one. 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, . .

We have infinite points in the coordinate plane. A line passes through a point (x, y). The different forms of equations of straight lines are equations of horizontal and vertical lines, Point-slope form, two-point form, slope-intercept form, intercept form, and normal form. Get to know more about the Equation of a Line Parallel to the x-axis in the following sections of this page.

Also, Read:

• Types of Lines
• Angle Between Two Straight Lines
• Lines and Angles

Equation of a Line Parallel to X-Axis

A line can be defined as a straight one-dimensional geometric figure that doesn’t have the thickness and extends endlessly in both directions. A straight line on the coordinate plane can be described by an equation is called the equation of a line. If all the points the straight line having the same y-coordinate or ordinate values, then it is called the equation of a line parallel to the x-axis.

The general form of the equation of a line that is parallel to the x-ais is y = k. Here, k is the distance between the x-axis and the line. If a point P(x, y) lies on the line, then y = b. The equation of x-ais is y = 0 as the x-axis is parallel to itself at a distance of 0 from it.

Different Forms of the Equation of Line

The various forms of the equations of a straight line are long the lines.

Slope Intercept Form

The equation of a slope-intercept form of a straight line is y = mx + b.

Here, m is the slope,

b is the y-intercept.

Point Slope Form

The point-slope form of a line is y – y₁ = m(x – x₁)

Here, m is the slope of the line

(x₁, y₁) is a point on the line.

Two Point Form

The two points form of a line is \(\frac { y – y₁ }{ y₂ – y₁ } = \frac { x – x₁ }{ x₂ – x₁ }\)

Here, (x₁, y₁), (x₂, y₂) are the two points on the line

Slope of the line = \(\frac { y₂ – y₁ }{ x₂ – x₁ } \)

Intercept Form

Intercept form of a line is \(\frac { x }{ a } +\frac { y }{ b }\) = 1

Here, a is the x-intercept

b is the y-intercept

Equation of x-axis

The equation of x-axis is y = 0. Because the value of “ordinate” in all the points on the x-axis is zero.

Equation of y-axis

The equation of y-axis is x = 0. Why because the value of abscissa in all points on the y-axis is zero.

General Equation

The general equation of a straight line is ax + by + c = 0.

Equation of a Line Parallel to X-Axis Examples

Example 1:

Find the equation of a line parallel to the x-axis at a distance of 7 units above the x-axis?

Solution:

We know that the equation of a line parallel to the x-axis at a distance b from it is y = b.

Therefore, the equation of a straight line parallel to the x-axis at a distance 7 units above the x-axis is y = 7.

Example 2:

Find the equation of a line parallel to the x-axis at a distance of 5 units below the x-axis?

Solution:

We know that If a straight line is parallel and below to x-axis at a distance b, then its equation is y = -b.

Therefore, the equation of a line parallel to the x-axis at a distance of 5 units below the x-axis is y = -5.

Example 3:

Find the equation of a straight line parallel to the x-axis at a distance of 10 units above the x-axis?

Solution:

We know that the equation of a line parallel to the x-axis at a distance b from it is y = b.

Therefore, the equation of a straight line parallel to the x-axis at a distance 10 units above the x-axis is y = 10.

FAQs on Equation of Line Parallel to X-Axis

1. How do you find the equation of a line?

The general form of equation of a line is ax + by + c = 0. Any equation in this form is called the equation of a straight line.

2. What is the equation of the line parallel to the x-axis?

The equation of a straight line parallel to the x-axis is y = b as all the points on that line have y-coordinate values as zero’s. Here, b is the distance between the line and the x-axis.

3. How do you write an equation of a line parallel to a line?

The slope-intercept form of a line is y = mx + c. If two lines are parallel, then their slopes are equal and the y-intercept depends on the line points. So, if you know one line, then it is easy to find the equation of a line parallel to the given line and passes through one point.

Decimal numbers are the part of numbers that have two parts. The decimal numbers are in the standard form representing integer and non-integer numbers. Generally, the decimal points are written in a fraction which consists of 10, 100, 1000 in the denominators. The numbers that are expressed in the decimal form are called decimal numbers. Let us check the complete concept on decimal in the below article.

Also, Read:

• Correct to Two Decimal Places
• Converting Decimals to Fractions
• Conversion of a Decimal Fraction into a Fractional Number

Decimals – Definition

A decimal number is a number that has two parts. One part has a whole number and the other part is a fractional part. Both parts are separated by decimal points. If 2.48 is a decimal number, then 2 is the whole number and 48 is the fractional part. “.” is the decimal point.

• The digits having in the whole number part are called ones, then tens, then hundreds, then thousands, and so on.
• The places after the decimal point begin with tenths, then hundredths, then thousandths, and so on………

Examples:

(i) In the decimal number 47.25, the whole number part is 47 and the decimal part is .25
(ii) In the decimal number 89.063, the whole number part is 89 and the decimal part is .063
(iii) Take the decimal number 11.056 where the whole number part is 11 and the decimal part is .056

Types of Decimal Numbers

Decimal numbers are classifieds into different types. They are given with definitions and examples explained in detail.

Recurring Decimal Numbers – Recurring Decimal Numbers are Repeating or Non-Terminating Decimals.
Examples of Recurring Decimal Numbers are 2.123123 (Finite) and 4.252525252525… (Infinite)

Non-Recurring Decimal Numbers – Non-Recurring Decimal Numbers are Non Repeating or Terminating Decimals.
Examples of Non-Recurring Decimal Numbers are 5.14812 (Finite) and 2.5428454845…. (Infinite)

Decimal Fraction – Decimal Fraction is the fraction that consists of the denominator as powers of ten.
Examples are 72.66 = 7266/100 and 43.536 = 43536/1000.

Converting a Decimal Number into Decimal Fraction
To convert the Decimal Number into Decimal Fraction, place the 1 in the denominator and remove the decimal point from the given number. The 1 is followed by the number of zeros that are equal to the number of digits given after the decimal point.

Examples:
1. 47.59
The given decimal number is 47.59
47.59 = 4759/100
4 represents the power of 101 that is the tenths position.
7 represents the power of 100 that is the unit’s position.
5 represents the power of 10-1 that is the one-tenth position.
9 represents the power of 10-2 that is the one-hundredths position.
So that is how each digit is represented by a particular power of 10 in the decimal number.
2. 61.27
The given decimal number is 61.27
61.27 = 6127/100
6 represents the power of 101 that is the tenths position.
1 represents the power of 100 that is the unit’s position.
2 represents the power of 10-1 that is the one-tenth position.
7 represents the power of 10-2 that is the one-hundredths position.
So that is how each digit is represented by a particular power of 10 in the decimal number.

Place Value in Decimals

Place value of a number in decimals is the position of every digit that helps to find its value. The position of each digit that before and after the decimal point is different. Check out the below examples to know each digit’s place value.

Examples:
Let us take a number 296.

• The position of “2” is in One’s place, which means 2 ones (i.e. 2).
• The position of “9” is in the Ten’s place, which means 9 tens (i.e. ninety).
• The position of “6” is in the Hundred’s place, which means 6 hundred.
• As we go left, each position becomes ten times greater.
• Hence, we read it as “two hundred ninety-six”.

As each digit, we move to the left side the value becomes 10 times greater than the previous value.

• The tens place digit is 10 times bigger than Ones.
• The hundreds place digit is 10 times bigger than Tens.

If we consider a decimal number, the digits after the decimal points will become 10 times smaller than other digits. The digits present on the left side of the decimal are multiplied with the positive powers of ten in increasing order from right to left. The digits present in the right of the decimal point are multiplied with the negative powers of 10 in increasing order from left to right.

Example:
1. 61.28
The decimal expansion of the given number 61.28 is
[(6 * 10) + (1 * 1)] + [(2 * 0.1) + (5 * 0.01)]

Properties of Decimals

We have given the main properties of the decimal numbers below those are under multiplication and division operations. Check out all the properties of decimals given below.

• When two decimal numbers are multiplied with each other, then the result will be a decimal number.
• When a decimal number and a whole number are multiplied with each other, then the result will be a decimal number.
• If any decimal fraction is multiplied by 1, the product remains the same decimal fraction by itself.
• If any decimal fraction is multiplied by 0, then the product becomes 0.
• When a decimal number divided by 1, then the quotient must be a decimal number.
• Also, when a decimal number is divided by the same number, then the quotient becomes 1.
• If in case, 0 divided by any decimal number, the quotient becomes 0.
• The division of a decimal number by 0 is not applicable and possible as the reciprocal of 0 does not exist.

