Common Core Math

Do you find it difficult to understand the Integers Multiplication? Here is the best solution for you all. We are providing Examples on Multiplication of Integers here. Integers Multiplication is an important concept which helps you to score more marks in the exam. Questions on Multiplication of Integers with Answers are available so that you can practice them regularly. Learn How to Multiply Integers by referring to the further modules.

Worked out Multiplication of Integers Problems

Before going to solve the Integers Multiplication problems, know all the definitions, rules, formulae etc. In the upcoming sections, you will find all the details and also problem-solving techniques and tips. To be more precise, you can only solve the problems if you know all the details regarding integers.  Know various properties of integers beforehand and how they work while multiplying the integers.

Key Points to Remember on Properties of Integers Multiplication

• The Closure Integer Property of multiplication defines that the product value of two or three integer numbers will be an integer number.
• The commutative Integer Multiplication property defines that swapping two or three integers will not differ the value of the final result.
• The associative Integer Multiplication property defines that the grouping of integer values together will not affect the final result.
• The distributive Integer property of multiplication defines that the distribution concept of 1 operation value on other mathematical integer values within the given braces.
• Multiplication by zero defines the product value of any negative or positive integer number by zero
• Multiplicative Integer defines the final result as 1 when any integer number is multiplied with 1.

Integer Multiplication Rules on Problems

Question 1:

The temperature in an area drops by 4 degrees for 4 hours. How much is the total drop in the temperature?

Solution:

As given in the question, the temperature drops by 4 degrees. Therefore, the temperature is a negative factor.

Also. given that it decreases for 4 hours.

The total drop in temperature is (-4) * (4) = -16

Therefore, the total drop in temperature is 16 degrees C

Thus, the final result is -16 degree C

Question 2:

Jason borrowed $2 a day to buy a launch. She now owes $60. How many days did Jason borrow $2?

Solution:

As given in the question, Jason borrowed to buy a launch = $2

After buying she owes $60

No of days Jason borrowed money = 60/2 = 30

Therefore, the total days = 30 days

Thus, the final result is 30 days.

Question 3: A football team 12 yards on each of the four consecutive plays. What was the team’s total change in position for four plays?

Solution: 

As given in the question, A football team lost yards = -12 yards

No of plays = 4

Team total change in position for 4 plays = (-12) * (4) = 48

Therefore, the total change = -48 yards

Thus, the final answer is -48 yards.

Question 4: On a certain day, the temperature changed at a rate of -2 degrees F per hour. If this happened for continuous 5 days. For how many days there was a change in temperature?

Solution:

As per the given question, The temperature changed at the rate = -2 degree F

The change happened for days = 5 days

No of days there was a change = (-2)*5 = -10

Therefore, there was a change for days = 10 days

Thus, the final solution is 10 days.

Question 5: Flora made 6 deposits $ 7 each from her bank account. What was the overall change in her account?

Solution:

As per the given question,

Flora made no of deposits = 6

Amount of deposited money = $7

The overall change in the account = 6 * ($7) = 42

Therefore, the change in money = 42

Thus, the final solution is $42

Questions on Multiplication of Integers

Question 6: A winter coat was priced at $200. Each month for three months, the price was reduced by $15. How much was the coat reduced in price?

Solution:

As per the given question,

The price of the winter coat is reduced by $15, Therefore it is negative = -$15

No of times it is reduced = 3

The absolute values of |3| and |-15| are 3 and 15

The coat reduced in price = 3*15 = 45

Therefore, the total change in price = $45

Thus the final answer = -$45

Question 7:

Netflix charges $9 per month for their streaming plan to watch movies. If they automatically bill a customer for 6 months, How much will be deducted from the customer’s bank account?

Solution:

As per the given question,

Netflix charges $9 per month. Therefore, it is negative.

Given, the bill will be deducted automatically for months = 6

The absolute values of |6| and |9| are 6 and 9.

The amount of money deducted from customers bank account = 6*9 = 54

Therefore, the total amount after determining the signs = -$54

Hence, the final solution is -$54

Question 8: Lisa decided her hair was too long. In June and again in July. she cut 3 inches off. Then, in August, September, and October she cut off 2 inches. Write an equation to represent the change in the length of her hair?

Solution:

As given in the question,

In 2 months, she cut 3 inches off her hair. Cutting her hair made the length shorter, therefore it is negative.

In 3 months, she cut 2 inches off her hair. This is also negative.

For the month of June and July, the length of the hair she cut = 2 * (-3) = -6

For the months August, September, and October, the length of the hair she cut = 3 * (-2) = -6

Therefore, the total length = (-6) + (-6) = -12 inches

Thus. the complete length she cut = 12 inches

Hence, the final solution is -12 inches

Question 9: The depth of the water in a pool decreases an average of two inches each week during the summer. What will be the change in the depth of water of four weeks?

Solution:

As given in the question,

The depth of the water in a pool decreases each week = 2 inches

As the water level decreases, it will be negative.

The decrease in water for weeks = 4

The change in depth of water = (-2)*4 = -8

Therefore, the water level decreases by 8 inches.

Thus, the final solution is -8 inches

Question 10:

For every 1000 feet, you gain in elevation, the temperature drops by 3 degrees. If you increase your elevation by 5000 feet, How would the temperature change?

Solution:

As per the given question,

The temperature drops by 3 degrees. Therefore, it will be negative.

Also given for every 1000 feet it is 3 degrees. Thus for every 5000 feet, it is 5 degrees.

The temperature change = (-3)*5 = -15

Thus, for every 5000 feet, the temperature changes by -15 degrees.

Hence, the final solution is -15 degrees.

Lines and Angles concepts are explained by experts in order to help students to get excellent marks in their examination. The line is a combination of infinite points that has both directions infinitely. There are different lines available such as intersecting lines, perpendicular lines, transversal lines, etc. An angle is nothing but a combination of two rays with a common endpoint.

Based on the operations performed on a line, then the lines are classified as Parallel Lines, Perpendicular Lines, Transversal, etc. In the same way, the angles are classified as Complementary Angles, Supplementary Angles, Adjacent Angles, Vertically Opposite Angles based on their operation.

Quick Links of Lines and Angles

Working on Lines and Angles problems will help students to get an idea about how to solve the problems. All the concept links are given below. Just check out the concepts and open the topic that is difficult for you. After that solve all the problems available on that topic to get a grip on that entire topic.

• Basic Geometrical Concepts
• Angles
• Classification of Angles
• Related Angles
• Some Geometric Terms and Results
• Complementary Angles
• Supplementary Angles
• Complementary and Supplementary Angles
• Adjacent Angles
• Linear Pair of Angles
• Vertically Opposite Angles
• Parallel Lines
• Transversal Line
• Parallel and Transversal Lines

Line Segment

The line segment is a line that has two endpoints with a fixed length.

Ray

A ray is a line that has only one endpoint and extends infinitely in another direction. From the below figure, point A is the and point.

Perpendicular Lines

Two lines are said to be Perpendicular Lines when they meet at a single point and forms a right angle. From the below figure, PQ and RS are two lines and they intersect at a point x. Also, the two lines PQ and RS form a right angle 90º.

Parallel Lines

Two lines are said to be Parallel Lines when they do not intersect each other and they have the same distance throughout the plane. From the below figure, A and B are two parallel lines.

Transversal Line

A transversal Line is a line that intersects two lines at distinct points. From the below figure, C is the Transversal Line, that intersects the lines A and B at two points D and E.

Properties of Lines

• The points those are not lie on the same line are called non-collinear points.
• The points that lie on the same line are called Collinear points

Acute Angle

The angle that is less than the right angle is called Acute Angle.

Obtuse Angle

The angle that is more than the right angle is called Acute Angle.

Right Angle

The angle with 90 degrees is called as Right Angle.

Straight Angle

The angle that is equal to 180 degrees is called Straight Angle.

Complementary Angles

If the sum of the two angles is 90 degrees, then they are called Complementary Angles.

