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NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

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NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals - FREE PDF Download

The NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals covers all the chapter's questions (All Exercises). These NCERT Solutions for Class 8 Maths have been carefully compiled and created in accordance with the most recent CBSE Syllabus 2024-25 updates. Students can use these NCERT Solutions for Class 8 to reinforce their foundations. Subject experts at Vedantu have created the continuity and differentiability class 8 NCERT solutions to ensure they match the current curriculum and help students while solving or practising problems.

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Exercises Under NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

  • Exercise 3.1 introduces polygons, covering their basic definition and classification based on the number of sides, such as triangles, quadrilaterals, pentagons, etc. It also distinguishes between convex and concave polygons, helping students understand the differences between these types of polygons.

  • Exercise 3.2 delves into the properties of quadrilaterals, exploring various types like trapeziums, kites, and parallelograms. This exercise helps students learn to identify different quadrilaterals and understand their specific properties.

  • Exercise 3.3 examines the properties of parallelograms, such as opposite sides being equal and parallel, opposite angles being equal, and diagonals bisecting each other. It includes problems for identifying parallelograms based on these properties and proving certain properties using theorems.

  • Exercise 3.4 looks at special parallelograms like rhombuses, rectangles, and squares. It highlights their unique properties, such as all sides being equal in a rhombus and all angles being 90 degrees in a rectangle. This exercise helps students understand and differentiate between these specific types of parallelograms.


List of Formulas

There are two major kinds of formulas related to quadrilaterals - Area and Perimeter. The following tables depict the formulas related to the areas and perimeters of different kinds of quadrilaterals.


Area of Quadrilaterals

Area of a Square

Side x Side

Area of a Rectangle

Length x Width

Area of a Parallelogram

Base x Height

Area of a Rhombus

1/2 x 1st Diagonal x 2nd Diagonal

Area of a Kite

1/2 x 1st Diagonal x 2nd Diagonal



Perimeter of Quadrilaterals

Perimeter of any quadrilateral is equal to the sum of all its sides, that is, AB + BC + CD + AD.


Name of the Quadrilateral

Perimeter

Perimeter of a Square

4 x Side

Perimeter of a Rectangle

2 (Length + Breadth)

Perimeter of a Parallelogram

2 (Base + Side)

Perimeter of a Rhombus

4 x Side

Perimeter of a Kite

2 (a + b), where a and b are the adjacent pairs 



Access NCERT Solutions for Class 8 Maths Chapter 3 – Understanding Quadrilaterals

Exercise 3.1

1. Given here are some figures.


Some figures



Classify each of them on the basis of following.

  1. Simple Curve

Ans: Given: the figures $(1)$to $(8)$

We need to classify the given figures as simple curves.

We know that a curve that does not cross itself is referred to as a simple curve.

Therefore, simple curves are $1,2,5,6,7$.

  1. Simple Closed Curve

Ans: Given: the figures $(1)$to $(8)$

We need to classify the given figures as simple closed curves.

We know that a simple closed curve is one that begins and ends at the same point without crossing itself.

Therefore, simple closed curves are $1,2,5,6,7$.

  1. Polygon

Ans: Given: the figures $(1)$to $(8)$

We need to classify the given figures as polygon.

We know that any closed curve consisting of a set of sides joined in such a way that no two segments

cross is known as a polygon.

Therefore, the polygons are $1,2$.

  1. Convex Polygon

Ans: Given: the figures $(1)$to $(8)$

We need to classify the given figures as convex polygon.

We know that a closed shape with no vertices pointing inward is called a convex polygon.

Therefore, the convex polygon is $2$.

  1. Concave Polygon

Ans: Given: the figures $(1)$to $(8)$

We need to classify the given figures as concave polygon.

We know that a polygon with at least one interior angle greater than 180 degrees is called a concave

polygon.

Therefore, the concave polygon is $1$.


2. What is a regular polygon?

State the name of a regular polygon of

(i) 3 sides (ii) 4 sides (iii) 6 sides


Solution: A regular polygon is a flat shape with all sides of equal length and all interior 5angles equal in measure.  In simpler terms, all the sides are the same size and all the corners look the same.