Arithmetic Operations on Decimals

We can perform addition, subtraction, multiplication, and division operations on Decimals easily. Check out the below concepts to understand different Arithmetic Operations on Decimals.

Addition Operation on Decimals
When you add two decimal numbers, line up the decimal points of the given numbers and add them. If you don’t see a decimal point, then that is only a whole number.

Subtraction Operation on Decimals
Subtraction Operation on Decimals is also similar to the Addition Operation on Decimals. You need to line up the decimal point of the given numbers and subtract the values.

Multiplication Operation on Decimals
Multiplication Operation on Decimals is like integers as if the decimal point not present. Firstly, find out the product and count the number after the decimal point in the given numbers. The count will let you know how many numbers present after the decimal point in the result.

Division Operation on Decimals
The Division Operation on Decimals is simply dividing the given two decimal numbers. Move the decimal points to make them whole numbers. Then, perform the division operation like normal integers.

Decimal to Fraction Conversion

We consider the digits after the decimal point as the tenths, hundredths, thousandths, and so on. Write down the decimal numbers in the expanded form and simplify the values.
Example:
Let us consider a decimal number 5.21
Conver it into a fraction number.
The expanded form of 5.21 is 521 x (1/100) = 521/100.

Fraction to Decimal Conversion

To convert the fraction number into a decimal number, divide the numerator by denominator.
Example: 9/7 is a fraction. If it is divided, we get 1.285714

Decimal Problems with Solutions

Example 1:

Convert 6/10 in decimal form?
Solution:
Given fraction number is 6/10.
To convert fraction to decimal, divide 6 by 10, we get the decimal form.
Thus, 6/10 = 0.6
Hence, the decimal form of 6/10 is 0.6.

Example 2:

Express 2.36 in fraction form?
Solution:
The given decimal number is 2.36
The expanded form of 2.36 is
= 236 x (1/100)
= 236 /100
= 118/50
= 59/25
Hence, the equivalent fraction for 2.36 is 59/25.

Frequently Asked Questions on Decimals

1. What is meant by Decimal?

A decimal is a number that mainly has two parts named as a whole number part and a fractional part separated by a decimal point.

2. What are the different types of decimals?

There are two types of decimals considered. They are

• Terminating decimals (or) Non-recurring decimals
• Non-terminating decimals (or) Recurring decimals

3. Write the expanded form of 85.3?

The expanded form of 85.3 is 80 + 5 + (3/10)

4. How to convert fractions to decimals?

Factions to decimals are converted by dividing the numerator by the denominator value.

A polygon is a two-dimensional figure that consists of a finite number of sides. The sides of a polygon are connected by line segments. The end connected by the line segments is called the vertex. There are Different Types of Polygons available in geometry. Poly means many and gon means an angle. The area and perimeter of the polygon completely depend on their shape. Any polygon can be classified depends on its sides and vertices. If a polygon has four sides and four vertices, then that polygon is named a quadrilateral.

Also, Read:

• Convex and Concave Polygons
• Regular and Irregular Polygon

Types of Polygons and their Properties

Based on the sides and vertices, the polygons are classified into different types. Check out the below polygons and their properties along with the examples and images.

• Regular Polygons
• Irregular Polygons
• Concave Polygons
• Convex Polygons
• Trigons
• Quadrilateral Polygons
• Pentagon Polygons
• Hexagon Polygons
• Equilateral Polygons
• Equiangular Polygons

Regular Polygon

A Polygon is said to be a Regular Polygon when all of its sides are equal. Also, a regular polygon has all the interior angles are the same. For example, a regular hexagon consists of equal six sides, and also its interior angles give a total of 120 degrees.

Examples:

• A square consists of all its sides equal to 4cm, and also all the angles are at 90°.
• A regular pentagon that has 5 equal sides. All the interior angles measure 108 degrees.
• An equilateral triangle has all three sides equal to 8cm and angles measure to 60°.

Irregular Polygon

A Polygon is said to be an irregular polygon when all of its sides are not equal. They have irregular shapes. Also, the angles of irregular polygons are not equal.

Examples:

• A triangle with unequal sides.
• A quadrilateral with unequal sides.

Convex Polygon

A Polygon is said to be a convex polygon when the measure of the interior angle is less than 180 degrees. The corners of a convex polygon are always outwards. Convex Polygon is completely opposite to the concave polygon.

Example:

An irregular hexagon that vertices are completely outwards.

Concave Polygon

A Polygon is said to be a concave polygon when there are at least one angle measures more than 180 degrees. The corners of a concave polygon present inwards and also outwards.

Trigons

A Polygon is said to be a trigon that has three sides. Trigons are also known as triangles. The triangles are classifieds into different categories. They are
Scalene Triangle: It has all sides that are unequal.
Isosceles Triangle: It has two sides that are equal.
Equilateral Triangle: All sides are equal in length and also all angles are 60 degrees.

Quadrilateral Polygon

A Polygon is said to be a Quadrilateral Polygon when it has four sides. Examples of Quadrilateral Polygon are square, rectangle, rhombus, and parallelogram.

Pentagon Polygon

A Polygon is said to be a Pentagon Polygon when it has five sides. If all the sides of a pentagon polygon are equal, then it is called a regular pentagon. If not it is called an irregular pentagon.

Hexagons

A polygon is called a hexagon when that has six sides and six vertices. If all the six sides of a hexagon are equal, then it is called a regular hexagon. Also, in a regular hexagon, all the interior and exterior angles are equal.

Equilateral Polygons

The polygons are called equilateral polygons when it has all sides are equal. An equilateral triangle, a square, etc., are examples of Equilateral Polygons.

Equiangular Polygons

The polygons are called Equiangular Polygons when all of their interior angles are equal. An example is a rectangle.

Types of Polygons with Sides 3-20

Below is the Classification of Polygons according to the number of sides and the angle measure. They are as follows

FAQs on Different Types of Polygons?

1. What is a Polygon?

A polygon is a two-dimensional figure that consists of a finite number of sides.

2. How many different types of Polygons are there?

There are two different types of Polygons namely Regular and Irregular. However, depending on the sides and vertices they are furthermore classified as Concave Polygons, Convex Polygons, Trigons, Quadrilateral Polygons, Pentagon Polygons, Hexagon Polygons, Equilateral Polygons, Equiangular Polygons, etc.

3. What is a 9 sided shape called?

9 Sided Shape is Called an Enneagon or Nonagon.

Worksheet on Decimal Word Problems will help the students to explore their knowledge of decimal word problems. Solve all the problems to learn the depth concept of Decimal Word Problems. Know the definition, properties, different operations performed on decimals by visiting our website. We have given the complete decimal concepts along with examples. Check out the Decimal Word Problems Worksheet and know the various strategies to solve problems in an easy way.

Also, Read:

• Decimals
• Converting Decimals to Fractions

What are Decimals?

The decimal is a number that has the whole number and the fractional part separated by a decimal point. The point between the whole number and fractions part is known as the decimal point. For example, 24.9 is the decimal number where 24 is a whole number and 9 is the fractional part after the decimal. We have included addition, subtraction, multiplication, and division decimal word problems below.

Word Problems on Decimals

Have a look at all the problems given below and get a grip on all types of problems on decimals.

1. Anshika scored 521 marks out of 600 in the final examination. How many marks did she lose?

2. Arun had 0.24 liters of cold drink. Kiran had 0.68 liters more cold drink. How much cold drink did they have together?

3. The weight of a baby lion was 169.24 kg. After two years, his weight increased by 107.64 kg. Find the weight of the lion after two years?

4. Ganesh had a rope of 54.28 m. He cut the rope into two pieces. If the length of one piece was 13.26 m, what was the length of the other piece?

5. Each side of a regular polygon is 21.7 m and its perimeter is 108.5 m. Find the number of sides of the polygon?

6. Fariha took 2.4 minutes to complete the race and Anil took 2.1 minutes to complete the race. Who won the race?

7. The annual rainfall received by Arunachal Pradesh is 261.5 cm and that by Assam is 297.4 cm. Who received less rainfall?

8. Rishu’s height is 134.51 cm. She stands on a tool of height 9.40 cm. What is the combined height now?

9. The milkman delivers 6.03 liter of milk to a house in the morning and 3.230 liters in the evening. What is the total quantity of milk delivered by the milkman?