Supplementary Angles

If the sum of the angles is 180 degrees, then those angles are called Supplementary Angles.

Properties of Angles

• An angle is formed at a point where two rays intersecting.
• If the angle is more than 180 degrees and less than 360 degrees, then it is called a reflex angle.
• If the two angles are adjacent angles and add up to 180 degrees then they form linear pair.

Quick Access to Lines and Angles

We have given solutions for all questions with step by step explanation in understandable language. Therefore, students can get great knowledge on the entire concepts of Lines and Angles. All the questions and answers that are present in the updated syllabus are included on our website with individual concept explanation. So, students who want to learn a single topic can go through the required link available below and learn it from the concerned link.

Are you searching for Lines and Angles Material? You have free access to Lines and Angles Material on our website. You can read and practice all problems either online or offline. So, immediately bookmark our website and practice all the math problems at your comfortable times. Do hard work and get complete knowledge on the Lines and Angles concept. Clear your doubts by contacting us through the comment section.

When two triangles lie on the same base and between the same parallels, then those triangles measure equal area. Check the proof for Triangles on the same base and between the same parallel are equal in Area. Get the relationship between the areas of triangles having the same base in the following sections. For the reference of students, we have also included the solved examples on Triangle on the Same Base and between Same Parallels.

Proof for Area of Triangles are Equal Having the Same Base

Get a detailed explanation for how the areas of triangles are equal when their bases are the same and between the same parallels.

In the following figure, ∆ABD and ∆DEF having equal base length and between the same parallels BF and AD.

In this case, Area of △ ABC = Area of △ DEF

To prove that Triangle on the Same Base and between Same Parallels have the same area, follow these instructions.

Let us take △ ABC and △ ABD on the same base AB and between the same parallels AB and CD.

Draw a parallelogram ABPQ with AB base and lying between the same parallels AB and Q.

The relationship between the area of the triangle and parallelogram says that when they lie on the same base and between the same parallels, then the triangle is half of the parallelogram area.

So, Area of ∆ ABC = 1/2 x Area of parallelogram ABPQ

In the same way, Area of ∆ ABD = 1/2 x Area of parallelogram ABPQ

Therefore, Area of ∆ ABC = Area of ∆ ABD.

Hence proved.

Questions on Triangles on the Same Base and between Same Parallels

Example 1.

AD is the median of ∆ABC and ∆ADC. E is any point on AD. Show that area of ∆ABE = area of ∆ACE.

Solution:

Here, AD is the median for △ ABC.

So, BD = CD

In the above figure, △ ABD and △ ADC are having equal bases BD and CD and they are between the same parallels BC and U.

So, Area of △ ADC = Area of △ ABD —– (i)

Since E lies on AD

Therefore, Ed is the median for △ BCE

Now, BDE and CDE are having the equal base length BD = CD and between the same paralleks BC and m, then

Area of △ CDE = Area of △ BDE —— (ii)

Subtracting equation (i) and equation (ii), we get

Area of △ ADC – Area of △ CDE = Area of ∆ABD – Area of ∆BED

Area of △ ABE = Area of △ ACE

Hence, shown.

Example 2.

Prove that the medians of the triangle divide it into two triangles having equal area?

Solution:

AD is the median of the ∆ BCA and AE is the altitude of ∆ BCA and also ∆ ADC.

And AE is perpendicular to BC

Here, AD is the median of ∆ ABC

Therefore, BD = CD

On multiplying both sides of the equation by AE, we get

BD x AE = CD x AE

Again multiply both sides by 1/2

1/2 x BD x AE = 1/2 x CD x AE

Area of ∆ABD = Area of ∆ADC

Hence shown.

If a triangle and parallelogram having the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram. Get the example problems and solutions on Triangle and Parallelogram on the Same Base and between Same Parallels. Also, find the relationship between them in the following sections explained step by step.

Proof for Area of Triangle Equal to Half of the Parallelogram having the Same Base

In the following figure, parallelogram ABCD and ∆ABD are on the same base AB and between the same parallels AF and DC. Get the area of the triangle ABD and area of the parallelogram ABCD and check their areas.

Area of the parallelogram ABCD = base x height

= AB x CF

Here, CF and DE area parallel lines CF = DE.

Area of the parallelogram = AB x DE = AB x CF.

Area of △ ABD = 1/2 x base x height

= 1/2 x AB x DE

= 1/2 x Area of the parallelogram ABCD

Therefore, Area of the parallelogram ABCD = 1/2 x Area of △ ABD

Hence proved.

Here are various case studies for the area of the parallelogram is half of the area of the triangle when they lie on the same base.

1. When a triangle and a parallelogram lies on the same base and have the same altitude, then the area of the triangle is equal to half of the parallelogram.
2. If they lie between the same parallels and have the same altitude, then the area of the triangle is equal to half of the area of the parallelogram.
3. When the triangle and rectangle have the same base and between the same parallels, then the area of the triangle is half of the area of the rectangle.

Example Questions for Triangle and Parallelogram on Same Base and between Same Parallels

Example 1:

∆ ABD and parallelogram ABCD are on the same base AB. If the base and altitude of the parallelogram are 10 cm and 6 cm, find the area of the triangle.

Solution:

Given that,

Base of the parallelogram = 10 cm

Altitude of the parallelogram = 6 cm

Area of the parallelogram = base x altitude

= 10 x 6 = 60 cm²

We know that when a triangle and parallelogram having the same base and between the same parallels then the area of the triangle is equal to half of the area of the parallelogram.

So, area of △ ABD = 1/2x Area of the parallelogram ABCD

= 1/2 x 60 = 30 cm²

Therefore, triangle area is 30 cm².

Example 2:

∆ ABD and parallelogram ABCD are on the same base AB. If the base and height of the triangle is 16 cm and 12 cm, find the area of the parallelogram.

Solution:

Given that,

The base length of the triangle = 16 cm

Height of the triangle = 12 cm

Area of ∆ ABD = 1/2 x base x height

= 1/2 x 16 x 12

= 1/2 x 192 = 96 cm²

We know that when a triangle and parallelogram having the same base and between the same parallels then the area of the parallelogram is equal to double the area of the parallelogram.

Area of the parallelogram ABCD = 2 x Area of ∆ ABD

= 2 x 96 = 192 cm²

Therefore, parallelogram area is 192 cm².

Example 3:

In the adjacent figure, BCDE is a rectangle and ABC is a triangle show that area of the triangle ABC is equal to half of the area of the rectangle BCDE.

Solution:

Given that,

Triangle ABC and rectangle BCDE lies on the same base BC and between the same parallels BC and ED>

Area of triangle ABC = 1/2 x base x height

= 1/2 x BC x AP

AP and Cd are parallel lines and having the same length. So, AP = CD.

△ ABC area = 1/2 x BC x AP = 1/2 x BC x CD

Area of the rectangle = length x breadth

= BC x CD

= 1/2 x Area of △ ABC.

Therefore, the Area of the rectangle BCDE = 1/2 x Area of △ ABC

Hence, shown.

When one parallelogram and a rectangle lies on the same base and between the same parallels then they measure equal area. Here we are providing the example questions, answers for Parallelograms, and Rectangles on Same Base and between the Same Parallels. We are also offering proof that shows parallelogram and rectangle on same between same parallels have the same area.

Proof for Rectangles and Parallelograms on the Same Base have Equal Area

Below-mentioned is the steps for the parallelogram and rectangles having the same base and between the same parallels having equal area.

In the following figure, ABCD is a parallelogram and ABEF is the rectangle that lies on the same base AB and the same parallels AB and DF.

Area of parallelogram ABCD = base x altitude

AF is the altitude of the parallelogram.

So, Parallelogram area = AB x AF.

Area of the rectangle ABEF = base x height

= AB x AF

Therefore, the area of the parallelogram ABCD = Area of the rectangle ABEF.

