case study on quadrilaterals class 8

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the NCERT book.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals are prepared based on Class 8 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 8 Solutions Maths Chapter 3 are in accordance with the latest CBSE guidelines and marking schemes.

Class 8 Maths Chapter 3 Exercise 3.1 Solutions

Class 8 Maths Chapter 3 Exercise 3.2 Solutions

Class 8 Maths Chapter 3 Exercise 3.3 Solutions

Class 8 Maths Chapter 3 Exercise 3.4 Solutions

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NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals

NCERT solutions for class 8 maths chapter 3 understanding quadrilaterals define a polygon as a simple closed curve that is made up of straight lines. Thus, a quadrilateral can be defined as a polygon that has four sides, four angles, and four vertices. This chapter starts by introducing children to some very important concepts that they need to learn before moving on to studying quadrilaterals . These topics include the classification of polygons on the basis of sides, examining diagonals , concave, convex, regular, and irregular polygons as well as the angle sum property. The scope of NCERT solutions class 8 maths chapter 3 is very vast as there are several properties and types of quadrilaterals available. However, the explanation given in these solutions helps to simplify the learning process ensuring that students can build a strong geometrical foundation. 

Class 8 maths NCERT solutions chapter 3 elaborates on special quadrilaterals such as squares , rectangles , parallelograms , kites , and rhombuses . They show kids how to solve problems based on these figures and intelligently utilize the associated properties to remove the complexities from such questions. In the NCERT solutions Chapter 3 Understanding Quadrilaterals we will take an in-depth look at the basic elements and theories of these four-sided polygons and also you can find some of these in the exercises given below.

• NCERT Solutions Class 8 Maths Chapter 3 Ex 3.1
• NCERT Solutions Class 8 Maths Chapter 3 Ex 3.2
• NCERT Solutions Class 8 Maths Chapter 3 Ex 3.3
• NCERT Solutions Class 8 Maths Chapter 3 Ex 3.4

NCERT Solutions for Class 8 Maths Chapter 3 PDF

Using the NCERT solutions class 8 maths children can solidify several concepts of quadrilaterals. They understand the conditions under which a special quadrilateral such as a parallelogram becomes a square, how to find the measure of an interior or exterior angle , and so on. The links to all these brief and precise solutions are given below and kids can use them to improve their mathematical acumen.

☛ Download Class 8 Maths NCERT Solutions Chapter 3 Understanding Quadrilaterals

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NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

Quadrilaterals form a vital shape contributing to geometrical studies. Thus, children need to develop a robust conceptual foundation as they will require it in higher classes for solving more complicated problems and constructing this figure. They can do this by revising the solutions given above regularly. The following sections deal with an exercise-wise detailed analysis of NCERT Solutions Class 8 Maths Chapter 3 understanding quadrilaterals.

• Class 8 Maths Chapter 3 Ex 3.1 - 7 Questions
• Class 8 Maths Chapter 3 Ex 3.2 - 6 Questions
• Class 8 Maths Chapter 3 Ex 3.3 - 12 Questions
• Class 8 Maths Chapter 3 Ex 3.4 - 6 Questions

☛ Download Class 8 Maths Chapter 3 NCERT Book

Topics Covered: Identifying the polygon, finding the measure of angles, and verifying the exterior angles of a polygon are topics under class 8 maths NCERT solutions chapter 3. Apart from this, there are many sections dealing with the various elements of trapeziums , parallelograms, rectangles, squares, etc.

Total Questions: There are a total of 31 fantastic sums in Class 8 maths chapter 3 Understanding Quadrilaterals. 7 are simple theory-based problems, 16 are in-between and 8 are higher-order thinking sums.

List of Formulas in NCERT Solutions Class 8 Maths Chapter 3

The questions in the NCERT solutions class 8 maths chapter 3 are not only based on some formulas but also see the use of various vital properties. The sum of interior and exterior angles , along with theorems give the keys to attempting these sums. The angle sum property states that the sum of all the interior angles of a polygon is a multiple of the number of triangles that make up that polygon. Such pointers covered in NCERT solutions for class 8 maths chapter 3 make up the crux of this lesson and are given below.

• Angle Sum Property of a Quadrilateral: a + b + c + d = 360°. (a, b, c, d are the interior angles).
• The opposite sides and opposite angles of a parallelogram are equal in length.
• The adjacent angles in a parallelogram are supplementary.
• The diagonals of a parallelogram bisect each other.
• The diagonals of a rhombus are perpendicular bisectors of one another.