10. Rebecca‘s kite is flying at a height of 28.3 m and Shelly’s at a height of 32.6 m from the ground. Whose kite is flying high and by how much?

11. A bike travels 39.2 km in 7 hours. How much distance will it travel in 1 hour?

12. Ron jogged 3.3 km, Mike jogged 4.8 times more distance than Ron. Find the distance covered by Mike?

13. The daily consumption of milk in a house is 4.36 liters. How much milk will be consumed in 30 days?

14. A tin contains 13.4 liters of oil. How many such tin contains 80.4 liters of oil?

15. Find the cost of 58.3 m cloth if the cost of 1 m cloth is $44.80.

16. Shruti bought a bag for $387.05. She gave the shopkeeper 2 notes of $200. How much money will she get back?

17. A tailor needs 46.36 m of cloth for the shirts and 56.90 m for trousers. How much cloth does the tailor need in all?

18. A spool of thread has a thread measuring 97.60 m. If 53.44 m thread has been cut, what length of thread is still left in the spool?

19. The cost of a chair is $3056.94. Tania wants to buy 7 chairs for her house. How much money will she pay to the shopkeeper?

20. David has a jug full of milk. He pours the complete milk in 5 glasses, each glass of capacity 0.6 l. How much milk was there in the jug?

21. Find the area of a square whose side is 2.50 m.

22. The weight of 1 bag of sugar is 11.4 kg. What is the weight of 14 such bags?

23. A vehicle covers a distance of 10.2 km in 3.4 liters of petrol. How much distance will it cover in 1 liter of petrol?

24. Ron has 5.60 l of juice. He pours it into 8 glasses equally. How much juice is there in each glass?

25. Shelly has a ribbon of length 44.44 m. She cuts it into 11 equal parts. What is the length of each equal part?

26. The cost of 4 pens is $49.24. What is the cost of 1 pen?

27. The weight of a box is 141.029 kg. What will be the weight of 23 such boxes?

28. Sonia has 18.48 l of juice. She pours it into 8 jars equally. How much juice is there in each jar?

Multiplying numbers is so easy compared to Multiplying Fractions. The fraction is represented as the division of the whole. The fraction is in the form of “x/y” where “x” is the numerator and “y” is the denominator. One can apply the fraction concept to real-time examples easily after reading the entire concept here. If you have an apple and you made it into 4 equal parts, then it can represent as \(\frac { 1 }{ 4 } \). Or else if it cut into 7 pieces then it can represented as \(\frac { 1 }{ 7 } \).

How to Multiply Fractions?

We have provided simple steps to multiply fractions and find the solution of multiplying fractions. It is easy to find out the solution using the below procedure. Multiplying fractions can be defined as the product of a fraction with another fraction or with the variables or with an integer. Follow the below process to multiply fractions

• Multiply the numerator with numerator
• Multiply the denominator with the denominator
• Simplify the fractions, if needed

Example:
1. Multiply \(\frac { 2 }{ 3 } \) × \(\frac { 1 }{ 5 } \)

Solution:
Given that \(\frac { 2 }{ 3 } \) × \(\frac { 1 }{ 5 } \)
To multiply the above fractions, firstly multiply the numerators
2 × 1 = 2
multiply the denominators
3 × 5 = 15
Now, simplify the fraction, we get \(\frac { 2 }{ 15 } \)
If \(\frac { x }{ y } \) and \(\frac { m }{ n } \) are the multiplicand and multiplier, then the output is \(\frac { xm }{ yn } \)
Product of Fraction = Product of Numerator/Product of Denominator

Fractions Parts and Types

A fraction consists of two parts. One is the numerator and another one is the denominator. If \(\frac { a }{ b } \) is a fraction, then the two parts are a and b where a is the numerator and b is the denominator. Or else if \(\frac { 3 }{ 4 } \) is a fraction, then the two parts are 3 and 4 where 3 is the numerator and 4 is the denominator.

Mainly, there are three types of fractions considered. They are proper fractions, improper fractions, and mixed fractions.
Proper fractions: A fraction is said to be a proper fraction when the numerator of a fraction is less than the denominator.
Examples: \(\frac { 1 }{ 4 } \), \(\frac { 5 }{ 6 } \), \(\frac { 7 }{ 11 } \)
Improper fractions: A fraction is said to be an improper fraction when the numerator is greater than the denominator.
Examples: \(\frac { 5 }{ 4 } \), \(\frac { 7 }{ 6 } \), \(\frac { 13 }{ 11 } \)
Mixed Fraction: A fraction is said to be a mixed fraction when we write the improper fraction in the combination of a whole number and a fraction.
Examples: 1 \(\frac { 5 }{ 4 } \), 3 \(\frac { 9 }{ 6 } \), 2 \(\frac { 6 }{ 7 } \)

Also, Read:

• Worksheet on Multiplication of Fractions
• Multiplication of Fractions
• Addition and Subtraction of Fractions

Fractional Simplification

Generally, the multiplication of fractions can be finished by multiplying numerators with numerators and multiplying denominators with denominators. To make the fractional multiplication simpler, we can reduce the fraction by canceling the common factors. By canceling out the common factors from the given factor, it becomes easier to find the exact output.

Example: \(\frac { 9 }{ 4 } \) and \(\frac { 2 }{ 3 } \)
\(\frac { 9 }{ 4 } \) can written as \(\frac { 3 × 3 }{ 2×2 } \)
\(\frac { 3 × 3 }{ 2×2 } \) × \(\frac { 2 }{ 3 } \) = \(\frac { 3 }{ 2 } \)
If there is no common factors, then the numerators and denominators are multiplied directly.

Types of Fraction Multiplication

There are different types of Fraction Multiplication available. They are

• Multiplication of Fraction with Whole Numbers
• Multiplication of Fraction with another Fraction
• Multiplication of Fraction with Variables

Multiplication of Fractional Number by a Whole Number

In Multiplication of Fractional Number by a Whole Number, we multiply the numerator with the numerator and the denominator remains the same. Before you multiply, reduce the fraction to the lowest terms. Check out different problems on the Multiplication of Fractional Number by a Whole Number below.

1. Multiply 2 \(\frac { 2 }{ 3 } \) by 9

Solution:
Given that multiply 2 \(\frac { 2 }{ 3 } \) by 9.
Firstly, convert given mixed fraction 2 \(\frac { 2 }{ 3 } \) to fraction.
2 \(\frac { 2 }{ 3 } \) = \(\frac { 8 }{ 3 } \)
Now, multiply \(\frac { 8 }{ 3 } \) by 9.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
8 × 9 = 72.
So, the fraction is \(\frac { 72 }{ 3 } \)
Simplify the fraction to get the final answer.
\(\frac { 72 }{ 3 } \) = 24.

The final answer is 24.

(ii) Multiply \(\frac { 3 }{ 4 } \) by 6

Solution:
Given that multiply \(\frac { 3 }{ 4 } \) by 6
multiply \(\frac { 3 }{ 4 } \) by 6.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
3 × 6 = 18.
So, the fraction is \(\frac { 18 }{ 4 } \)
Simplify the fraction to get the final answer.
\(\frac { 18 }{ 4 } \) = \(\frac { 9 }{ 2 } \).

The final answer is \(\frac { 9 }{ 2 } \).

(iii) Multiply 3 \(\frac { 3 }{ 2 } \) by 6

Solution:
Given that multiply 3 \(\frac { 3 }{ 2 } \) by 6.
Firstly, convert given mixed fraction 3 \(\frac { 3 }{ 2 } \) to fraction.
3 \(\frac { 3 }{ 2 } \) = \(\frac { 9 }{ 2 } \)
Now, multiply \(\frac { 9 }{ 2 } \) by 6.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
9 × 6 = 54.
So, the fraction is \(\frac { 54 }{ 2 } \)
Simplify the fraction to get the final answer.
\(\frac { 54 }{ 2 } \) = 27.

The final answer is 27.

Multiplication of Fractional Number by Another Fractional Number

Multiplying Fraction with Another Fraction is explained with the below examples.