Here are the names of regular polygons based on the number of sides:

(i) 3 sides -  Equilateral Triangle (all three angles are also 60 degrees each)

(ii) 4 sides - Square

(iii) 6 sides - Hexagon


Exercise-3.2

1. Find ${\text{x}}$in the following figures.


${\text{x + y + z + w}}$




Some figures to find ${\text{x}}$in


Ans:

Given:


A triangle with 125*


We need to find the value of ${\text{x}}{\text{.}}$

We know that the sum of all exterior angles of a polygon is ${360^ \circ }.$

Thus.

$  {\text{x}} + {125^ \circ } + {125^ \circ } = {360^ \circ } $

$   \Rightarrow {\text{x}} + {250^ \circ } = {360^ \circ } $

$   \Rightarrow {\text{x}} = {360^ \circ } - {250^ \circ } $

$   \Rightarrow {\text{x}} = {110^ \circ } $ 



A polygon


Ans:

Given:


A polygon


We need to find the value of ${\text{x}}{\text{.}}$

We know that the sum of all exterior angles of a polygon is ${360^ \circ }.$

Thus,

$  {\text{x}} + {90^ \circ } + {60^ \circ } + {90^ \circ } + {70^ \circ } = {360^ \circ } $

$  \Rightarrow {\text{x}} + {310^ \circ } = {360^ \circ } $

$   \Rightarrow {\text{x}} = {360^ \circ } - {310^ \circ } $

 $  \Rightarrow {\text{x}} = {50^ \circ } $ 


2. Find the measure of each exterior angle of a regular polygon of 

  1. 9 Sides

Ans:

Given: a regular polygon with $9$ sides

We need to find the measure of each exterior angle of the given polygon.

We know that all the exterior angles of a regular polygon are equal.

The sum of all exterior angle of a polygon is ${360^ \circ }$.

Formula Used: ${\text{Exterior}}\;{\text{angle}} = \dfrac{{{{360}^ \circ }}}{{{\text{Number}}\;{\text{of}}\;{\text{sides}}}}$

Therefore,

Sum of all angles of given regular polygon $ = {360^ \circ }$

Number of sides $ = 9$

Therefore, measure of each exterior angle will be

$   = \dfrac{{{{360}^ \circ }}}{9} $

 $  = {40^ \circ } $ 

  1. 15 Sides

Ans:

Given: a regular polygon with $15$ sides

We need to find the measure of each exterior angle of the given polygon.

We know that all the exterior angles of a regular polygon are equal.

The sum of all exterior angle of a polygon is ${360^ \circ }$.

Therefore,

Sum of all angles of given regular polygon $ = {360^ \circ }$

Number of sides $ = 15$

Formula Used: ${\text{Exterior}}\;{\text{angle}} = \dfrac{{{{360}^ \circ }}}{{{\text{Number}}\;{\text{of}}\;{\text{sides}}}}$

Therefore, measure of each exterior angle will be

$   = \dfrac{{{{360}^ \circ }}}{{15}} $

$ = {24^ \circ } $ 


3. How many sides does a regular polygon have if the measure of an exterior angle is ${24^ \circ }$?

Ans: Given: A regular polygon with each exterior angle ${24^ \circ }$

We need to find the number of sides of given polygon.

We know that sum of all exterior angle of a polygon is ${360^ \circ }$.

Formula Used: ${\text{Number}}\;{\text{of}}\;{\text{sides}} = \dfrac{{{{360}^ \circ }}}{{{\text{Exterior}}\;{\text{angle}}}}$

Thus,

Sum of all angles of given regular polygon $ = {360^ \circ }$

Each angle measure $ = {24^ \circ }$

Therefore, number of sides of given polygon will be

$   = \dfrac{{{{360}^ \circ }}}{{{{24}^ \circ }}} $

 $  = 15 $ 


4. How many sides does a regular polygon have if each of its interior angles is ${165^ \circ }$?

Ans: Given: A regular polygon with each interior angle ${165^ \circ }$

We need to find the sides of the given regular polygon.

We know that sum of all exterior angle of a polygon is ${360^ \circ }$.

Formula Used: ${\text{Number}}\;{\text{of}}\;{\text{sides}} = \dfrac{{{{360}^ \circ }}}{{{\text{Exterior}}\;{\text{angle}}}}$

${\text{Exterior}}\;{\text{angle}} = {180^ \circ } - {\text{Interior}}\;{\text{angle}}$

Thus,

Each interior angle $ = {165^ \circ }$

So, measure of each exterior angle will be

$   = {180^ \circ } - {165^ \circ } $

$   = {15^ \circ } $ 

Therefore, number of sides of polygon will be

$   = \dfrac{{{{360}^ \circ }}}{{{{15}^ \circ }}} $

$   = 24 $ 


5. 