Important Questions for Class 8 Maths NCERT Solutions Chapter 3

Ncert solutions for class 8 maths video chapter 3, faqs on ncert solutions class 8 maths chapter 3, do i need to practice all questions provided in ncert solutions class 8 maths understanding quadrilaterals.

All the sums in the NCERT Solutions Class 8 Maths Understanding Quadrilaterals cover different subtopics of the lesson. These sums also pave a foundation for the geometrical topics in grades that are to follow. Thus, it is crucial for kids to practice all questions so as to get a clear idea of all the components in a quadrilateral.

What are the Important Topics Covered in Class 8 Maths NCERT Solutions Chapter 3?

Each exercise is based on a different topic such as angles of a polygon, rhombus, square, and rectangles; thus, each section that falls under the NCERT Solutions Class 8 Maths Chapter 3 must be given equal importance. Kids need to strategize their studies to focus more on learning properties and then applying them to questions.

How Many Questions are there in NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals?

There are a total of 31 questions in the NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals that are distributed among 4 exercises. There are different types of questions such as true and false sums, identifying the type of shape based on certain properties, and finding the measure of a particular angle using formulas.

What are the Important Formulas in Class 8 Maths NCERT Solutions Chapter 3?

Formulas such as the angle sum property of a quadrilateral, exterior angle property of a polygon, and other associated theories form the foundation of the NCERT Solutions Class 8 Maths Chapter 3. Students must spend a good amount of time practicing questions so as to get a good understanding of their application.

How CBSE Students can utilize NCERT Solutions Class 8 Maths Chapter 3 effectively?

To effectively utilize NCERT Solutions Class 8 Maths Chapter 3 it is advised that students go through the theory and solved examples associated with each exercise. They should then try to attempt the problem on their own. Finally, to get the best out of these solutions kids should cross-check their answers and go through the steps so that they can organize their answers in a well-structured manner.

Why Should I Practice NCERT Solutions Class 8 Maths Understanding Quadrilaterals Chapter 3?

The only way to ensure that a student has perfected his knowledge of a chapter is by practicing the questions periodically. The NCERT Solutions Class 8 Maths Understanding Quadrilaterals Chapter 3 has been given by experts with certain tips included to simplify the problems. By regular revision, kids will be confident with the topic and can get an amazing score in their examination.

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

Students can easily download the free PDF available of NCERT Solutions for Class 8 Maths chapter 3 understanding quadrilaterals from the website. All questions are discussed by the experts of maths teachers and according to the guidelines of NCERT (CBSE). While answering the exercise questions, students will understand the topic in a more comfortable and better way. We try to keep all the answers exciting and straightforward so that students can easily understand. Other than this, you can also download the NCERT Solutions for Class 8 Science . It will help to improve the syllabus and secure good marks in the examinations. NCERT Solutions for all classes and subjects are also available on Vedantu.

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

Exercise 3.1.

1. Given here are some figures.

Classify each of them on the basis of following.

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 $1,2,5,6,7$.

Simple Closed Curve

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 $1,2,5,6,7$.

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 $1,2$.

Convex Polygon

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 $2$.

Concave Polygon

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

Therefore, the concave polygon is $1$.

2. How many diagonals does each of the following have? 

A Convex Quadrilateral

Ans: Given: a convex quadrilateral

We need to find the number of diagonals in the given quadrilateral

We know that a four-sided closed shape with no vertices pointing inward is called a convex quadrilateral.

Consider, a convex quadrilateral

Now, make diagonals

Therefore, a convex quadrilateral has 2 diagonals.

A Regular Hexagon

Ans: Given: A regular hexagon

We need to find the number of diagonals of a regular hexagon.

We know that a regular hexagon is a closed curve with six equal sides.

Consider, a regular hexagon

Therefore, a regular hexagon has $9$ diagonals.

Ans:  Given: A triangle

We need to find the number of diagonals of a triangle.

We know that a triangle is a closed curve having three sides.

Consider, a triangle

Therefore, a triangle does not have any diagonal.

3. What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)

Ans: Given: A convex quadrilateral

We need to find the sum of the measures of the angles of a convex quadrilateral. Will this property hold if the quadrilateral is not convex?