(i). Multiply \(\frac { 3 }{ 5 } \) by \(\frac { 6 }{ 5 } \)

Solution:
Given that Multiply \(\frac { 3 }{ 5 } \) by \(\frac { 6 }{ 5 } \).
Firstly, multiply the numerators with numerators.
3 × 6 = 18.
Next, multiply the denominators with denominators.
5 × 5 = 25.
Finally, write the fraction in the simplest form.
\(\frac { 18 }{ 25 } \)

The final answer is \(\frac { 18 }{ 25 } \)

(ii) Multiply \(\frac { 3 }{ 4 } \) by \(\frac { 7 }{ 2 } \)

Solution:
Given that Multiply \(\frac { 3 }{ 4 } \) by \(\frac { 7 }{ 2 } \).
Firstly, multiply the numerators with numerators.
3 × 7 = 21.
Next, multiply the denominators with denominators.
4 × 2 = 8.
Finally, write the fraction in the simplest form.
\(\frac { 21 }{ 8 } \)

The final answer is \(\frac { 21 }{ 8 } \).

(iii) Multiply \(\frac { 2 }{ 3 } \), \(\frac { 2 }{ 5 } \), and \(\frac { 2 }{ 7 } \)

Solution:
Given that Multiply \(\frac { 2 }{ 3 } \), \(\frac { 2 }{ 5 } \), and \(\frac { 2 }{ 7 } \).
Firstly, multiply the numerators with numerators.
2 × 2 × 2 = 8.
Next, multiply the denominators with denominators.
3 × 5 × 7 = 105.
Finally, write the fraction in the simplest form.
\(\frac { 8 }{ 105 } \)

The final answer is \(\frac { 8 }{ 105 } \).

Multiplication of a Mixed number by Another Mixed Number

To find the multiplication of a mixed number with another mixed number, we need to change the mixed fractions to fractions and multiply them.

(i) Multiply 3 \(\frac { 2 }{ 5 } \) and 2 \(\frac { 3 }{ 7 } \)

Solution:
Given that multiply 3 \(\frac { 2 }{ 5 } \) and 2 \(\frac { 3 }{ 7 } \).
Firstly, convert given mixed fractions 3 \(\frac { 2 }{ 5 } \) and 2 \(\frac { 3 }{ 7 } \) to fractions.
3 \(\frac { 2 }{ 5 } \) = \(\frac { 17 }{ 5 } \)
2 \(\frac { 3 }{ 7 } \) = \(\frac { 17 }{ 7 } \)
Now, multiply \(\frac { 17 }{ 5 } \) by \(\frac { 17 }{ 7 } \).
Firstly, multiply the numerators with numerators.
17 × 17 = 289.
Next, multiply the denominators with denominators.
5 × 7 = 35.
Finally, write the fraction in the simplest form.
\(\frac { 289 }{ 35 } \)

The final answer is \(\frac { 289 }{ 35 } \).

(ii) Multiply 2 \(\frac { 4 }{ 3 } \) and 1 \(\frac { 6 }{ 5 } \)

Solution:
Given that multiply 2 \(\frac { 4 }{ 3 } \) and 1 \(\frac { 6 }{ 5 } \).
Firstly, convert given mixed fractions 2 \(\frac { 4 }{ 3 } \) and 1 \(\frac { 6 }{ 5 } \) to fractions.
2 \(\frac { 4 }{ 3 } \) = \(\frac { 10 }{ 3 } \)
1 \(\frac { 6 }{ 5 } \) = \(\frac { 11 }{ 5 } \)
Now, multiply \(\frac { 10 }{ 3 } \) by \(\frac { 11 }{ 5 } \).
Firstly, multiply the numerators with numerators.
10 × 11 = 110.
Next, multiply the denominators with denominators.
5 × 3 = 15.
Finally, write the fraction in the simplest form.
\(\frac { 110 }{ 15 } \) = \(\frac { 22 }{ 3 } \)

The final answer is \(\frac { 22 }{ 3 } \).

(i) Multiply 4 \(\frac { 2 }{ 9 } \) and 5 \(\frac { 1 }{ 4 } \)

Solution:
Given that multiply 4 \(\frac { 2 }{ 9 } \) and 5 \(\frac { 1 }{ 4 } \).
Firstly, convert given mixed fractions 4 \(\frac { 2 }{ 9 } \) and 5 \(\frac { 1 }{ 4 } \) to fractions.
4 \(\frac { 2 }{ 9 } \) = \(\frac { 38 }{ 9 } \)
5 \(\frac { 1 }{ 4 } \) = \(\frac { 21 }{ 4 } \)
Now, multiply\(\frac { 38 }{ 9 } \) by \(\frac { 21 }{ 4 } \).
Firstly, multiply the numerators with numerators.
38 × 21 = 798.
Next, multiply the denominators with denominators.
9 × 4 = 36.
Finally, write the fraction in the simplest form.
\(\frac { 798 }{ 36 } \)

The final answer is \(\frac { 798 }{ 36 } \).

Multiplying Fractions Examples

I. Find the product
(i) \(\frac { 5 }{ 4 } \) × 1
(ii) \(\frac { 3 }{ 5 } \) × 6
(iii) \(\frac { 10 }{ 15 } \) × 7
(iv) \(\frac { 2 }{ 3 } \) × 0
(v) \(\frac { 1 }{ 4 } \) × \(\frac { 2 }{ 7 } \)
(vi) 2\(\frac { 9 }{ 13 } \) × 4
(vii) \(\frac { 1 }{ 6 } \) × \(\frac { 7 }{ 1 } \)
(viii) \(\frac { 1 }{ 4 } \) × \(\frac { 8 }{ 6 } \) × \(\frac { 3 }{ 10 } \)
(ix) \(\frac { 5 }{ 16 } \) × \(\frac { 11 }{ 23 } \)
(x) \(\frac { 1 }{ 2 } \) of 50
(xi) \(\frac { 1 }{ 3 } \) of 90
(xii) \(\frac { 5 }{ 6 } \) of \(\frac { 9 }{ 12 } \)

(i) \(\frac { 5 }{ 4 } \) × 1
Solution:
Given that \(\frac { 5 }{ 4 } \) × 1
Now, multiply \(\frac { 5 }{ 4 } \) by 1.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
5 × 1 = 5.
So, the fraction is \(\frac { 5 }{ 4 } \)

The final answer is \(\frac { 5 }{ 4 } \).

(ii) \(\frac { 3 }{ 5 } \) × 6
Solution:
Given that \(\frac { 3 }{ 5 } \) × 6
multiply \(\frac { 3 }{ 5 } \) by 6
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
3 × 6 = 18.
So, the fraction is \(\frac { 18 }{ 5 } \)
Simplify the fraction to get the final answer.
\(\frac { 18 }{ 5 } \).

The final answer is \(\frac { 18 }{ 5 } \).

(iii) \(\frac { 10 }{ 15 } \) × 7
Solution:
Given that \(\frac { 10 }{ 15 } \) × 7
multiply \(\frac { 10 }{ 15 } \) by 7
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
10 × 7 = 70.
So, the fraction is \(\frac { 70 }{ 15 } \)
Simplify the fraction to get the final answer.
\(\frac { 70 }{ 15 } \) = \(\frac { 14 }{ 3 } \).

The final answer is \(\frac { 14 }{ 3 } \).

(iv) \(\frac { 2 }{ 3 } \) × 0
Solution:
Any fraction that multiplies with 0 gives 0.
Therefore, the answer is 0.

(v) \(\frac { 1 }{ 4 } \) × \(\frac { 2 }{ 7 } \)
Solution:
Given that Multiply \(\frac { 1 }{ 4 } \) by \(\frac { 2 }{ 7 } \).
Firstly, multiply the numerators with numerators.
1 × 2 = 2.
Next, multiply the denominators with denominators.
4 × 7 = 28.
Finally, write the fraction in the simplest form.
\(\frac { 2 }{ 28 } \)
Simplify the fraction to get the final answer.
\(\frac { 1 }{ 14 } \).

The final answer is \(\frac { 1 }{ 14 } \).

(vi) 2\(\frac { 9 }{ 13 } \) × 4
Solution:
Given that multiply 2\(\frac { 9 }{ 13 } \) × 4.
Firstly, convert given mixed fraction 2\(\frac { 9 }{ 13 } \) to fraction.
2\(\frac { 9 }{ 13 } \) = \(\frac { 35 }{ 13 } \)
Now, multiply \(\frac { 35 }{ 13 } \) by 4.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
35 × 4 = 140.
So, the fraction is \(\frac { 140 }{ 13 } \)
Simplify the fraction to get the final answer.
\(\frac { 140 }{ 13 } \).

The final answer is \(\frac { 140 }{ 13 } \).