  1. Is it possible to have a regular polygon with measure of each exterior angle as ${22^ \circ }$?

Ans:

Given: A regular polygon with each exterior angle ${22^ \circ }$

We need to find if it is possible to have a regular polygon with given angle measure.

We know that sum of all exterior angle of a polygon is ${360^ \circ }$. The polygon will be possible if ${360^ \circ }$ is a perfect multiple of exterior angle.

Thus,

$\dfrac{{{{360}^ \circ }}}{{{{22}^ \circ }}}$ does not give a perfect quotient. 

Thus, ${360^ \circ }$ is not a perfect multiple of exterior angle. So, the polygon will not be possible.

  1. Can it be an interior angle of a regular polygon? Why?

Ans: Given: Interior angle of a regular polygon $ = {22^ \circ }$

We need to state if it can be the interior angle of a regular polygon.

We know that sum of all exterior angle of a polygon is ${360^ \circ }$. The polygon will be possible if ${360^ \circ }$ is a perfect multiple of exterior angle.

And, ${\text{Exterior}}\;{\text{angle}} = {180^ \circ } - {\text{Interior}}\;{\text{angle}}$

Thus, Exterior angle will be

$   = {180^ \circ } - {22^ \circ } $

 $  = {158^ \circ } $ 

$\dfrac{{{{158}^ \circ }}}{{{{22}^ \circ }}}$ does not give a perfect quotient. 

Thus, ${158^ \circ }$ is not a perfect multiple of exterior angle. So, the polygon will not be possible.


6. 

  1. What is the minimum interior angle possible for a regular polygon?

Ans:  Given: A regular polygon

We need to find the minimum interior angle possible for a regular polygon.

A polygon with minimum number of sides is an equilateral triangle.

So, number of sides $ = 3$

We know that sum of all exterior angle of a polygon is ${360^ \circ }$.

And, 

${\text{Exterior}}\;{\text{angle}} = \dfrac{{{{360}^ \circ }}}{{{\text{Number}}\;{\text{of}}\;{\text{sides}}}}$

Thus, Maximum Exterior angle will be

$   = \dfrac{{{{360}^ \circ }}}{3} $

$   = {120^ \circ } $ 

We know, ${\text{Interior}}\;{\text{angle}} = {180^ \circ } - {\text{Exterior}}\;{\text{angle}}$

Therefore, minimum interior angle will be

$   = {180^ \circ } - {120^ \circ } $

$   = {60^ \circ } $ 

  1. What is the maximum exterior angel possible for a regular polygon?

Ans: Given: A regular polygon

We need to find the maximum exterior angle possible for a regular polygon.

A polygon with minimum number of sides is an equilateral triangle.

So, number of sides $ = 3$

We know that sum of all exterior angle of a polygon is ${360^ \circ }$.

And, 

${\text{Exterior}}\;{\text{angle}} = \dfrac{{{{360}^ \circ }}}{{{\text{Number}}\;{\text{of}}\;{\text{sides}}}}$

Therefore, Maximum Exterior angle possible will be

$   = \dfrac{{{{360}^ \circ }}}{3} $

$ = {120^ \circ } $

 Exercise 3.3

1. Given a parallelogram ABCD. Complete each statement along with the definition or property used.


A parallelogram ABCD


  1. $\;{\text{AD}}$ = $...$

Ans:

Given: A parallelogram ${\text{ABCD}}$ 

We need to complete each statement along with the definition or property used.

We know that opposite sides of a parallelogram are equal.

Hence, ${\text{AD}}$ = ${\text{BC}}$ 

  1. $\;\angle {\text{DCB }} = $ $...$

Ans:

Given: A parallelogram ${\text{ABCD}}$.

We need to complete each statement along with the definition or property used.

${\text{ABCD}}$ is a parallelogram, and we know that opposite angles of a parallelogram are equal.

Hence, $\angle {\text{DCB   =  }}\angle {\text{DAB}}$

  1. ${\text{OC}} = ...$ 

Ans:

Given: A parallelogram ${\text{ABCD}}$.

We need to complete each statement along with the definition or property used.

${\text{ABCD}}$ is a parallelogram, and we know that diagonals of parallelogram bisect each other.