Solved Examples for Parallelograms and Rectangles on Same Base and between Same Parallels

Example 1:

Parallelogram ABCD and rectangle ABFE have the same AB and the length and breadth of the rectangle are 15 cm and 9 cm. Find the area of the parallelogram.

Solution:

Given that,

Length of the rectangle l = 15 cm

The breadth of the rectangle b = 9 cm

Parallelogram ABCD and rectangle ABFE have the same AB.

So, area of the parallelogram = Area of the rectangle.

Area of the rectangle = length x breadth

= 15 x 9 = 135 cm²

Therefore, area of the parallelogram is 135 cm².

Example 2:

In the adjacent figure, ABCD is a parallelogram and EFCD is a rectangle. Also, AL ⊥ DC. Prove that

(a) Area (ABCD) = Area (EFCD)

(b) Area (ABCD) = DC × AL

Solution:

(a)

Rectangle is CDEF.

Parallelogram is ABCD.

In the given figure, AL and DE are parallel and have the same length. So, DE = AL.

Area of ABCD = CD x AL

= CD x DE square units.

Area of EFCD = CD x DE

So, Area of EFCD = Area of ABCD

Hence proved.

(b)

As ABCD is a parallelogram.

Area of ABCD = base x altitude

The base is CD and the altitude is AL.

Therefore, Area of ABCD = CD x AL

Hence proved.

Example 3:

Parallelogram ABCD and rectangle ABFE have the same AB and the base and height of the parallelogram are 28 cm and 16 cm. Find the area of the rectangle.

Solution:

Given that,

Base of the parallelogram = 28 cm

Height of the parallelogram = 16 cm

Area of the parallelogram = base x height

= 28 x 16 = 448 cm²

We know, that when Parallelograms and Rectangles on Same Base and between Same Parallels, then the area of the parallelogram = area of the rectangle.

So, the area of the rectangle = 448 cm².

Students can check how parallelograms having the same base and between the same parallels are equal in area and proof for that. When two or more parallelogram lies on the same base and between same parallels then they have the same area. Also, find the solved example questions here explaining the concept and learn the concept behind them.

Proof for Theorem: Parallelogram on the Same Base & Between Same Parallels are Equal in Area

Let us draw a parallelogram ABCD on a cardboard sheet or thick sheet of paper.

Now, draw a line segment DE as shown in the figure.

Now, cut the triangle ADE congruent to triangle ADE in a seperate sheet with the help of a tracing paper and place the triangle ADE in such a way that AD coincides BC. The entire process is as shown below.

Note that there are two parallelograms ABCD and EE’CD on the same base CD and between the same parallel AE’ and CD.

∆ADE ≅ ∆ A’ D’ E’

Therefore, Area of ADE = Area of A’D’E’

Also, Area of ABCD = Area of ADE + Area of EBCE

= Area of A’D’E’ + Area of EBCD

= Area of EE’CD

So, the two parallelograms are in an equal area.

Parallelograms on Same Base and Between Same Parallels Examples

Example 1:

A farmer was having a field in the form of a parallelogram ABCD. She took any point P on AB and joined it to points C and D. In how many parts the fields are divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Solution:

If DP and CP are joined parallelogram ABCD is divided into three triangles.

They are △ APD, △ CDP, △ BCP

△ CDP and parallelogram ABCD are on the same base i.e CD and between parallel sides CD ∥ AC

∴ Area of △ CDP = 1/2 x area of parallelogram ABCD —- (i)

∴ Area of △ APD + Area of △ BCP = 1/2 x area of parallelogram ABCD —- (ii)

From equation (i) and equation (ii)

Area of △ CDP = Area of △ APD + Area of △ BCP

∴ The farmer can use the part of the field to sow wheat, i.e. Area of △ APD + Area of △ BCP, and in the same area he can use the Area of △ CDP to sow pulses.

Example 2:

Parallelograms ABCD and ABEF are situated on the opposite sides of AB in such a way that D, A, F are not collinear. Prove that DCEF is a parallelogram, and parallelogram ABCD + parallelogram ABEF = parallelogram DCEF.

Solution:

AB and DC are two opposite sides of parallelogram ABCD,

Therefore, AB ∥ DC and AB = DC

Again, AB and EF are two opposite sides of parallelogram ABEF

Therefore, AB ∥ EF and AB ∥ EF

Therefore, DC ∥ EF and DC = EF

Therefore, DCEF is a parallelogram.

Therefore, ∆ADF and ∆BCE, we get

AD = BC (opposite sides of parallelogram ABCD)

AF = BE (opposite sides of parallelogram ABEF)

And DF = CE (opposite sides of parallelogram CDEF)

Therefore, ∆ADF ≅ ∆BCE (side – side – side)

Therefore, ∆ADF = ∆BCE

Therefore, polygon AFECD – ∆BCE = polygon AFCED – ∆ADF

Parallelogram ABCD + Parallelogram ABEF = Parallelogram DCEF

Two geometric shapes are said to be on the same base and between the same parallels, if they have a common side called the base and vertices opposite to the common base on the line parallel to the base. In the below sections, you can learn about the How to Identify figure on the same base and between the same parallels. We know the measure of the plane region enclosed by a closed figure called the area.

The area is measured in cm², m², and other square units. We also know how to calculate the area of different figures using various formulas. Here you will use the formulas to study the relationship between the area of figures when they lie on the same base and between the same parallels.

Some Geometric Figures on the Same Base and between Same Parallels

Below-mentioned is the geometric figures for which you can change the same base and between the same parallels.

The common base for triangles ABC, BCD is BC. And both triangles lie on the same base.

Parallelograms ABCD and triangle CDE lie on the same base CD.

Parallelograms ABCD and EFCD are on the same base DC.

Trapezium ABCD and parallelogram EFCD have a common side DC. We say that trapezium ABCD and parallelogram EFCD are on the same base DC.

Example Questions on Figures that Lie on the Same Base and Between Same Parallels

Example 1.

ABC is right-angled at B with BC = 10 cm and AC = 15 cm. also ∆ABC and ∆BCD are on the same base BC. Find the area of ∆BCD.

Solution:

In right angled ∆ ABC, AC = 15 cm and BC = 10 cm. using Pythagoras theorem, we get

AC² = AB² + BC²

15² = x² + 10²

x² = 15² – 10²

x² = 225 – 100

x² = 125

x = √125

x = 11.18 cm

Now, since ∆ ABC and ∆BCD are on the same base BC.

Therefore, area of ∆ ABC = Area of ∆BCD

1/2 × base × height = Area of ∆BCD

1/2 × 10 × 11.18 = Area of ∆BCD

Area of ∆BCD = 5 x 11.18 = 55.9 cm²

Therefore, area of ∆BCD is 55.9 cm².

Example 2.

AD is the median of ∆ABC. E is any point on AD. Show that area of ∆ABE = area of ∆ACE

Solution:

The common base for ∆ABE, ∆ACE is AE.

ABC is a triangle with AD as the median.

i.e BD = CD

E is any point on AD.

In ∆ABC,

Given that,

D is the midpoint of BC.

∴ AD is median

Since median divides the triangle into two triangles of equal area.

So, Area of ∆ABD = Area of ∆ABC —– (i)

In ∆EBC

Given that, D is the midpoint of BC.

∴ ED is median.

Since median divides the triangle into two triangles of equal area.

So, Area of ∆EBD = Area of ∆EDC —— (ii)

Subtracting equation (1) from equation (2)

Area of ∆ABD – Area of ∆EBD = Area of ∆ADC – Area of ∆EDC

Area of ∆ABE = Area of ∆ACE

Hence, proved.

Example 3.

Parallelogram PQRS and PQTU are on the same base PQ and between the same parallels PQ and UR. Area of parallelogram PQRS = 56 cm² and the altitude of the parallelogram PQTU = 7 cm. Find the length of the common side of two parallelograms.

Solution:

Given that,

Area of parallelogram PQRS = 56 cm²

The altitude of the parallelogram PQTU = 7 cm

Area of parallelogram PQRS = base x height

56 = PQ x 7

PQ = 56/7

PQ = 8 cm

Therefore, the length of the common side of the two parallelograms is 8 cm.