(vii) \(\frac { 1 }{ 6 } \) × \(\frac { 7 }{ 1 } \)
Solution:
Given that Multiply \(\frac { 1 }{ 6 } \) × \(\frac { 7 }{ 1 } \).
Firstly, multiply the numerators with numerators.
1 × 7 = 7.
Next, multiply the denominators with denominators.
6 × 1 = 6.
Finally, write the fraction in the simplest form.
\(\frac { 7 }{ 6 } \)
Simplify the fraction to get the final answer.
\(\frac { 7 }{ 6 } \).

The final answer is \(\frac { 7 }{ 6 } \).

(viii) \(\frac { 1 }{ 4 } \) × \(\frac { 8 }{ 6 } \) × \(\frac { 3 }{ 10 } \)
Solution:
Given that Multiply \(\frac { 1 }{ 4 } \) × \(\frac { 8 }{ 6 } \) × \(\frac { 3 }{ 10 } \).
Firstly, multiply the numerators with numerators.
1 × 8 × 3 = 24.
Next, multiply the denominators with denominators.
4 × 6 × 10 = 240.
Finally, write the fraction in the simplest form.
\(\frac { 24 }{ 240 } \)
Simplify the fraction to get the final answer.
\(\frac { 1 }{ 10 } \).

The final answer is \(\frac { 1 }{ 10 } \).

(ix) \(\frac { 5 }{ 16 } \) × \(\frac { 11 }{ 23 } \)
Solution:
Given that Multiply \(\frac { 5 }{ 16 } \) × \(\frac { 11 }{ 23 } \).
Firstly, multiply the numerators with numerators.
5 × 11 = 55.
Next, multiply the denominators with denominators.
16 × 23 = 368.
Finally, write the fraction in the simplest form.
\(\frac { 55 }{ 368 } \)
Simplify the fraction to get the final answer.
\(\frac { 55 }{ 368 } \).

The final answer is \(\frac { 55 }{ 368 } \).

(x) \(\frac { 1 }{ 2 } \) of 50
Solution:
Given that \(\frac { 1 }{ 2 } \) of 50
multiply \(\frac { 1 }{ 2 } \) by 50
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
1 × 50 = 50.
So, the fraction is \(\frac { 50 }{ 2 } \)
Simplify the fraction to get the final answer.
\(\frac { 50 }{ 2 } \) = 25.

The final answer is 25.

(xi) \(\frac { 1 }{ 3 } \) of 90

Solution:
Given that \(\frac { 1 }{ 3 } \) of 90
multiply \(\frac { 1 }{ 3 } \) by 90
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
1 × 90 = 90.
So, the fraction is \(\frac { 90 }{ 3 } \)
Simplify the fraction to get the final answer.
\(\frac { 90 }{ 3 } \) = 30.

The final answer is 30.

II. Multiply and write the product in the lowest terms.

(i) \(\frac { 1 }{ 2 } \) × 60
(ii) \(\frac { 1 }{ 3 } \) × 18
(iii) \(\frac { 2 }{ 5 } \) × 25
(iv) \(\frac { 4 }{ 3 } \) × 0
(v) \(\frac { 7 }{ 29 } \) × 1
(vi) 6 × \(\frac { 7 }{ 36 } \)
(vii) \(\frac { 5 }{ 34 } \) × \(\frac { 34 }{ 8 } \)
(viii) \(\frac { 12 }{ 25 } \) × \(\frac { 5 }{ 6 } \)
(ix) \(\frac { 6 }{ 14 } \) × \(\frac { 56 }{ 7 } \)
(x) \(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 5 } \) × \(\frac { 5 }{ 8 } \)
(xi) \(\frac { 6 }{ 3 } \) × \(\frac { 1 }{ 2 } \) × \(\frac { 3 }{ 3 } \)
(xii) 3\(\frac { 8 }{ 5 } \) × \(\frac { 5 }{ 4 } \)

(i) \(\frac { 1 }{ 2 } \) × 60
Solution:
Given that \(\frac { 1 }{ 2 } \) × 60
Now, multiply \(\frac { 1 }{ 2 } \) by 60.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
1 × 60 = 60.
So, the fraction is \(\frac { 60 }{ 2 } \)
Simplify the fraction to get the final answer.
\(\frac { 60 }{ 2 } \) = 30.

The final answer is 30.

(ii) \(\frac { 1 }{ 3 } \) × 18
Solution:
Given that \(\frac { 1 }{ 3 } \) × 18
Now, multiply \(\frac { 1 }{ 3 } \) by 18.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
1 × 18 = 18.
So, the fraction is \(\frac { 18 }{ 3 } \)
Simplify the fraction to get the final answer.
\(\frac { 18 }{ 3 } \) = 6.

The final answer is 6.

(iii) \(\frac { 2 }{ 5 } \) × 25
Solution:
Given that \(\frac { 2 }{ 5 } \) × 25
Now, multiply \(\frac { 2 }{ 5 } \) by 25.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
2 × 25 = 50.
So, the fraction is \(\frac { 50 }{ 5 } \)
Simplify the fraction to get the final answer.
\(\frac { 50 }{ 5 } \) = 10.

The final answer is 10.

(iv) \(\frac { 4 }{ 3 } \) × 0
Solution:
Any fraction that multiplies with 0 gives 0.
Therefore, the answer is 0.

(v) \(\frac { 7 }{ 29 } \) × 1
Solution:
Any fraction that multiplies with 1 gives the same output.
Therefore, the answer is \(\frac { 7 }{ 29 } \).

(vi) 6 × \(\frac { 7 }{ 36 } \)
Solution:
Given that 6 × \(\frac { 7 }{ 36 } \)
Now, multiply 6 by \(\frac { 7 }{ 36 } \).
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
6 × 7 = 42.
So, the fraction is \(\frac { 42 }{ 36 } \)
Simplify the fraction to get the final answer.
\(\frac { 42 }{ 36 } \) = \(\frac { 21 }{ 18 } \).

The final answer is \(\frac { 21 }{ 18 } \).

(vii) \(\frac { 5 }{ 34 } \) × \(\frac { 34 }{ 8 } \)
Solution:
Given that Multiply \(\frac { 5 }{ 34 } \) × \(\frac { 34 }{ 8 } \).
Firstly, multiply the numerators with numerators.
5 × 34 = 170.
Next, multiply the denominators with denominators.
34 × 8 = 272.
Finally, write the fraction in the simplest form.
\(\frac { 170 }{ 272 } \)
Simplify the fraction to get the final answer.
\(\frac { 5 }{ 8 } \).

The final answer is \(\frac { 5 }{ 8 } \).

(viii) \(\frac { 12 }{ 25 } \) × \(\frac { 5 }{ 6 } \)
Solution:
Given that Multiply \(\frac { 12 }{ 25 } \) × \(\frac { 5 }{ 6 } \).
Firstly, multiply the numerators with numerators.
12 × 5 = 60.
Next, multiply the denominators with denominators.
25 × 6 = 150.
Finally, write the fraction in the simplest form.
\(\frac { 60 }{ 150 } \)
Simplify the fraction to get the final answer.
\(\frac { 60 }{ 150 } \) = \(\frac { 2 }{ 5 } \).

The final answer is \(\frac { 2 }{ 5 } \).

(ix) \(\frac { 6 }{ 14 } \) × \(\frac { 56 }{ 7 } \)
Solution:
Given that Multiply \(\frac { 6 }{ 14 } \) × \(\frac { 56 }{ 7 } \).
Firstly, multiply the numerators with numerators.
6 × 56 = 336.
Next, multiply the denominators with denominators.
14 × 7 = 98.
Finally, write the fraction in the simplest form.
\(\frac { 336 }{ 98 } \)
Simplify the fraction to get the final answer.
\(\frac { 336 }{ 98 } \) = 168.

The final answer is 168.

(x) \(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 5 } \) × \(\frac { 5 }{ 8 } \)
Solution:
Given that Multiply \(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 5 } \) × \(\frac { 5 }{ 8 } \).
Firstly, multiply the numerators with numerators.
1 × 4 × 5 = 20.
Next, multiply the denominators with denominators.
3 × 5 × 8 = 120.
Finally, write the fraction in the simplest form.
\(\frac { 20 }{ 120 } \)
Simplify the fraction to get the final answer.
\(\frac { 20 }{ 120 } \) = \(\frac { 1 }{ 6 } \).

The final answer is \(\frac { 1 }{ 6 } \).