Hence, ${\text{OC  =  OA}}$

  1. $m\angle DAB\; + \;m\angle CDA\; = \;...$

Ans:

Given: A parallelogram ${\text{ABCD}}$.

We need to complete each statement along with the definition or property used.

${\text{ABCD}}$ is a parallelogram, and we know that adjacent angles of a parallelogram are supplementary to each other.

Hence, $m\angle DAB\; + \;m\angle CDA\; = \;180^\circ $


 2. Consider the following parallelograms. Find the values of the unknowns x, y, z.


Parallelograms


(i)

Ans:

Given: A parallelogram ${\text{ABCD}}$

We need to find the unknowns ${\text{x,y,z}}$

The adjacent angles of a parallelogram are supplementary.

Therefore, ${\text{x} + 100^\circ  = 180^\circ }$

${\text{x} = 80^\circ }$ 

Also, the opposite angles of a parallelogram are equal.

Hence, ${\text{z}} = {\text{x}} = 80^\circ $ and ${\text{y}} = 100^\circ $

(ii) 

Ans:

Given: A parallelogram.

We need to find the values of ${\text{x,y,z}}$

The adjacent pairs of a parallelogram are supplementary.

Hence, $50^\circ  + {\text{y}} = 180^\circ $

${\text{y}} = 130^\circ $

Also, ${\text{x}} = {\text{y}} = 130^\circ $(opposite angles of a parallelogram are equal)

And, ${\text{z}} = {\text{x}} = 130^\circ $ (corresponding angles)

(iii)  

Ans:

Given: A parallelogram 

We need to find the values of ${\text{x,y,z}}$

${\text{x}} = 90^\circ $(Vertically opposite angles)

Also, by angle sum property of triangles

${\text{x}} + {\text{y}} + 30^\circ  = 180^\circ $

${\text{y}} = 60^\circ $

Also,${\text{z}} = {\text{y}} = 60^\circ $(alternate interior angles)

(iv)

Ans:

Given: A parallelogram

We need to find the values of ${\text{x,y,z}}$

Corresponding angles between two parallel lines are equal.

Hence, ${\text{z}} = 80^\circ $
Also,${\text{y}} = 80^\circ $ (opposite angles of parallelogram are equal)

In a parallelogram, adjacent angles are supplementary

Hence,${\text{x}} + {\text{y}} = 180^\circ $

$  {\text{x}} = 180^\circ  - 80^\circ  $

$  {\text{x}} = 100^\circ  $ 

(v) 

Ans:

Given: A parallelogram

We need to find the values of ${\text{x,y,z}}$

As the opposite angles of a parallelogram are equal, therefore,${\text{y}} = 112^\circ $ 

Also, by using angle sum property of triangles

$  {\text{x}} + {\text{y}} + 40^\circ  = 180^\circ  $

$  {\text{x}} + 152^\circ  = 180^\circ  $

$  {\text{x}} = 28^\circ  $ 

And ${\text{z}} = {\text{x}} = 28^\circ $(alternate interior angles)


3. Can a quadrilateral ${\text{ABCD}}$be a parallelogram if 

(i) $\angle {\text{D}}\;{\text{ + }}\angle {\text{B}} = 180^\circ ?$

Given: A quadrilateral ${\text{ABCD}}$

We need to find whether the given quadrilateral is a parallelogram.

For the given condition, quadrilateral ${\text{ABCD}}$ may or may not be a parallelogram.

For a quadrilateral to be parallelogram, the sum of measures of adjacent angles should be $180^\circ $ and the opposite angles should be of same measures.

(ii) ${\text{AB}} = {\text{DC}} = 8\;{\text{cm}},\;{\text{AD}} = 4\;{\text{cm}}\;$and ${\text{BC}} = 4.4\;{\text{cm}}$

Ans:

Given: A quadrilateral ${\text{ABCD}}$

We need to find whether the given quadrilateral is a parallelogram.

As, the opposite sides ${\text{AD}}$and ${\text{BC}}$are of different lengths, hence the given quadrilateral is not a parallelogram.

(iii) $\angle {\text{A}} = 70^\circ $and $\angle {\text{C}} = 65^\circ $

Ans:

Given: A quadrilateral ${\text{ABCD}}$

We need to find whether the given quadrilateral is a parallelogram.

As, the opposite angles have different measures, hence, the given quadrilateral is a parallelogram.


4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

Ans:

Given: A quadrilateral.