Properties of Division of Integers are here. Check all the rules, methods, and explanations of integer division properties. Follow simple tricks and tips to remember all rules of division of integer properties. With the Dividing Integers Properties, you can easily simplify and answer the series of operations. Refer to various properties of the division of integers and apply them while solving the problems. Scroll to the below sections to know more details regarding the division of integer properties.

Properties of Division of Integers – Importance

Knowing the properties of the division of integers is important because most of the candidates often get confused between the properties of multiplication and division. Both the operations have different properties and are mandatory to follow all the rules. The properties help to solve many problems in an easy manner. To recall, integer numbers are positive or negative numbers, including zero. All the properties for multiplication, subtraction, division, and addition are applicable to integers. Integers are denoted by the letter “Z”.

Rules for Division of Integers

Rule 1: The quotient of 2 positive(+) integers is positive(+).

Rule 2: The quotient of a positive(+) integer and a negative(-) integer is negative(-).

Rule 3: The quotient of 2 negative(-) integers is positive(+).

In brief, if the signs of the two integers are the same, then the result will be positive. If the signs of the two integers are different, then the result will be negative.

Integers Division Properties

The division is the inverse operation of multiplication.

Let us take the example of whole numbers,

24/4 which means dividing 24 by 4 is nothing but finding an integer when multiplied with 4 that gives us 24, such integer is 6.

So, 24/6=4 and 24/4=6

Hence, for each whole-number multiplication statement, there are two division statements.

1. Division of negative integer by positive integer

Whenever a negative number is divided by a positive number, the result will always be negative.

Steps:

1. Divide the given number as the whole number first.
2. Then add, minus symbol before the quotient. Thus we get the final result as a negative integer.

Example:

(-10)/(2) = (-5)

(-32)/(8) = (4)

2. Division of positive integer by negative integer

Whenever a positive integer is divided by a negative integer, the result will always be negative.

Steps:

1. Divide the given number as the whole number first.
2. Then add, minus symbol before the quotient. Thus we get the final result as a negative integer.

Example:

81/(–9) = –9

60/(–10) = –6

3. Divide a negative integer by a negative integer

Whenever a negative integer is divided by a negative integer, then the result will always be positive.

Steps:

1. Divide the given number as the whole number first.
2. Then add, plus symbol before the quotient. Thus we get the final result as a positive integer.

Example: 

(–15)/(– 3)=5

(–21)/(– 7)=3

4. Closure Under Division Property

Generally, the closure property is, if there are 2 integers, then the addition or subtraction of those integer results in an integer. But integers division does not follow closure property.

Example:

Let us consider the pair of integers.

(-12)/(-6) = 2 (Result is an integer)

(-5)/(-10) = -1/2 (Result is not an integer)

From the above examples, we conclude that integers are not closed under division.

5. Commutative Property of Division

The commutative property states that swapping or changing the order of integers does not affect the final result. Integers division does not follow commutative property also.

Let us consider the pairs of integers.

(– 14)/(– 7)=2

(– 7)/(– 14)=1/2

(– 14)/(– 7)≠(– 7)/(– 14)

From the above examples, we conclude that integers are not commutative for integers.

6. Division of Integer by Zero

Any integer divided by zero gives no result or meaningless result.

Example:

5 ÷ 0 = not defined

When zero is divided by an integer other than zero it results in zero.

Example:

0 ÷ 6 = 0

7. Division of Integer by 1

When an integer is divided by 1, it gives the result as 1.

Example:

(– 7)/1=(– 7)

The above example shows that when a negative integer is divided by 1, it gives the same negative integer.

Dividing Integers Properties Examples

Question 1:

Verify that a/(b+c)≠(a/b)+(a/c) when a = 8, b = – 2, c = 4.

Solution:

L.H.S=a/(b + c)

= 8/(-2+4)

= 8/2=4

R.H.S=(a/b)+(a/c)

= [8/(-2)]+(8/4)

= (-4)+2

= -2

Therefore, L.H.S≠R.H.S

Hence the above equation is verified.

Question 2:

(– 80)/(4) is not the same as 80/(–4). True/False

Solution:

(– 80)/(4)=-20

80/(–4)=-20

As (– 80)/(4) = 80/(–4), therefore the above statement is false.

Question 3:

Evaluate [(– 8)+4)]/[(–5)+1]

Solution: [(– 8)+4)]/[(–5)+1]

= (-4)/(-4)

= 1

Question 4:

Find the quotient of

(-15625)/(-125)

Solution:

From the question, (-15625)/(-125)

= (-15625/-125)

= (15625/125)

= 125

Question 5:

Find the value of [32+2*17+(-6)]/15

Solution:

[32+2*17+(-6)]/15

= [32+34+(-6)]/15

= (66 – 6) ÷ 15

= 60/15

= 4

How to Divide Whole Numbers?

1. After dividing the first digit of the dividend by divisor, if the divisor is a larger number than the first digit of the dividend, then divide the first 2 digits of the dividend by divisor and so on.
2. Always, write the quotient above the dividend.
3. Multiply the quotient value by the divisor and write the product value under the dividend.
4. Subtract the product value from the dividend and bring down the next digit of that dividend.
5. Repeat solving from Step 1 until there are no digits left in the dividend.
6. Finally, verify the solution by multiplying the quotient times the divisor.

Hope you can now get the complete information on the Properties of the Division of Integers. Get the latest updates on all types of mathematical concepts like Integers, Time and Work, Pipes & Cisterns, Ratios and Proportions, Variations, etc. Follow all the articles to get complete clarity on the Integers topic. Stay tuned to our site to get the complete data or information regarding mathematical concepts.

Wanna start your preparation from basics? Then here is the most basic and important concept. Learn all the Fundamental Operations and their usage in day to day life. Solve all the important problems involved in fundamental algebra operations. Know the definitions, rules, strategies, tips, tricks, and problems. Refer to the upcoming sections to get more information regarding mathematical expressions and operations.

Fundamental Operations – Introduction

Fundamental Operations are the basic concept to solve any type of problem. We perform the operations at one time and generally, it starts from the left towards the right. If the expression contains more than 1 fundamental operation, then you can’t perform them in the order they appear in the question. There are certain rules to follow to perform various operations when more than 1 fundamental value is available. The precedence order has to be followed to solve the Fundamental Operation. The precedence order will be shown in the below sections.

Order of Expression – DMAS Rule

The fundamental operations are expressed in an order. Generally, the order will be in the format of “DMAS” where “D” stands for “Division”, “M” stands for “Multiplication”, “A” stands for “Addition”, “S” stands for “Subtraction”. This order is considered and sequentially performed from left to right.

As there are fundamental operations in arithmetic, there are in a wat that three pairs of arithmetic operations, and for each pair of operations is the reversal of co-operation within the pairs. As addition and subtraction are reverse operations of each other. It is similar for other operations too. In regard to Algebra, everyone must understand that it is a set of rules based on fundamental operations which helps for faster calculations and an easy approach for variable problem-solving.

Basic Fundamental Operations to Simplify Mathematical Expressions

Addition

The addition is defined by the operator (+). It is the fundamental operation of the arithmetic operators. It implies the combination of distinct quantities or sets. It involves counting numbers from one to one by incrementing the values. The result value after adding all the numbers is called the “sum”. The numbers and the initial number are called addends.

The addition of natural numbers involves 2 important properties i.e., associativity and commutativity.

Addition Operation on Positive and Negative Integers

Positive (+) + Positive (+) = Positive (+)
Negative (-) + Positive (+) = Negative (-)
Positive (+) + Negative (-) = Sign of the largest Number
Negative (-) + Positive (+) = Sign of the largest number

Examples:
5+4 = 9
(-5) + (-4) = -9
(-5) + 4 = -1
4 + (-5) = -1
(-4) + 5 = -1
5 + (-4) = 1

Example Problem:

An empty water tank of 8 feet high is available. A monkey is sitting at the bottom of that tank. The monkey tried jumping to the top of the tank. While jumping it jumps 3 ft up and slides down to 2 ft. How many jumps will the monkey take to reach the top of the water tank?