(xi) \(\frac { 6 }{ 3 } \) × \(\frac { 1 }{ 2 } \) × \(\frac { 3 }{ 3 } \)
Solution:
Given that Multiply \(\frac { 6 }{ 3 } \) × \(\frac { 1 }{ 2 } \) × \(\frac { 3 }{ 3 } \).
Firstly, multiply the numerators with numerators.
6 × 1 × 3 = 18.
Next, multiply the denominators with denominators.
3 × 2 × 3 = 18.
Finally, write the fraction in the simplest form.
\(\frac { 18 }{ 18 } \)
Simplify the fraction to get the final answer.
\(\frac { 18 }{ 18 } \) = 1.

The final answer is 1.

(xii) 3\(\frac { 8 }{ 5 } \) × \(\frac { 5 }{ 4 } \)
Solution:
Given that multiply 3\(\frac { 8 }{ 5 } \) × \(\frac { 5 }{ 4 } \).
Firstly, convert given mixed fraction 3\(\frac { 8 }{ 5 } \) to fraction.
3\(\frac { 8 }{ 5 } \) = \(\frac { 23 }{ 5 } \)
Multiply \(\frac { 23 }{ 5 } \) × \(\frac { 5 }{ 4 } \).
Firstly, multiply the numerators with numerators.
23 × 5 = 115.
Next, multiply the denominators with denominators.
5 × 4 = 20.
Finally, write the fraction in the simplest form.
\(\frac { 115 }{ 20 } \)
Simplify the fraction to get the final answer.
\(\frac { 115 }{ 20 } \) = \(\frac { 23 }{ 4 } \).

The final answer is \(\frac { 23 }{ 4 } \).

III. Find the given quantity.

(i) \(\frac { 1 }{ 6 } \) of 48 kg apples
Solution:
Given that \(\frac { 1 }{ 6 } \) of 48 kg apples.
\(\frac { 1 }{ 6 } \) × 48 kg
8 kg

The answer is 8 kg.

(ii) \(\frac { 1 }{ 7 } \) of $280
Solution:
Given that \(\frac { 1 }{ 7 } \) of $280.
\(\frac { 1 }{ 7 } \) × $280
$40

The answer is $40.

(iii) \(\frac { 6 }{ 3 } \) of 54 km
Solution:
Given that \(\frac { 6 }{ 3 } \) of 54 km.
\(\frac { 6 }{ 3 } \) × 54 km
108 km

The answer is 108 km.

(iv) \(\frac { 2 }{ 8 } \) of 40 chairs
Solution:
Given that \(\frac { 2 }{ 8 } \) of 40 chairs.
\(\frac { 2 }{ 8 } \) × 40 chairs
10 chairs

The answer is 10 chairs.

Word problems on Multiplying Fractions

1. 3\(\frac { 5 }{ 8 } \) m of cloth is required to make a shirt. Sam wants to make 32 shirts, what length of cloth does he need?

Solution:
Given that 3\(\frac { 5 }{ 8 } \) m of cloth is required to make a shirt.
Sam wants to make 32 shirts.
3\(\frac { 5 }{ 8 } \) m of 32.
Firstly, convert given mixed fraction 3\(\frac { 5 }{ 8 } \) to fraction.
3\(\frac { 5 }{ 8 } \) = \(\frac { 29 }{ 8 } \)
\(\frac { 29 }{ 8 } \) × 32
Now, multiply \(\frac { 29 }{ 8 } \) by 32.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
29 × 32 = 928.
So, the fraction is \(\frac { 928 }{ 8 } \)
Simplify the fraction to get the final answer.
\(\frac { 928 }{ 8 } \) = 116.

The final answer is 116.

2. \(\frac { 5 }{ 2 } \) cups of milk is required to make a cake of 1 kg. How many cups of milk is required to make a cake of \(\frac { 2 }{ 5 } \) kg?

Solution:
Given that \(\frac { 5 }{ 2 } \) cups of milk is required to make a cake of 1 kg.
To make a cake of \(\frac { 2 }{ 5 } \) kg, multiply \(\frac { 5 }{ 2 } \) with \(\frac { 2 }{ 5 } \).
Firstly, multiply the numerators with numerators.
2 × 5 = 10.
Next, multiply the denominators with denominators.
5 × 2 = 10.
Finally, write the fraction in the simplest form.
\(\frac { 10 }{ 10 } \)
Simplify the fraction to get the final answer.
\(\frac { 10 }{ 10 } \) = 1.

The final answer is 1.

3. Shelly bought \(\frac { 11 }{ 9 } \) liters of juice. If the cost of 1-liter juice is $36, find the total cost of juice?

Solution:
Given that Shelly bought \(\frac { 11 }{ 9 } \) liters of juice.
If the cost of 1-liter juice is $36, multiply \(\frac { 11 }{ 9 } \) with $36.
\(\frac { 11 }{ 9 } \) × $36 = $44.

The final answer is $44.

4. The weight of each bag is 7\(\frac { 1 }{ 9 } \) Kg. What would be the weight of 36 such bags?

Solution:
Given that the weight of each bag is 7\(\frac { 1 }{ 9 } \) Kg.
If the weight of 36 such bags is 7\(\frac { 1 }{ 9 } \) Kg × 36.
Firstly, convert given mixed fraction 7\(\frac { 1 }{ 9 } \) to fraction.
7\(\frac { 1 }{ 9 } \) = \(\frac { 64 }{ 9 } \)
Now, multiply \(\frac { 64 }{ 9 } \) by 36.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
64 × 36 = 2304.
So, the fraction is \(\frac { 2304 }{ 9 } \)
Simplify the fraction to get the final answer.
\(\frac { 2304 }{ 9 } \) = 256.

The final answer is 256.

5. Sam works for 1 \(\frac { 5 }{ 6 } \) hours each day. For how much time will she work in a month if she works for 24 days in a month?

Solution:
Given that Sam works for 1 \(\frac { 5 }{ 6 } \) hours each day.
If she works for 24 days in a month, 1 \(\frac { 5 }{ 6 } \) hours each day × 24
Firstly, convert given mixed fraction 1 \(\frac { 5 }{ 6 } \) to fraction.
1 \(\frac { 5 }{ 6 } \) = \(\frac { 11 }{ 6 } \)
Now, multiply \(\frac { 11 }{ 6 } \) by 24.
Multiply the numerator of the fractional number by the whole number. The denominator remains the same.
11 × 24 = 264.
So, the fraction is \(\frac { 264 }{ 6 } \)
Simplify the fraction to get the final answer.
\(\frac { 264 }{ 6 } \) = 44.

The final answer is 44.

Fraction as a part of a whole definition, rules, facts, and examples here. Get various problems and solutions involved in whole fractions. Know the tips to solve various problems and find the definition here. Follow the step-by-step procedure to solve part-whole fraction problems. Refer to the models that are present in whole fractions. Check the below sections to know the various concepts involved in fraction whole parts.

What is a Fraction of a Whole?

In the interpretation of part-whole, the denominator value shows equal parts number in whole and the numerator represents the number of parts that are included in a particular fraction. The construct of the whole-part is often represented with the model area like dividing the shape into equal parts. The fraction represents the part of the object. Therefore, the fraction is the whole object part. It is a collection of objects or part of the collection. If the number is not a whole number, then it is a fraction number.

The fraction is a whole number part like 1, 2, 3, 4, 5, 6, ………. 200 …….. etc.

Example of fraction numbers: \(\frac { 1 }{ 3 } \), \(\frac { 1 }{ 5 } \), \(\frac { 1 }{ 7 } \), \(\frac { 1 }{ 9 } \), \(\frac { 1 }{ 11 } \), \(\frac { 1 }{ 13 } \), etc.

How do you write a Fraction as a Whole?

The way to represent whole parts is a fraction. If the fraction is represented with a/b, then the numerator “a” represents the equal number of parts where the whole number is divided into parts. The denominator “b” represents the way that represents the whole number. The denominator value cannot be zero, as division by zero is the undefined value.

Important Concepts of Whole Parts

Property of one

The number which is divided by itself and which is not divided by zero is called the property of one.