We need to draw a rough figure of a quadrilateral that is not a paralleloghram but has exactly two opposite angles of equal measure.

A kite is a figure which has two of its interior angles, $\angle {\text{B}}$and $\angle {\text{D}}$of same measures. But the quadrilateral ${\text{ABCD}}$is not a parallelogram as the measures of the remaining pair of opposite angles are not equal.


A quadrilateral


5. The measures of two adjacent angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram.

Ans: Given: A parallelogram with adjacent angles in the ratio $3:2$

We need to find the measure of each of the angles of the parallelogram.

Let the angles be $\angle {\text{A}} = 3{\text{x}}$and $\angle {\text{B}} = 2{\text{x}}$

As the sum of measures of adjacent angles is $180^\circ $ for a parallelogram.

$  \angle {\text{A}} + \angle {\text{B}} = 180^\circ  $

 $ 3{\text{x}} + 2{\text{x}} = 180^\circ  $

 $ 5{\text{x}} = 180^\circ  $

 $ {\text{x}} = 36^\circ  $ 

$~\angle A=$ $\angle {\text{C}}$ $= 3{\text{x}} = 108^\circ$and $~\angle B=$ $\angle {\text{D}}$ $= 2{\text{x}} = 72^\circ$(Opposite angles of a parallelogram are equal).

Hence, the angles of a parallelogram are $108^\circ ,72^\circ ,108^\circ $and $72^\circ $.


6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

Ans:

Given: A parallelogram with two equal adjacent angles.

We need to find the measure of each of the angles of the parallelogram.

The sum of adjacent angles of a parallelogram are supplementary.

$  \angle {\text{A}} + \;\angle {\text{B}} = 180^\circ  $

$  2\angle {\text{A}}\;{\text{ =  180}}^\circ  $

$  \angle {\text{A}}\;{\text{ = }}\;{\text{90}}^\circ  $

$  \angle {\text{B}}\;{\text{ = }}\angle {\text{A}}\;{\text{ = }}\;{\text{90}}^\circ  $

Also, opposite angles of a parallelogram are equal

Therefore,

$  \angle {\text{C}} = \angle {\text{A}} = 90^\circ  $

$  \angle {\text{D}} = \angle {\text{B}} = 90^\circ  $ 

Hence, each angle of the parallelogram measures $90^\circ $.


7. The adjacent figure ${\text{HOPE}}$is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.

Ans:

Given: A parallelogram ${\text{HOPE}}$.

We need to find the measures of angles ${\text{x,y,z}}$and also state the properties used to find these angles.

$\angle {\text{y}} = 40^\circ $(Alternate interior angles)

And $\angle {\text{z}} + 40^\circ  = 70^\circ $(corresponding angles are equal)

$\angle {\text{z}} = 30^\circ $

Also, ${\text{x}} + {\text{z}} + 40^\circ  = 180^\circ $(adjacent pair of angles)

${\text{x}} = 110^\circ $


8. The following figures ${\text{GUNS}}$and ${\text{RUNS}}$are parallelograms. Find ${\text{x}}$and${\text{y}}$. (Lengths are in cm).

(i) 


Parallelograms


Ans:

Given: Parallelogram ${\text{GUNS}}$.

We need to find the measures of ${\text{x}}$and ${\text{y}}$.

${\text{GU = SN}}$(Opposite sides of a parallelogram are equal).

$  3{\text{y }} - {\text{ }}1{\text{ }} = {\text{ }}26{\text{ }} $

$  3{\text{y }} = {\text{ }}27{\text{ }} $

$  {\text{y }} = {\text{ }}9{\text{ }} $ 

Also,${\text{SG = NU}}$

Therefore, 

$  3{\text{x}} = 18 $

$  {\text{x}} = 3 $ 

(ii) 


Parallelogram ${\text{RUNS}}$


Ans:

Given: Parallelogram ${\text{RUNS}}$

We need to find the value of ${\text{x}}$and ${\text{y}}{\text{.}}$

The diagonals of a parallelogram bisect each other, therefore, 

$  {\text{y }} + {\text{ }}7{\text{ }} = {\text{ }}20{\text{ }} $

$  {\text{y }} = {\text{ }}13 $

 $ {\text{x }} + {\text{ y }} = {\text{ }}16 $

$  {\text{x }} + {\text{ }}13{\text{ }} = {\text{ }}16 $

 $ {\text{x }} = {\text{ }}3{\text{ }} $ 


9. In the above figure both ${\text{RISK}}$and ${\text{CLUE}}$are parallelograms. Find the value of ${\text{x}}{\text{.}}$