Solution:

As per the given question,

Monkey’s first jump = 3 ft up and 2 ft down

= (+3)-2=1

2nd Jump of the Monkey = 3 ft up and 2 ft down = (1+3) = 4

= (+4)-2=2

Therefore, the monkey covers 2 ft after the 2nd jump

3rd Jump of the Monkey = 3 ft up and 2 ft down = (2+3) = 5

=(+5)-2 = 3

Therefore, the monkey covers 3 ft after the 3rd jump

4th Jump of the Monkey = 3 ft up and 2 ft down = (3+3) = 6

=(+6)-2 = 4

Hence, the monkey covers 4 ft after the 4th jump

5th Jump of the Monkey = 3 ft up and 2 ft down = (4+3)=7

=(+7)-2 = 5

Therefore, the monkey covers 5 ft after the 5th jump.

6th Jump of the Monkey = 3 ft up = (5+3) = 8

As the water tank is 8 feet high, the monkey will reach the top of the water tank in the 6th jump.

Subtraction

The subtraction is defined by the operator (-). Subtraction is the inverse property of addition. It means that removing one quantity from another quantity. It involves incrementing down from the given number. The result value after subtracting all the numbers is called the “difference”. The number from which other number is subtracted is called subtrahend. The subtracted number is called minuend.

Subtraction Operation on Positive and Negative Integers

Negative (-) – Positive (+) = Negative (-)
Positive (+) – Negative (-) = Positive (+)
Negative (-) – Negative (-) = Sign of the largest Number
Negative (-) + Positive (+) = Sign of the largest number

Examples:

(-5)-4 = (-5) + (-4) =- 9
5 – (-4) = 5 +4 = 9
(-5) – (-4) = (-5) + 4 = -1
(-4) – (-5) = (-4) + 5 = 1

Example Problem:

At a height of 3000 feet above sea level, a plane is flying. At some time, there is a submarine that is exactly above and floating 700 feet below sea level. Calculate the vertical distance using the concept of subtraction of integers?

Solution:

Height of the plane that is flying = 3000 feet

Depth of the submarine = -700 feet ( It is negative, as it is below the sea level).

To calculate the vertical distance, we use the subtraction of integers operation.

3000-(-700) = 3000 + 700 = 3700 feet

Therefore, the vertical distance is 3700 feet.

Multiplication

The multiplication is defined by the operator (“x” o “*”). Multiplication is the repeated property of addition. It means that a particular number is being repeatedly added to itself several times. It involves adding the number with the same number. The result value after multiplying all the numbers is called the “product”. The number to which another number is multiplied is called multiplicand. The multiplied number is called a multiplier.

Multiplication Operation on Positive and Negative Integers

Negative (-) x Positive (+) = Negative (-)
Positive (+) x Negative (-) = Negative (-)
Negative (-) x Negative (-) = Negative (-)
Positive (+) + Positive (+) = Positive (+)

Examples:

5×4 = 12

(-3) x (-4) = 12

(-3) x 4 = – 12

3 x (-4) = -12

Example Problems:

The submarine descends 40 feet per minute from sea level. Find the relation of the submarine with the sea level 5 minutes after it starts descending?

Solution:

The submarine below sea level =40 feet

As it is below sea level, it is negative. Therefore, it is -40.

After 5 minutes, the submarine is at = -40 x 5 = -200 feet

Therefore, the final solution is -200 feet.

Division

The division is defined by the operator (÷). The division is the inverse property of multiplication. It means that a particular number or quantity is split into equal or the same parts. It involves the splitting of numbers in the same proportions. The result value after dividing the numbers is called the “quotient”. The original number is called “dividend”. The number which is used for dividing into groups is called “divisor”. The final result is “quotient”.

Division Operation on Positive and Negative Integers

Negative (-) ÷ Positive (+) = Negative (-)

Positive (+) ÷ Negative (-) = Negative (-)

Negative (-) ÷ Negative (-) = Positive (+)

Positive (+) ÷ Positive (+) = Positive (+)

Examples:

20÷4 = 5

(-20)÷(-4) = 5

(-20) ÷ 4 = -5

20÷(-4) = -5

Example Problem:

In a living room, there are 120 books in total and they are placed on 6 shelves. Consider that each shelf has an equal number of books, Calculate the number of books on each shelf?

Solution:

As per the given question,

The total no of books = 120

No of seats placed on shelves = 6

To determine the number of books on each shelf = 120/6 =20

Therefore, the number of books on each shelf = 20 books.

Thus the final solution is 20 books.

Practice Test on Parallelogram will help you to test your knowledge. Answer on your own for every question given below. Before you take the practice test, make sure you have read the concept completely and solved all problems. It becomes easy if you have a perfect grip on the entire concept. Also, it will avoid you to confuse while choosing the answers. Complete concept and Objective Questions on Parallelogram are given on our website for free of cost.

Tick (✔) the correct answer in each of the following

1. From the given options, which parallelogram two diagonals are not necessarily equal ………………..

(a) square
(b) rectangle
(c) isosceles trapezium
(d) rhombus

2. A rhombus has diagonals of 16 cm and 12 cm. Find the length of each side?

(a) 8cm
(b) 12cm
(c) 10cm
(d) 9cm

3. Two adjacent angles of a parallelogram are (4b + 15)° and (2b – 10)°. The value of b is ………………. .

(a) 29.16
(b) 32
(c) 42
(d) 36

4. From the given options, which parallelogram two diagonals do not necessarily intersect at right angles ………………..

(a) rhombus
(b) rectangle
(c) kite
(d) parallelogram

5. The length and breadth of a rectangle are in the ratio 8 : 6. If the diagonal measures 50 cm. Find the perimeter of a rectangle?

(a) 800 cm
(b) 700 cm
(c) 600 cm
(d) 560 cm

6. The bisectors of any two adjacent angles of a parallelogram intersect at ………………..

(a) 90°
(b) 30°
(c) 60°
(d) 45°

7. If an angle of a parallelogram is 2/3 of its adjacent angle find the angle of a parallelogram.

(a) 72°
(b) 54°
(c) 108°
(d) 81°

8. The diagonals do not necessarily bisect the interior angles at the vertices in a ………………..

(a) square
(b) rectangle
(c) rhombus
(d) all of these

9. In a square PQRS, PQ = (3a + 5) cm and RS = (2a – 2) cm. Find the value of a?

(a) 4
(b) 5
(c) 7
(d) 8

10. If one angle if a parallelogram is 24º less than twice the smallest angle then the largest angle if parallelogram is?

(a) 68°
(b) 112°
(c) 102°
(d) 176°

Uses of Brackets are given here. Find the various uses and types of brackets. Before knowing the complete details regarding the brackets, know the fundamental operations like addition, subtraction, multiplication, and division. Know the history, rules, and precedence order of brackets. Go through the below sections to know the complete details regarding the usage of brackets.

History of Brackets

The word bracket is derived from the French word “Braguette” which the “piece around”. Anything that is written in brackets is defined as the piece of that bracket. Most brackets are used to enclose notes, references, explanations, etc. which are called crotchets as per the typographical brackets. Later, brackets were used as the group bracketed together for equal standing in some of the graded systems which are mostly used for sports brackets.

Brackets Usage

Mathematical brackets are known as symbols and parentheses which are most often used to create groups or that which clarify the order in which operations are to be done in the given algebraic expression.

Brackets symbol             Name

            ( )                         Parentheses or common brackets

            { }                        Braces or Curly brackets

            [ ]                         Brackets or square brackets or box brackets

             _                          Vinculum

The Left Part of the Bracket indicates the start of the bracket and the right part indicates the end of the bracket.

While Writing mathematical expressions having more than one bracket, parenthesis is used in the innermost part followed by braces, and these two are covered by square brackets. You need to know about the Uses of brackets to perform a  set of operations prior to the others.