Example: 3/3 = 1 where 3 ≠ 0

Mixed Numbers

The mixed number is nothing but the whole number (a) along with the fraction value (b/c) where c ≠ 0 and it ca be written as a b/c where c ≠ 0

Example: 4¾ is the mixed fraction

Proper and Improper Fractions

If the proper fraction is ab, then a<b and if the improper fraction is ab, then a≥b

Conversion of an improper fraction into a mixed fraction

1. The numerator has to be divided with denominator
2. Check remainder, quotient, divisor
3. Note down the mixed number as the quotient i.e., remainder/divisor

Conversion of a mixed fraction into an improper fraction

1. The denominator has to be multiplied by the whole number
2. The product we get in Step 1 has to be added to the numerator
3. In the last step, add the final product and put the denominator as it is

Fractions of equivalent property

If the numbers are a, b and c where c ≠ 0, b ≠ 0 then a/b = a-c/b-c

Also, Check:

• Fractions
• Types of Fractions
• Dividing Fractions

Problems on Fraction as a Part of a Whole

Problem 1:

Mary had some stamps. She gave 7 stamps to her younger brothers. Mary then had 14 stamps. How many stamps did Mary have at first?

Solution:

As given in the question,

No of stamps she gave to her younger brother = 7

No of stamps she had = 14

To find the number of stamps at first, we have to add the values

Therefore 7 + 14 = 21

Thus, Mary had 21 stamps at first

Problem 2:

Rahul spent \(\frac { 3 }{ 4 } \) of an hour for 2 days working on his science project. Kite spent \(\frac { 1 }{ 4 } \)of an hour for 6 days working on his science project. Find the one who spent most f his time working on the science project?

Solution:

As given in the question,

Amount of time Rahul spent for an hour = \(\frac { 3 }{ 4 } \)

No of working days = 2

Amount of time Kite spent for an hour = \(\frac { 1 }{ 4 } \)

No of working days = 6

To find the total time they spent on the science project, we apply the law of multiplication here

Therefore, amount of time Rahul spent = \(\frac { 3 }{ 4 } \) * 2

It can be written as \(\frac { 3 }{ 4 } \) + \(\frac { 3 }{ 4 } \) = \(\frac { 6 }{ 4 } \) hours

Hence, Rahul takes \(\frac { 6 }{ 4 } \) hours to complete the science project

The amount of time Kite spent on his project = \(\frac { 1 }{ 4 } \) * 6

It can be written as \(\frac { 1 }{ 4 } \) + \(\frac { 1 }{ 4 } \) +\(\frac { 1 }{ 4 } \) +\(\frac { 1 }{ 4 } \) +\(\frac { 1 }{ 4 } \) +\(\frac { 1 }{ 4 } \) = \(\frac { 6 }{ 4 } \)

Hence, Kite takes \(\frac { 6 }{ 4 } \) hours to complete the science project

From, the above simplification, we came to know that both of them completes the project in some time.

Therefore, they both complete the science project in \(\frac { 6 }{ 4 } \) hours

Problem 3:

Baine worked 3\(\frac { 1 }{ 5 } \) hrs before work and 2\(\frac { 2 }{ 3 } \) hrs after lunch. How many hours did she work altogether? How many hours did she leave in her 8 days of work?

Solution:

As given in the question,

Baine worked before work = 3\(\frac { 1 }{ 5 } \) hrs

Baine worked after lunch = 2\(\frac { 2 }{ 3 } \)

Therefore, the total amount of time he worked = 3\(\frac { 1 }{ 5 } \) hrs + 2\(\frac { 2 }{ 3 } \)

We have to find the common denominator for 3 and 5

To find the no of hours she worked altogether, we have to add both the fraction values

3\(\frac { 1 }{ 5 } \) hrs + 2\(\frac { 2 }{ 3 } \)

The result is 5\(\frac { 13 }{ 15 } \) hours

To find the number of hours left in her day, we have to subtract the fraction values

Now, we just can’t drop 5\(\frac { 13 }{ 15 } \) as it is \(\frac { 0 }{ 15 } \)

The way we can write 8 is 7 plus 1 and \(\frac { 0 }{ 15 } \)

Thus, \(\frac { 0 }{ 15 } \) is 1 times 15 is 15, plus 0 is just 15

So, I have 7 and \(\frac { 15 }{ 15 } \) minus 5 and \(\frac { 13 }{ 15 } \)

Now, I am ready to subtract 15 minus 13 is 2, and 7 minus 5 is 2

She has two hours and \(\frac { 2 }{ 15 } \)

That is what she has left

Hence, she has worked so far is \(\frac { 2 }{ 15 } \)

Therefore, 5\(\frac { 13 }{ 15 } \) hours they work altogether

\(\frac { 2 }{ 15 } \) hours she left in her 8 days of work

Joint Variation definition, rules, methods and formulae are here. Check the joint variation problems and solutions to prepare for the exam. Refer to problems of direct and inverse variations and the relationship between the variables. Know the different type of variations like inverse, direct, combined and joint variation. Go through the below sections to check definition, various properties, example problems, value tables, concepts etc.

Joint Variation – Introduction

Joint Variation refers to the scenario where the value of 1 variable depends on 2 or more and other variables that are held constant. For example, if C varies jointly as A and B, then C = ABX for which constant “X”. The joint variation will be useful to represent interactions of multiple variables at one time.

Most of the situations are complicated than the basic inverse or direct variation model. One or the other variables depends on the multiple other variables. Joint Variation is nothing but the variable depending on 2 or more variables quotient or product. To understand clearly with an example, The amount of busing candidates for each of the school trip varies with the no of candidates attending the distance from the school. The variable c (cost) varies jointly with n (number of students) and d (distance).

Joint Variation problems are very easy once you get the perfection of the lingo. These problems involve simple formulae or relationships which involves one variable which is equal to the “one” term which may be linear (with just an “x” axis), a quadratic equation (like “x²) where more than one variable (like “hr²”), and square root (like “\sqrt{4 – r^2\,}4−r2​”) etc.

Functions of 2 or More Variables

It is very uncommon for the output variable to depend on 2 or more inputs. Most of the familiar formulas describe the several variables functions. For suppose, if the rectangle perimeter depends on the length and width. The cylinder volume depends on its height and radius. The travelled distance depends on the time and speed while travelling. The function notation of the formulas can be written as

P = f(l,w) = 2l + 2w where P is the perimeter and is a function of width and length

V = f(r,h) = Πr²h where V is the volume and is a function of radius and height

d = f(r,t) = rt where d is the distance and is a function of time and rate.

Tables of Values

Just for the single variable functions, we use the tables to describe two-variable functions. The heading of the table shows row and column and it shows the value if two input variables and the complete table shows the values of the output variable.

Graphs

You can easily make graphs in three dimensions for two-variable functions. Instead of representing graphs, we represent functions by holding two or one variable constants.

Also, Read:

• What is Variation
• Practice Test on Ratio and Proportion

How to Solve Joint Variation Problems?

Follow the step by step procedure provided below to solve problems involving Joint Variation and arrive at the solution easily. They are along the lines

Step 1: Write the exact equation. The problems of joint variation can be solved using the equation y =kxz. While dealing with the word problems. you should also consider using variables other than x,y and z. Use the variables which are relevant to the problem being solved. Read the problem carefully and determine the changes in the equation of joint variation such as cubes, squares or square roots.

Step 2: With the help of the information in the problem, you have to find the value of k which is called the constant of proportionality and variation.

Step 3: Rewrite the equation starting with 1 substituting the value of k and found in step 2.

Step 4: Use the equation in step 3 and the information in the problem to answer the question. While solving the word problems, remember including the units in the final answer.

Joint Variation Problems with Solutions

Problem 1:

The area of a triangle varies jointly as the base and the height. Area = 12m² when base = 6m and height = 4m. Find base when Area = 36m² and height = 8m?

Solution:

The area of the triangle is represented with A

The base is represented with b

Height is represented with h

As given in the question,

A = 12m² when B = 6m and H = 4m

We know the equation,

A = kbh where k is the constant value

12 = k(6)(4)

12 = k(24)

Divide by 24 on both sides, we get

12/24 = k(24)/24

1/2 = k

The value of k = 1/2

As the equation is

A = kbh

A = 1/2bh

To find the base of the triangle of A = 36m² and H = 8m

A = 1/2bh

36 = 1/2(b)(8)

36 = 4b

Dividing both sides by 4, we get

36/4 = 4b/4

9 = b

The value of base = 9m

Hence, the base of the triangle when A = 36m² and H = 8m is 9m

Problem 2:

Wind resistance varies jointly as an object’s surface velocity and area. If the object travels at 80 miles per hour and has a surface area of 30 square feet which experiences 540 newtons wind resistance. How much fast will the car move with 40 square feet of the surface area in order to experience a wind resistance of 495 newtons?