Consider, a convex quadrilateral ABCD and then make a diagonal AD.

We know that the sum of angles of a triangle is ${180^ \circ }$.

So, In $\vartriangle {\text{ACD}}$

Sum of angles of $\vartriangle {\text{ACD}}$ is ${180^ \circ }$

Now, In $\vartriangle {\text{ABD}}$ 

Sum of angles of $\vartriangle {\text{ABD}}$ is ${180^ \circ }$.

Therefore, sum of angles of a convex quadrilateral will be sum of angles of $\vartriangle {\text{ACD}}$ and $\vartriangle {\text{ABD}}$

$= {180^ \circ } + {180^ \circ } $

$= {360^ \circ } $ 

Now, consider a concave quadrilateral ABCD, and then make a diagonal AC. The quadrilateral ABCD is made of two triangles, $\vartriangle {\text{ACD}}$ and $\vartriangle {\text{ABC}}$.

Consider, $\vartriangle {\text{ACD}}$

The sum of angles of the triangle are ${180^ \circ }$.

Now, consider $\vartriangle {\text{ABC}}$

The sum of the angles of triangle are ${180^ \circ }$.

Therefore, sum of the angles of quadrilateral ABCD will be

$ = {180^ \circ } + {180^ \circ } $

$   = {360^ \circ } $ 

Thus, we can say that the property hold true for a quadrilateral which is not convex because a quadrilateral can be divided into two triangles.

4. Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)

What can you say about the angle sum of a convex polygon with number of sides?

Given: the table

We need to observe table and make a statement about the angle sum of a convex polygon with number of sides.

From table, we can observe that the angle sum of a convex polygon with ${\text{n}}$sides is $({\text{n}} - 2) \times {180^ \circ }.$

Therefore, the angle sum of a convex polygon with $7$ number of sides will be

$   = (7 - 2) \times {180^ \circ } $

$   = 5 \times {180^ \circ } $

$ = {900^ \circ } $ 

Ans: Given: the table

Therefore, the angle sum of a convex polygon with $8$ number of sides will be

$   = (8 - 2) \times {180^ \circ } $

$   = 6 \times {180^ \circ } $

$   = {1080^ \circ } $ 

Therefore, the angle sum of a convex polygon with $10$ number of sides will be

$ = (10 - 2) \times {180^ \circ } $

$   = 8 \times {180^ \circ } $

$   = {1440^ \circ } $ 

From table, we can observe that the angle sum of a convex polygon with ${\text{n}}$sides will be

$ = ({\text{n}} - 2) \times {180^ \circ }$

5. What is a regular polygon? State the name of a regular polygon of 

Given: $3$ sides

We need to write the statement of a regular polygon and then state the name of the regular polygon with given number of sides.

A regular polygon is a polygon having all angles equal and all sides equal.

We know that a polygon with three equal sides and each ${60^ \circ }$ angle is a triangle.

So, it will be an equilateral triangle. The diagram will be

Ans: Given: $4$ sides

We know that a polygon with four equal sides and each ${90^ \circ }$ angle is called a square.

So, the diagram will be

Ans: Given: $6$ sides

We know that a polygon with six equal sides and each ${120^ \circ }$ angle is called a regular hexagon.

6.  Find the angle measure ${\text{'x'}}$in the following figures.

Given: A quadrilateral with angles ${50^ \circ },{130^ \circ },{120^ \circ },{\text{x}}$

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

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

${50^ \circ } + {130^ \circ } + {120^ \circ }{\text{ + x}} = {360^ \circ } $

 $  \Rightarrow {300^ \circ } + {\text{x}} = {360^ \circ } $

 $  \Rightarrow {\text{x}} = {360^ \circ } - {300^ \circ } $

 $  \Rightarrow {\text{x}} = {60^ \circ } $ 

Given: A quadrilateral with angles ${70^ \circ },{60^ \circ },{\text{x}}$

From given figure, 

$  {90^ \circ } + y = {180^ \circ } $

 $  \Rightarrow y = {180^ \circ } - {90^ \circ } $

 $  \Rightarrow y = {90^ \circ } $ 

Now, the quadrilateral has angles, ${70^ \circ },{60^ \circ },{90^ \circ }{\text{,x}}$