Types of Brackets, Braces, Paranthesis in Math

Mathematical brackets are used for grouping. These brackets can include:

• ( )
• [ ]
• { }

In the grouping of numbers, brackets come in pairs. There will be a pair of sets i.e., the opening bracket and a closing bracket. Brackets are generally used to give clarity in the order of operations.

Suppose that there is an expression: 3+5*7-2. You cannot understand which operation to perform at the beginning. Therefore, we include brackets to understand the precedence of operations. If the problem is given as (3+5)*(7-2) = 8*5 = 40.

In the above problem, the parentheses will tell you the usual order of operations and will give you visual clarity.

“( ) Brackets”

The symbols “(” “)” are known as parentheses. These are called Brackets or Round Brackets. They are called Round Brackets as they are not curly or square braces. The input of the function is enclosed in parentheses. Parenthesis means “to put beside” in Geek language. Things like additional information, asides, clarifications, citations are defined by Paranteseis. Any type of information written in parentheses can be as short as a word or number or a few sentences. If something is given in parenthesis, then that sentence must have the capability of standing on its own.

Example: The little puppy (Rocky) skipped across the garden to her mother.

“Square Brackets [ ]”

The square brackets are denoted inside the parentheses to define something to the sub-ordinate clause. Square brackets are defined by the symbol “[ ]”. The main purpose of using square brackets is writing in the conjunction or to insert a name or word for clarification.

Square brackets are needed to add clarifications.

Example: She [Rosy] hit the policeman

They are also used to add additional information

Example: Two teams in the FIFA football final match were from South America [ Argentina and Uruguay]

Missing words can also be added with the help of square brackets.

Example: It is [a] good answer.

Authorial Comments or editorial can also be added using square brackets.

Example: There are not present [my emphasis]

Square brackets are also used to modify the direct quotation.

Example: The direct quotation is “I love traveling” which can be modified as ” He love[s] traveling”.

These can also be used for nesting.

Example: (We use square brackets [in this way] inside the round brackets).

“Curly Brackets { }”

Mostly curly brackets are not used for general purposes. They are mostly used in programming or math concepts. They are mainly used to hold terms or terms and hold a list of items.

Example:

3{2+[5(3+1)+4]}

This is the example of lists

Example of programming languages:

$value=0;

do{

$value++;

if ($value >=20)

{

Print(“Value is equal to or greater than 10. Ending.”);

exit;

}

}

until ($value >=200);

Curly brackets can also be used in sets in mathematics.

Grouping of numbers resembles Sets.

Example: {1,2,4,6,9}

Difference between ( ) and [ ] in math

Square brackets [ ] are used for commands whereas parenthesis or round brackets represent braces and to describe some special words.

Multiple Level of Grouping

If you planning for an equation, then grouping within other grouping is a little bit confusing. To avoid that confusion, we use various brackets and group the numbers with order precedence. If we are using the brackets, then we can define the level of operations within the equation.

Example:

2 + {1+[2+3*(5+4)]}

In the above example, first, we go for the inner calculations.

Therefore, first, we have to go for the grouping of (5+4) = 9

Then, we go for the multiplication of 3 and 9 i.e., 18

For the next order, we go for the addition of 3 with 18, the result will be 21.

The next precedence will be the addition of 1 and 21, the result will be 22.

Then, we go for the last grouping of 2 and 22, Therefore, the result of the addition is 24.

With the above example, it is clear that finding a way of solving the equation will be easy.

Questions on How to Use Brackets

Problem 1:

Simplify the expression : [(3*2) + (4*5)]/(7-3)

Solution:

Start solving the problem by simplifying the equation that is inside the parentheses i.e., 4 *5 = 20

Then, solve the equation of other parentheses, i.e., 3 * 2 = 6

Then simplify the equation within the square brackets i.e., [6 + 20]= 26

Now, solve the equation outside the parenthesis i.e., 7-3 = 4

Now finish the division of the two values = 26/4 = 6.5

Therefore, the final solution is 6.5

Problem 2:

Simplify the equation, [(3+2)*4 – (5*6)]/[1-(3+4)*2]

Solution:

Start solving the problem by simplifying the equation inside the parenthesis i.e., 5 *6 = 30

Then, solve the equation (3+2)*4 = %*4 = 20

Then, go for the next equation of 3+4 = 7, 7*4 = 28

Now, go for the other equation in the brackets i.e., 5*6=30

The second part of the equation is 3 + -7, then 7*2 = 14

Now, solve the equation of 1-14 = -13

Finally, divide the values -10/-13 = 0.769

Therefore, the final solution is -0.769

Thus, the three types of brackets are helpful for solving various equations. Check all the uses and importance of using the brackets and also know the preceding order.

Depending upon the length and angles, quadrilaterals are classifieds in different ways. Let us check Different Types of Quadrilaterals and their definition, properties along with their diagrams. A quadrilateral can be explained using the below properties

• The sum of the interior angles is 360 degrees in a quadrilateral.
• A quadrilateral consists of 4 sides and 4 vertices, and also 4 angles.
• The sum of the interior angle from the formula of the polygon using (n – 2) × 180 where n is equal to the number of sides of the polygon.

The main types in a quadrilateral are squares and rectangles, etc., with the same angles and sides.

Various Types of Quadrilaterals

Mainly quadrilaterals are classified into six types. They are

1. Parallelogram
2. Rhombus
3. Rectangle
4. Square
5. Trapezium
6. Kite

Parallelogram

A quadrilateral is said to be a parallelogram when it has two pairs of parallel sides and opposite sides are parallel and equal in length. Also, the opposite angles are equal in a parallelogram. Let us take a parallelogram PQRS, then the side PQ is parallel to the side RS. Also, the side PS is parallel to a side QR.

Two diagonals are present in the parallelogram and they intersect each other at a midpoint. From the figure, PR and QS are two diagonals. Also, the diagonals are equal in length from the midpoint.

PQ ∥ RS and PS ∥ QR.

Rhombus

Rhombus is a quadrilateral when all the four sides of a quadrilateral having equal lengths. In a rhombus, opposite sides are parallel and opposite angles are equal.

From the above figure, PQRS is a rhombus in which PQ ∥ RS, PS ∥ QR, and PQ = QR = RS = SP.

Rectangle

A quadrilateral is considered as a rectangle when all 4 angles of it are equal and each angle is 90 degrees. Also, both pairs of opposite sides of a rectangle are parallel and have equal lengths.

From the above figure, PQRS is a quadrilateral in which PQ ∥ RS, PS ∥ QR and ∠P = ∠Q = ∠R = ∠S = 90°.

So, PQRS is a rectangle.

Square

A square is a quadrilateral consists all the sides and angles are equal. Also, every angle of a square is 90 degrees. The pairs of opposite sides of a square are parallel to each other.

From the above figure, PQRS is a quadrilateral in which PQ ∥ RS, PS ∥ QR, PQ = QR = RS = SP and ∠P = ∠Q = ∠R = ∠S = 90°.

So, PQRS is a square.

Trapezium

A quadrilateral is called a trapezium when it has one pair of opposite parallel sides.

From the above figure, PQRS is a quadrilateral in which PQ ∥ RS. So, PQRS is a trapezium. A trapezium its non-parallel sides are equal is called an isosceles trapezium.

Kite

A quadrilateral is said to be a kite that has two pairs of equal-length sides and the sides are adjacent to each other.

From the above figure, PQRS is a quadrilateral. PQ = PS, QR = RS, PS ≠ QR, and PQ ≠ RS.

So, PQRS is a kite.

Important Points to Remember for Quadrilaterals

Look at some of the important points need to remember about a quadrilateral.

• A square is a rectangle and also it becomes a rhombus.
• The rectangle and rhombus do not become a square.
• A parallelogram is a trapezium.
• Square, rectangle, and rhombus are types of parallelograms.
• A trapezium is not a parallelogram.
• Kite is not a parallelogram.