Solution:

Let w be the wind resistance

Let s be the object’s surface area

Let v be the object velocity

The object’s surface area = 80 newtons

The wind resistance = 540 newtons

The object velocity = 30

We know the equation,

w = ksv where k is the constant

(540) = k (80) (30)

540 = k (2400)

540/2400 = k

9/40 = k

The value of k is 9/40

To find the velocity of the car with s = 40, w = 495 newtons and k = 9/40

Substitute the values in the equation

w = ksv

495 = (9/40) (40) v

495 = 9v

495/9 = v

v = 55 mph

The velocity of a car is 55mph for which the object’s surface area is 40 and wind resistance is 495 newtons

Hence, the final solution is 55mph

Problem 3:

For the given interest, SI (simple interest) varies jointly as principal and time. If 2,500 Rs left in an account for 5 years, then the interest of 625 Rs. How much interest would be earned, if you deposit 7,000 Rs for 9 years?

Solution:

Let i be the interest

Let p be the principal

Let t be the time

The interest is 625 Rs

The principal is 2500

The time is 5 hours

We know the equation,

i = kpt where k is the constant

Substituting the values in the equation,

(625) = k(2500)(5)

625 = k(12,500)

Dividing 12,500 on both the sides

625/12,500 = k (12,500)/12,500

1/20 = k

The value of k = 1/20

To find the interest where the deposit is 7000Rs for 9 years, use the equation

i = kpt

i = (1/20) (7000) (9)

i = (350) (9)

i = 3,150

Therefore, the interest is 3,150 Rs, if you deposit 7,000 Rs for 9 years

Thus, the final solution is Rs. 3,150

Problem 4:

The volume of a pyramid varies jointly as its height and the area of the base. A pyramid with a height of 21 feet and a base with an area of 24 square feet has a volume of 168 cubic feet. Find the volume of a pyramid with a height of 18 feet and a base with an area of 42 square feet?

Solution:

Let v be the volume of a pyramid

Let h be the height of a pyramid

Let a be the area of a pyramid

The volume v = 168 cubic feet

The height h = 21 feet

The area a = 24 square feet

We know the equation,

V = Kha where K is the constant,

Substitute the values in the equation

168 = k(21)(24)

168 = k(504)

Divide 504 on both sides

168/504 = k(504)/504

1/3 = k

The value of k = 1/3

To find the volume of a pyramid with a height of 18 feet and a base with an area of 42 square feet

Therefore,

h = 18 feet

a = 42 square feet

V = kha

V = (1/3) (18) (42)

V = (6) (42)

V = 252 ft³

The volume of the pyramid = 252 ft³ which has a height of 18 feet and a base with an area of 42 square feet

Therefore, the final solution is 252 ft³

Problem 5:

The amount of oil used by a ship travelling at a uniform speed varies jointly with the distance and the square of the speed. If the ship uses 200 barrels of oil in travelling 200 miles at 36 miles per hour, determine how many barrels of oil are used when the ship travels 360 miles at 18 miles per hour?

Solution:

As given in the question,

No of barrels of oil = 200

The distance at which the oil is travelling = 200 miles

The distance at which the ship is travelling = 36 miles per hour

We know the equation,

A = kds² where k is constant

200 = k.200.(36)²

Dividing both sides by 200

200/200 = k.200.(36)²/200

1 = k.(36)²

k = 1/1296

The value of k is 1/1296

To find the no of barrels when the ship travels 360 miles at 18 miles per hour

Substitute the values in the equation

A = kds²

A = 1/1296 * 360 * 18²

A = 90

Therefore, 90 barrels of oil is used when the ship travels 360 miles at 18 miles per hour

Thus, the final solution is 90 barrels

Round off is a type of estimation. Estimation is used in subjects like mathematics and physics. Round off means making a number simpler by keeping its value intact closer to the next number. Round off to nearest 10 is nothing but making the unit digits of the number to zero and getting the estimated nearest 10 for that number. Check the rules, detailed steps, and solved examples on rounding the numbers to the nearest 10.

Round off to Nearest 10 – Definition

Round off is a process of making a number simpler to read and remember. It is done for the whole numbers, decimals for various places of tens, thousands, hundreds, etc. Round off to Nearest 10 means writing the nearest 10 of the given number. By using the Rounding Numbers to the Nearest 10, you can easily estimate the answer quickly and easily. It is also used to get the average score of people in the class.

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Rules for Rounding Numbers to Nearest 10

We have two different rules for rounding numbers to the nearest 10. They are explained in the below modules

• Rule 1: When rounding the numbers to the nearest 10, if the digit in the unit’s place is less than 5 or between 0 and 4, then the unit’s place of the number is replaced by 0.
• Rule 2: If the digit in the unit’s place is greater than or equal to 5 or between 5 and 9, then the unit’s place is replaced by 0, and the tens place of the number is increased by 1.

How to Round Numbers to Nearest 10?

Follow the below-listed procedure to learn Round off to Nearest 10. They are along the lines

• Get a whole number for rounding.
• Identify the digit in the unit’s place.
• If the digit is between 0 and 4, then place zero in the unit’s place of the number.
• If the digit is or 6 or 7 or 8 or 9, then place zero in the unit place and add 1 to the tens place of the number.
• Now, write the new number as a rounded or estimated number.

Rounding to the Nearest Ten Examples

Example 1:

Round the following numbers to the nearest 10.

(i) 63

(ii) 578

(iii) 1052

Solution:

(i) The given number is 63

We see the digit in the unit place is 3 means that is less than 5. So, we round to the nearest multiple of a ten which is less than the number. So place zero in the unit place.

Therefore, rounding off 63 to the nearest 10 is 60.

(ii) The given number is 578

We see the digit in the unit place is 8 means that is more than 5. So, we round to the nearest multiple of a ten which is less than the number. So place zero in the unit place and increase digit in tens place by 1.

Therefore, rounding off 578 to the nearest 10 is 580.

(iii) The given number is 1052

We see the digit in the unit place is 2 means that is less than 5. So, we round to the nearest multiple of a ten which is less than the number. So place zero in the unit place.

Therefore, rounding off 1052 to the nearest 10 is 1050.

Example 2:

Round off the below-mentioned numbers to the nearest 10.

(i) 167

(ii) 55

(iii) 109

Solution:

(i) The given number is 167

We see the digit in the unit place is 7 means that is greater than 5. So, we round to the nearest multiple of a ten which is less than the number. So place zero in the unit place and increase digit in tens place by 1.

Therefore, the obtained number is 170.

(ii) The given number is 55

We see the digit in the unit place is 5 means that is equal to 5. So, we round to the nearest multiple of a ten which is less than the number. So place zero in the unit place and increase digit in tens place by 1.

Therefore, rounding off 55 to the nearest 10 is 60.

(iii) The given number is 109

We see the digit in the unit place is 9 means that is more than 5. So, we round to the nearest multiple of a ten which is less than the number. So place zero in the unit place and increase digit in tens place by 1.

Therefore, round off 109 to the nearest 10 is 110.

Example 3:

Round off to Nearest 10.

(i) 221

(ii) 854

(iii) 57

Solution:

(i) The given number is 221

We see the digit in the unit place is 1 means that is lesser than 5. So, we round to the nearest multiple of a ten which is less than the number. So place zero in the unit place.

Therefore, round off 221 to the nearest 10 is 220.

(ii) The given number is 854

We see the digit in the unit place is 4 means that is less than 5. So, we round to the nearest multiple of a ten which is less than the number. So place zero in the unit place.

Therefore, round off 854 to the nearest 10 is 850.

(iii) The given number is 57

We see the digit in the unit place is 7 means that is more than 5. So, we round to the nearest multiple of a ten which is less than the number. So place zero in the unit place and increase digit in tens place by 1.

Therefore, round off 57 to the nearest 10 is 60.

Example 4:

Round the following numbers to the nearest tens.

(i) 5526

(ii) 328

Solution:

(i) The given number is 5526

We see the digit in the unit place is 6 means that is more than 5. So, we round to the nearest multiple of a ten which is less than the number. So place zero in the unit place and increase digit in tens place by 1.

Therefore, round off 5526 to the nearest 10 is 5530.

(ii) The given number is 328

We see the digit in the unit place is 8 means that is more than 5. So, we round to the nearest multiple of a ten which is less than the number. So place zero in the unit place and increase digit in tens place by 1.

Therefore, round off 328 to the nearest 10 is 330.