Parallelograms


Ans:

Given: Parallelograms ${\text{RISK}}$and ${\text{CLUE}}$

We need to find the value of ${\text{x}}{\text{.}}$

As we know that the adjacent angles of a parallelogram are supplementary, therefore, 

In parallelogram ${\text{RISK}}$

$  \angle {\text{RKS  + }}\angle {\text{ISK}} = 180^\circ  $

 $ 120^\circ  + \angle {\text{ISK}} = 180^\circ  $ 

As the opposite angles of a parallelogram are equal, therefore,

In parallelogram ${\text{CLUE}}$,

$\angle {\text{ULC}} = \angle {\text{CEU}} = 70^\circ $

Also, the sum of all the interior angles of a triangle is $180^\circ $

Therefore,

$  {\text{x }} + {\text{ }}60^\circ {\text{ }} + {\text{ }}70^\circ {\text{ }} = {\text{ }}180^\circ  $

$  {\text{x }} = {\text{ }}50^\circ  $ 


10. Explain how this figure is a trapezium. Which of its two sides are parallel?


A trapezium


Ans:

Given:


A trapezium


We need to explain how the given figure is a trapezium and find its two sides that are parallel.

If a transversal line intersects two specified lines in such a way that the sum of the angles on the same side of the transversal equals $180^\circ $, the two lines will be parallel to each other.

Here, $\angle {\text{NML}} = \angle {\text{MLK}} = 180^\circ $

Hence, ${\text{NM}}||{\text{LK}}$

Hence, the given figure is a trapezium.


11. Find ${\text{m}}\angle {\text{C}}$in the following figure if ${\text{AB}}\parallel {\text{CD}}$${\text{AB}}\parallel {\text{CD}}$.


$\angle {\text{C}}$


Ans:

Given: ${\text{AB}}\parallel {\text{CD}}$ and quadrilateral


$\angle {\text{C}}$


We need to find the measure of $\angle {\text{C}}$

$\angle {\text{B}} + \angle {\text{C}} = 180^\circ $(Angles on the same side of transversal).

$  120^\circ  + \angle {\text{C}} = 180^\circ  $

$  \angle {\text{C}} = 60^\circ  $ 


12. Find the measure of $\angle {\text{P}}$and$\angle {\text{S}}$, if ${\text{SP}}\parallel {\text{RQ}}$in the following figure. (If you find${\text{m}}\angle {\text{R}}$, is there more than one method to find${\text{m}}\angle {\text{P}}$?)


$\angle {\text{P}}$and $\angle {\text{S}}$


Ans:

Given: ${\text{SP}}\parallel {\text{RQ}}$and 


$\angle {\text{P}}$and $\angle {\text{S}}$


We need to find the measure of $\angle {\text{P}}$and $\angle {\text{S}}$.

The sum of angles on the same side of transversal is $180^\circ .$

$\angle {\text{P}} + \angle {\text{Q}} = 180^\circ $

$  \angle {\text{P}} + 130^\circ  = 180^\circ  $

$  \angle {\text{P}} = 50^\circ  

Also,

 $\angle {\text{R }} + {\text{ }}\angle {\text{S }} = {\text{ }}180^\circ {\text{ }} $

$  {\text{ }}90^\circ {\text{ }} + {\text{ }}\angle {\text{S }} = {\text{ }}180^\circ  $

  ${\text{ }}\angle {\text{S }} = {\text{ }}90^\circ {\text{ }} $ 

Yes, we can find the measure of ${\text{m}}\angle {\text{P}}$ by using one more method.

In the question,${\text{m}}\angle {\text{R}}$and ${\text{m}}\angle {\text{Q}}$are given. After finding ${\text{m}}\angle {\text{S}}$ we can find ${\text{m}}\angle {\text{P}}$ by using angle sum property.


Exercise 3.4

1. State whether True or False.

(a) All rectangles are squares.

(b) All rhombuses are parallelograms.

(c) All squares are rhombuses and also rectangles.

(d) All squares are not parallelograms.

(e) All kites are rhombuses.

(f) All rhombuses are kites.

(g) All parallelograms are trapeziums.

(h) All squares are trapeziums.