We know that sum of all interior angles of a quadrilateral is ${360^ \circ }.$

Thus, 

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

$   \Rightarrow {220^ \circ } + {\text{x}} = {360^ \circ } $

 $  \Rightarrow {\text{x}} = {360^ \circ } - {220^ \circ } $

$   \Rightarrow {\text{x}} = {140^ \circ } $ 

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

From given figure,

$  {70^ \circ } + {\text{a}} = {180^ \circ } $

 $ \Rightarrow {\text{a}} = {180^ \circ } - {70^ \circ } $

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

$  {60^ \circ } + {\text{b}} = {180^ \circ } $

$   \Rightarrow {\text{b}} = {180^ \circ } - {60^ \circ } $

 $  \Rightarrow {\text{b}} = {120^ \circ } $ 

Therefore, the angles of the pentagon are ${30^ \circ }{\text{,x,}}{110^ \circ },{120^ \circ }{\text{,x}}$

We know that the sum of all interior angles of a pentagon is ${540^ \circ }.$

  ${30^ \circ } + {\text{x}} + {110^ \circ } + {120^ \circ } + {\text{x}} = {540^ \circ } $

  $ \Rightarrow {260^ \circ } + 2{\text{x}} = {540^ \circ } $

  $ \Rightarrow 2{\text{x}} = {540^ \circ } - {260^ \circ } $

 $  \Rightarrow 2{\text{x}} = {280^ \circ } $

  $ \Rightarrow {\text{x}} = \dfrac{{{{280}^ \circ }}}{2} $

 $  \Rightarrow {\text{x}} = {140^ \circ } $ 

Ans: Given: a regular pentagon with angle ${\text{x}}{\text{.}}$

$ 5{\text{x}} = {540^ \circ } $

 $  \Rightarrow {\text{x}} = \dfrac{{{{540}^ \circ }}}{5} $

$   \Rightarrow {\text{x}} = {108^ \circ } $ 

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

Given: 

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

Property Used:

A linear pair can be defined as two adjacent angles that add up to ${180^ \circ },$ or two angles that combine to form a line or right angle.

Exterior angle theorem: If a polygon is convex, the total of the exterior angle measures, one at each vertex, equals${360^ \circ }$.

Using Linear pair,

$  {\text{z}} + {30^ \circ } = {180^ \circ } $

$   \Rightarrow {\text{z}} = {180^ \circ } - {30^ \circ } $

 $  \Rightarrow {\text{z}} = {150^ \circ } $ 

Again, using Linear pair

$  {\text{x}} + {90^ \circ } = {180^ \circ } $

 $  \Rightarrow {\text{x}} = {180^ \circ } - {90^ \circ } $

$   \Rightarrow {\text{x}} = {90^ \circ } $ 

Using Exterior Angle Theorem,

$  {\text{y}} = {90^ \circ } + {30^ \circ } $

$   \Rightarrow {\text{y}} = {120^ \circ } $ 

$  {\text{x}} + {\text{y}} + {\text{z}} = {90^ \circ } + {120^ \circ } + {150^ \circ } $

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

We need to find the measure of ${\text{x + y + z + w}}$.

Sum of all the interior angles of a quadrilateral is ${360^ \circ }$.

$  {\text{a}} + {60^ \circ } + {80^ \circ } + {120^ \circ } = {360^ \circ } $

$   \Rightarrow {\text{a}} + {260^ \circ } = {360^ \circ } $

$   \Rightarrow {\text{a}} = {360^ \circ } - {260^ \circ } $

$   \Rightarrow {\text{a}} = {100^ \circ } $ 

$  {\text{a}} + {\text{w}} = {180^ \circ } $

 $  \Rightarrow {100^ \circ } + {\text{w}} = {180^ \circ } $

 $  \Rightarrow {\text{w}} = {180^ \circ } - {100^ \circ } $

$   \Rightarrow {\text{w}} = {80^ \circ } $ 

$  {\text{x}} + {120^ \circ } = {180^ \circ } $

$   \Rightarrow {\text{x}} = {180^ \circ } - {120^ \circ } $

$   \Rightarrow {\text{x}} = {60^ \circ } $ 

$  {\text{y}} + {80^ \circ } = {180^ \circ } $

$   \Rightarrow {\text{y}} = {180^ \circ } - {80^ \circ } $

 $  \Rightarrow {\text{y}} = {100^ \circ } $ 

$  {\text{z}} + {60^ \circ } = {180^ \circ } $

 $  \Rightarrow {\text{z}} = {180^ \circ } - {60^ \circ } $

  $ \Rightarrow {\text{z}} = {120^ \circ } $ 

$  {\text{x}} + {\text{y}} + {\text{z}} + {\text{w}} = {60^ \circ } + {100^ \circ } + {120^ \circ } + {80^ \circ } $

Exercise-3.2

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

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

$  {\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 } $ 

$  {\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 

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

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 } $ 

Given: a regular polygon with $15$ sides

Number of sides $ = 15$

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

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.