Parallelogram consists of a flat shape that has two opposite & parallel sides with equal length. The parallelogram has four sides and also it is called a quadrilateral. The pair of parallel sides are always equal in length of a parallelogram. Furthermore, the interior opposite angles also equal in measurement. While adding the adjacent angles of a parallelogram, you will get 180 degrees.

The area of the parallelogram always depends on its base and height. Also, the perimeter of the parallelogram depends on the length of its four sides.

List of Parallelogram Concepts

Have a glance at the list of Parallelogram Concepts available below and use them for your reference. All you need to do is simply tap on the quick links and avail the underlying concept within. Practice as much as you can and solve all the problems easily.

• Properties of a Rectangle Rhombus and Square
• Problems on Parallelogram
• Practice Test on Parallelogram

Parallelogram Definition

A parallelogram called a quadrilateral that has two pairs of parallel sides. The interior angles of the parallelogram on the same side of the transversal are supplementary. The sum of all the interior angles becomes 360 degrees in a parallelogram.

A rectangle and square also consist of similar properties of a parallelogram. If in case, the sides of the parallelogram become equal, then it treats as a rhombus. Or else, if a parallelogram has one parallel side and the other two sides are non-parallel, then it treats as a trapezium.

From the above figure, ABCD is a parallelogram, where CD || AB and BC || AD. Also, CD = AB and BC = AD.
And, ∠A = ∠C & ∠B = ∠D
Also, ∠A & ∠D are supplementary angles. Because ∠A & ∠D are interior angles present on the same side of the transversal. Similarly, ∠B & ∠C are supplementary angles.

Therefore, ∠B + ∠C = 180, ∠A + ∠D = 180.

Shape of Parallelogram

A parallelogram shape is a two-dimensional shape. It consists of four sides and two pairs of parallel sides. All the parallel sides of the Parallelogram are equal in length. Also, the interior angles of the parallelogram should always equal.

Angles of Parallelogram

The Parallelogram has four angles. Its opposite interior angles always equal and the angles on the same side of the transversal are always supplementary with each other. When you add the same side of the transversal angles, you can get 180 degrees. Furthermore, the sum of the interior angles is 360 degrees.

Properties of Parallelogram

Check the below properties of a parallelogram and solve the related problems easily by applying the same. They are as follows

• The opposite angles of a parallelogram are congruent.
• Also, the opposite sides are parallel and congruent
• The consecutive angles are supplementary
• Furthermore, The two diagonals bisect each other
• If one angle of a parallelogram is a right angle, then all other angles are right angles.
• Parallelogram law: The sum of squares of all the sides of a parallelogram is always equal to the sum of squares of its diagonals.
• Each diagonal bisects the parallelogram into two congruent triangles.

Area of Parallelogram

The area of the Parallelogram totally depends on the Base and height of the Parallelogram.
The area of the Parallelogram can be calculated as Area = Base × Height

Perimeter of Parallelogram

The perimeter of a parallelogram is calculated as the total distance of the boundaries of the parallelogram. By knowing the length and breadth of the parallelogram, we can get the perimeter of a parallelogram.

Perimeter = 2 (a+b) units where a and b are the length of the sides of the parallelogram.

Types of Parallelogram

Depends on the angles and sides of the Parallelogram, mainly four types of Parallelogram are considered.
Let ABCD is a Parallelogram,
1. If AB = BC = CD = DA are equal, then it is called a rhombus. The properties of the rhombus and parallelogram are equal.
2. Rectangle, Square are also types of the parallelogram.

Parallelogram Theorems

Theorem: Prove that in a parallelogram, the opposite sides are equal; the opposite angles are equal; diagonals bisect each other.

Proof:
Let ABCD be a parallelogram. Draw its diagonal AC.
In ∆ ABC and ∆ ACD, ∠1 = ∠4 (alternate angles)
∠3 = ∠2 (alternate angles)
and AC = CA (common)

Therefore, ∆ ABC ≅ ∆ ACD (by ASA congruence)
⇒ AB = CD, BC = AD and ∠B = ∠D.
Similarly, by drawing the diagonal BD, we can prove that
∆ ABD ≅ ∆ BCD
Therefore, ∠A = ∠C
Thus, AB = CD, BC = AD, ∠A = ∠C and ∠B = ∠D.
This proves (i) and (ii)
In order to prove (iii) consider parallelogram ABCD and draw its diagonals AC and BD, intersecting each other at O.
In ∆ OAB and ∆ OCD, we have
AB = CD [Opposite sides of a parallelogram]
∠AOB = ∠ COD [Vertically opposite angles]
∠OAB = ∠OCD [Alternate angles]
Therefore, ∆ OAB ≅ ∆ OCD [By ASA property]
⇒ OA = OC and OB = OD.
This shows that the diagonals of a parallelogram bisect each other.

Therefore, in a parallelogram, the opposite sides are equal; diagonals bisect each other; the opposite angles are equal.

Problems on Parallelogram are given in this article along with an explanation. It is easy to learn and understand the entire concept of a Parallelogram by solving every problem over here. There are various types of problems included according to the new updated syllabus. Get a good score in the exam and improve your preparation level immediately by working on your difficult topics.

1. Prove that any two adjacent angles of a parallelogram are supplementary?

Solution:
Let us take a parallelogram PQRS.

Then, PS ∥ QR and PQ is a transversal.
The sum of the interior angles on the same side of the transversal is 180°
Therefore, P + Q = 180°
Similarly, ∠R + ∠S = 180°, ∠Q + ∠R = 180°, and ∠S + ∠P = 180°.
Thus, the sum of any two adjacent angles of a parallelogram is 180°.

Hence, any two adjacent angles of a parallelogram are supplementary.

2. Two adjacent angles of a parallelogram PQRS are as 2 : 3. Find the measure of each of its angles?

Solution:
Let us take a parallelogram PQRS.

Then, ∠P and ∠Q are its adjacent angles.
Let ∠P = (2a)° and ∠Q = (3a)°.
The sum of adjacent angles of a parallelogram is 180°
Then, ∠P + ∠Q = 180°
⇒ 2a + 3a = 180
⇒ 5a = 180
⇒ a = 36.
Therefore, ∠P = (2 × 36)° = 72° and ∠Q = (3 × 36°) = 108°.
∠Q and ∠R are adjacent angles. By adding them, we get 180°
Also, ∠Q + ∠R = 180°
= 108° + ∠R = 180° [Since, ∠Q = 108°]
∠R = (180° – 108°) = 72°.
∠R and ∠S are adjacent angles and add up to 180°.
Also, ∠R + ∠S = 180°
⇒ 72° + ∠S = 180°
⇒ ∠S = (180° – 72°) 108°.

Therefore, ∠P = 72°, ∠Q = 108°, ∠R = 72°and ∠S = 108°.

3. In the adjoining figure, PQRS is a parallelogram in which ∠P = 75°. Find the measure of each of the angles ∠Q, ∠R, and ∠S.

Solution:
It is given that PQRS is a parallelogram in which ∠P = 75°.

Since the sum of any two adjacent angles of a parallelogram is 180°,
∠P + ∠Q = 180°
⇒ 75° + ∠Q = 180°
⇒∠Q = (180° – 75°) = 105°
∠Q and ∠R are adjacent angles and add up to 180º.
Also, ∠Q + ∠R = 180°
⇒ 105° + ∠R = 180°
⇒ ∠R = (180° – 105°) = 75°.
∠R and ∠S are adjacent angles
Further, ∠R + ∠S = 180°
⇒ 75° + ∠S = 180°
⇒ ∠S = (180° – 75°) = 105°.

Therefore, ∠Q = 105°, ∠R = 75° and ∠S = 105°.

4. In the adjoining figure, PQRS is a parallelogram in which ∠QPS = 75° and ∠SQR = 60°. Calculate:
(i) ∠RSQ and (ii) ∠PSQ.

Solution:
Let us draw a parallelogram PQRS.