Solution:

(a) False

Every square is indeed a type of rectangle, not every rectangle can be called a square.

(b) True

(c) True

(d) False

It's correct to say that all squares can be classified as parallelograms due to their shared characteristic of having opposite sides that are parallel and opposite angles that are equal.

(e) False.

Because, a kite shape is that its adjacent sides are not necessarily equal in length, unlike those of a square.

(f) True

(g) True

(h) True


2. Identify all the quadrilaterals that have.

(a) four sides of equal length

(b) four right angles

Solution:

(a) Rhombus and square have all four sides of equal length.

(b) Square and rectangles have four right angles. 


3. Explain how a square is

(i) a quadrilateral

(ii) a parallelogram

(iii) a rhombus

(iv) a rectangle

Solution:

(i) Square is a quadrilateral because it has four sides.

(ii) A square is a parallelogram because its opposite sides are parallel and opposite angles are equal.

(iii) Square is a rhombus because all four sides are of equal length and diagonals bisect at right angles.

(iv)Square is a rectangle because each interior angle, of the square, is 90° 


4. Name the quadrilaterals whose diagonals.

(i) bisect each other 

(ii) are perpendicular bisectors of each other 

(iii) are equal

Solution:

(i) Parallelogram, Rhombus, Square and Rectangle

(ii) Rhombus and Square

(iii)Rectangle and Square 


5. Explain why a rectangle is a convex quadrilateral.

Solution:

A rectangle is a convex quadrilateral because both of its diagonals lie inside the rectangle.


6. ABC is a right-angled triangle and O is the mid-point of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).


ABC is a right-angled triangle and O is the mid-point of the side opposite to the right angle


Solution:

AD and DC are drawn so that AD || BC and AB || DC

AD = BC and AB = DC

ABCD is a rectangle as opposite sides are equal and parallel to each other and all the interior angles are of 90°.

In a rectangle, diagonals are of equal length and also bisect each other.

Hence, AO = OC = BO = OD

Thus, O is equidistant from A, B and C.


Overview of Deleted Syllabus for CBSE Class 8 Maths Understanding Quadrilaterals

Chapter

Dropped Topics

Understanding Quadrilaterals

3.1 Introduction

3.2 Polygons

3.2.1 Classification of polygons

3.2.2 Diagonals

3.2.5 Angle sum property.



Class 8 Maths Chapter 3: Exercises Breakdown

Exercise

Number of Questions

Exercise 3.1

2 Questions & Solutions (1 Long Answer, 1 Short Answer)

Exercise 3.2

6 Questions & Solutions (6 Short Answers)

Exercise 3.3

12 Questions & Solutions (6 Long Answers, 6 Short Answers)

Exercise 3.4

6 Questions & Solutions (1 Long Answer, 5 Short Answers)



Conclusion

In conclusion, NCERT Solutions for Class 8 Maths Chapter 3 - Understanding Quadrilaterals provides a comprehensive and detailed understanding of the properties and characteristics of various types of quadrilaterals. By studying this chapter and using the NCERT solutions, students can enhance their knowledge of quadrilaterals and develop their problem-solving abilities. The chapter starts by introducing quadrilaterals and their diverse types, including parallelograms, rectangles, squares, rhombuses, and trapeziums. It goes on to explain each type, detailing their characteristic properties like side lengths, angles, diagonals, and symmetry. Students that practice these kinds of questions will gain confidence and perform well on tests.


Other Study Material for CBSE Class 8 Maths Chapter 3



Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



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FAQs on NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

1. What are the NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals?

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals provide step-by-step, CBSE-aligned answers to all textbook questions on polygons, quadrilaterals, their types (like parallelogram, rectangle, rhombus, square, kite, trapezium), and their properties. Each solution follows the correct methodology and in-depth explanations as per the latest CBSE 2025–26 syllabus, ensuring clarity for every exercise.

2. How do the NCERT Solutions help in solving Exercise 3.3 of Class 8 Maths Chapter 3?

The NCERT Solutions guide students through each question in Exercise 3.3 by clearly breaking down the proof or calculation steps involving properties of parallelograms, such as equal opposite angles, supplementary adjacent angles, and diagonal bisectors. Each answer follows CBSE marking schemes and includes reasoning for every property used.

3. What key properties of quadrilaterals should students remember for Class 8 Maths exams?

Students must remember the following properties for exams:

  • Sum of all interior angles of a quadrilateral is always 360°.
  • Opposite sides of parallelograms are equal and parallel.
  • Diagonals of a rectangle and square are equal and bisect each other.
  • All angles of a square and rectangle are right angles.
  • In rhombus and kite, diagonals intersect at right angles.

4. Why is understanding the difference between convex and concave polygons important in Chapter 3?

Knowing the difference between convex and concave polygons helps in classifying shapes correctly and understanding their angle properties, which is fundamental for accurate polygon identification, problem-solving, and avoiding confusion in competitive or board exams.

5. How can you find the number of sides in a regular polygon if each exterior angle is given?

Use the formula: Number of sides = 360° ÷ measure of each exterior angle. This formula is essential for Class 8 Chapter 3 and is frequently applied in NCERT exercise questions for regular polygons.

6. What is the significance of the angle sum property of quadrilaterals in CBSE Class 8 Maths Chapter 3?

The angle sum property states that the sum of all interior angles in any quadrilateral equals 360°. This concept is central for solving angle-finding questions and is highlighted in multiple exercises and higher-order thinking questions in Chapter 3.

7. Can a quadrilateral with two pairs of equal-length adjacent sides always be a parallelogram? Justify as per NCERT Solutions.

No, such a quadrilateral could be a kite but not necessarily a parallelogram. In parallelograms, opposite sides must be equal and parallel, whereas in a kite, only two pairs of adjacent sides are equal. NCERT solutions provide reasoning-based answers to distinguish these cases.

8. What formulas are recommended to memorize for perimeter and area of quadrilaterals in Class 8 Chapter 3?

Key formulas include:

  • Perimeter of Rectangle = 2 × (Length + Breadth)
  • Area of Rectangle = Length × Breadth
  • Area of Square = Side × Side
  • Area of Parallelogram = Base × Height
  • Area of Rhombus/Kite = (1/2) × Diagonal 1 × Diagonal 2
Mastery of these is crucial for quick solutions and CBSE scoring.

9. How do NCERT Solutions ensure CBSE board pattern compliance in Chapter 3 answers?

NCERT Solutions strictly follow CBSE board marking guidelines, provide logical stepwise explanations, state which property/theorem is being applied, and give concise, direct answers instead of only final results. This maximizes marks and strengthens conceptual clarity for board exams.

10. What common mistakes should students avoid when answering Class 8 Maths Chapter 3 Quadrilateral questions?

Common pitfalls include:

  • Confusing properties of similar-looking quadrilaterals (e.g., parallelogram vs. rhombus vs. kite).
  • Missing that angle sum is always 360° for any quadrilateral.
  • Not justifying each answer with a specific property or theorem, as required in NCERT Solutions.
  • Misapplying area or perimeter formulas to irregular quadrilaterals.

11. Why is it essential to solve all exercise questions from NCERT Solutions for Class 8 Maths Chapter 3?

Practicing all exercise questions builds a thorough understanding of each concept and prepares students for a wide variety of exam problems, reducing errors and improving speed. NCERT Solutions also expose students to CBSE board style and expected reasoning steps.

12. How is the concept of diagonal bisection used in solving parallelogram problems in NCERT Solutions for Class 8?

For parallelograms, the diagonals bisect each other, meaning they cut each other into two equal halves. This property is frequently used in NCERT exercises to set up equations and solve for unknown side lengths, coordinates, or angles.

13. What should a student do if they find certain quadrilateral questions tricky in NCERT Chapter 3?

Students should:

  • Revisit the solution steps and identify which property or theorem was applied.
  • Draw neat, labelled diagrams for visual clarity.
  • Practice similar questions from the NCERT Solutions to reinforce understanding.
  • Seek clarification from teachers or reliable explanations as per CBSE pattern.

14. How is Class 8 Chapter 3 on Understanding Quadrilaterals important for higher classes?

This chapter forms the basis for advanced topics in geometry such as congruence, similarity, coordinate geometry, and surface areas in later classes. Mastery here enables easier and deeper understanding of mathematical concepts in Class 9, 10, and beyond.

15. What real-life examples help in understanding quadrilaterals, as discussed in NCERT Solutions for Class 8?

Common objects like doors (rectangle), chessboards (square), kites (kite), tabletop (rectangle), diamond warning signs (rhombus), and trapezium-shaped benches help students relate quadrilaterals to daily life, aiding practical understanding and retention of properties.