We know that the opposite angles of a parallelogram are equal.
Therefore, ∠QRS = ∠QPS = 75°.
(i) Now, in ∆ QRS, we have
The sum of the angles of a triangle is 180°
∠RSQ + ∠SQR + ∠QRS = 180°
⇒ ∠RSQ + 60° + 75° = 180°
⇒ ∠RSQ + 135° = 180°
⇒ ∠RSQ = (180° – 135°) = 45°.
(ii) PS ∥ QR and QS are the transversals.
Therefore, ∠PSQ = ∠SQR = 60° [alternate interior angles]

Hence, ∠PSQ = 60°.

5. In the adjoining figure, PQRS is a parallelogram in which ∠RPS = 40°, ∠QPR = 35°, and ∠ROS = 65°.
Calculate: (i) ∠PQS (ii) ∠QSR (iii) ∠PRQ (iv) ∠RQS.

Solution:
(i) ∠POQ = ∠ROS = 65° (vertically opposite angles)
Now, from ∆OPQ, we can write as:
The sum of the angles of a triangle is 180°
∠OPQ + ∠PQO + ∠POQ =180°
⇒ 35°+ ∠PQO + 65° = 180°
⇒ ∠PQO + 100° = 180°
⇒ ∠PQO = (180° – 100°) = 80°
⇒ ∠PQS = ∠PQO = 80°.
(ii) PQ ∥ SR and QS is a transversal.
Therefore, ∠QSR = ∠PQS = 80° [alternate interior angles]
Hence, ∠QSR = 80°.
(iii) PS ∥ QR and PR is a transversal.
Therefore, ∠PRQ = ∠RPS = 40° [alternate interior angles]
Hence, ∠PRQ = 40°.
(iv) ∠QRS = ∠QPS = (35° + 40°) = 75° [opposite angles of a parallelogram]
Now, in ∆RQS, we have
The sum of the angles of a triangle is 180°.
∠QSR + ∠QRS + ∠RQS = 180°
⇒ 80° + 75° + ∠RQS = 180°
⇒ 155° + ∠RQS = 180°
⇒ ∠RQS = (180° – 155°) = 25°.
Hence, ∠RQS = 25°.

6. In the adjoining figure, PQRS is a parallelogram, PO and QO are the bisectors of ∠P and ∠Q respectively. Prove that ∠POQ = 90°.

Solution:
We know that the sum of two adjacent angles of a parallelogram is 180°
Therefore, ∠P + ∠Q = 180° ……………. (i)
Since PO and QO are the bisectors of ∠P and ∠Q, respectively, we have
∠OPQ = 1/2∠P and ∠PQO = 1/2∠Q.
From ∆OPQ, we have
The sum of the angles of a triangle is 180°
∠OPQ + ∠POQ + ∠PQO = 180°
⇒ ¹/₂∠P + ∠PQO + ¹/₂∠Q = 180°
⇒ ¹/₂(∠P + ∠Q) + ∠POQ = 180°
⇒ (¹/₂ × 180°) + ∠POQ = 180° [from equation (i)]
⇒ 90° + ∠POQ = 180°
⇒ ∠POQ = (180° – 90°) = 90°.

Hence, ∠POQ = 90°.

7. The ratio of two sides of a parallelogram is 5: 4. If its perimeter is 54 cm, find the lengths of its sides?

Solution:
Let the lengths of two sides of the parallelogram be 5a cm and 4a cm respectively.
Find the perimeter using given values.
Then, its perimeter = 2(5a + 4a) cm = 2 (9a) cm = 18a cm.
Therefore, 18a = 54 ⇔ a = 54/18 = 3.

Therefore, one side = (5 × 3) cm = 15 cm and other side = (4 × 3) cm = 12 cm.

8. The length of a rectangle is 16 cm and each of its diagonals measures 20 cm. Find its breadth?

Solution:
Let PQRS be the given rectangle in which length PQ = 16 cm and diagonal PR = 20 cm.

Since each angle of a rectangle is a right angle, we have
∠PQR = 90°.
From the right ∆PQR, we have
PQ² + QR² = PR² [From Pythagoras’ Theorem]
⇒ QR² = (PR² – PQ²) = {(20)² – (16)²} = (400 – 256) = 144

⇒ QR = √144 = 12 cm.

Hence, breadth = 12 cm.

9. In the below figure, PQRS is a rhombus whose diagonals PR and QS intersect at a point O. If side PQ = 20 cm and diagonal QS = 32 cm, find the length of diagonal PR.

Solution:
We know that the diagonals of a rhombus bisect each other at right angles.
Therefore, QO = ¹/₂QS = (¹/₂ × 32) cm = 16 cm, PQ = 20 cm and ∠POQ = 90°.
From right ∆OPQ, we have PQ² = PO² + QO²
⇒ PO² = (PQ² – QO²) = {(20) ² – (16)²} cm²
= (400 – 256) cm²
= 144 cm²
⇒ PO = √144 cm = 12 cm.

Therefore, PR = 2 × PO = (2 × 12) cm = 24 cm.

The quadrilateral is a figure formed by adding the four-line segments. It consists of 4 sides, 4 vertices, and two diagonals. By going through this entire article you will get a complete idea of Quadrilateral like Types, Sides, Angles, Vertices, Diagonals of it, its Properties etc. If P, Q, R, S are four points and where no three points are collinear and also the line segments PQ, QR, RS, and SP do not intersect at their endpoints. Check the below figure that is the quadrilateral PQRS.

In a quadrilateral PQRS
(i) The vertices in a quadrilateral are P, Q, R, S.
(ii) The sides of a quadrilateral are PQ, QR, RS, and SP.
(iii) Also, the angles of quadrilateral are ∠SPQ, ∠PQR, ∠QRS and ∠RSP.
(iv) The line segments are PQ and QS.

Convex Quadrilaterals and Concave Quadrilaterals

If each angle of a quadrilateral is less than 180°, then it is called a convex quadrilateral. Also, if one angle of the quadrilateral is more than 180°, then it is called a concave quadrilateral. A quadrilateral is not a simple closed figure.

Sides, Angles, Vertices, Diagonals of the Quadrilateral

Have a look at the complete details of a Quadrilateral below.

Adjacent Sides of a Quadrilateral

Adjacent Sides of a Quadrilateral are nothing but the sides that have a common endpoint. If PQRS is a Quadrilateral, then (PQ, QR), (QR, RS), (RS, SP), and (SP, PQ) are four pairs of adjacent sides of quadrilateral PQRS.

Opposite Sides of a Quadrilateral

In a given quadrilateral, the two sides are said to be opposite sides when they do not have a common endpoint. If PQRS is a quadrilateral, then (PQ, SR) and (PS, QR) are two pairs of opposite sides of quadrilateral PQRS.

Adjacent Angles of a Quadrilateral

An angle is formed when two rays meeting at a common endpoint. Two angles of a quadrilateral are said to be adjacent when they have a common arm. From the given figure, (∠P, ∠Q), (∠Q, ∠R), (∠R, ∠S), and (∠S, ∠P) are four pairs of adjacent angles of quadrilateral PQRS.

Opposite Angles of a Quadrilateral

Opposite Angles of a Quadrilateral are not adjacent angles. If you consider a quadrilateral PQRS, then (∠P, ∠R) and (∠Q, ∠S) are two pairs of opposite angles of quadrilateral PQRS.

Adjacent Vertices of a Quadrilateral

In a Quadrilateral, if two vertices have a common side are known as adjacent vertices. From the figure, the pairs of adjacent vertices are (P, Q); (Q, R); (R, S), and (S, P).

Opposite Vertices of a Quadrilateral

Opposite Vertices of a Quadrilateral are vertices that do not have a common side. From the figure, the pairs of opposite vertices are (P, R) and (Q, S).

Diagonal of a Quadrilateral

When a line segment of the opposite vertices of a quadrilateral is joined, then the diagonal of the quadrilateral is formed. From the given figure, the two diagonals are PR and QS.