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

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 $ 

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

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.

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

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

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.

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.

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$

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

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 } $ 

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.

Therefore, Maximum Exterior angle possible will be

$ = {120^ \circ } $

 Exercise 3.3

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

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

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, ${\text{AD}}$ = ${\text{BC}}$ 

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

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

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

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

${\text{OC}} = ...$ 

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

Hence, ${\text{OC  =  OA}}$

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

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

${\text{ABCD}}$ 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.

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 $

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)  

Given: A parallelogram 

${\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)

Given: A parallelogram

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  $ 

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 ${\text{BC}} = 4.4\;{\text{cm}}$

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 $

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.

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.

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.

Given: A parallelogram with two equal adjacent angles.

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

$  \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.

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).

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 $ 

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 ${\text{x}}{\text{.}}$

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

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 $

$  {\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?

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}}$.

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

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${\text{m}}\angle {\text{P}}$?)

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

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  

 $\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.

Different Types of Polygons, Their Sides, and Angle Sum

Polygons are closed figures having at least or more than three sides. They are made of line segments only. Polygons are classified according to the number of sides they have. Some of the most common polygons and their properties are given in the table below.

Understanding Quadrilaterals Class 8

According to geometry, a quadrilateral is a covered, two-dimensional shape that has four straight sides. The polygon has four vertices or corners. Quadrilaterals will typically imply approved forms with four sides like rectangle, square, Trapezoid, kite, or uneven and uncharacterized. From the polygon formula, we can also derive the Sum of interior angles, i.e. (n - 2) × 180, where n stands for the polygon's number of sides. However, squares, rectangles, etc., are particular types of quadrilaterals with some of their sides and equal angles.

Different Types of Quadrilaterals

There are five types of quadrilaterals based on their shape:

Parallelogram

A rectangle is a kind of quadrilateral having four right angles. Hence, every angle in a rectangle is equal (360°/4 = 90°). Moreover, the opposite planes of a rectangle are parallel and similar. Diagonals bisect each other. Letting the length of the rectangle L and breadth B then,

Area of a Rectangle = Length(L) × Breadth(B).

Perimeter = 2 × (L + B).

Properties:

Every angle of a rectangle are 90°.

Opposite sides are equal and Parallel.

Diagonals of a rectangle bisect each other.

Square is another quadrilateral having four equal sides and angles. It's also a normal quadrilateral as both its sides and angles are equal. Accurately like a rectangle, a square has four angles of 90 degrees each. We can also call it a rectangle whose two adjacent sides are equal. Letting the side of a square 'a' then,

Area = a × a = a².

Perimeter = 2 × (a + a) = 4a.

All the angles are 90°.

Each and every side is parallel and also equal to each other.

Diagonals bisect each other perpendicularly.

A parallelogram is a simple quadrilateral whose opposite sides are parallel, as we can understand by the name itself. Thus, it consists of two pairs of parallel sides. Besides, the opposite angles in a parallelogram are alike, and its diagonals divide each other.

Opposite angles are equal.

Opposite sides are equal and parallel.

Diagonals bisect each other.

The summation of any two adjacent angles is 180 degrees.

A rhombus is also a quadrilateral whose all four sides are identical in length and opposite sides parallel. However, the angles are not similar to 90°. A rhombus with right angles would match a square. We often call rhombus a diamond' as it looks similar to the diamond suit in playing cards. Letting the side of a rhombus is 'a' then, the perimeter = 4a.

Considering the length of two diagonals of the rhombus are d1 and d2, then the rhombus area = ½ × d1 × d2.

All planes are equal, and opposite planes are parallel.

The diagonals bisect each other at 90°.

A trapezium is also a quadrilateral having one parallel side pair. The parallel sides are known as 'bases,' and the rest are known as 'legs' or lateral sides. Letting the height of a trapezium 'h' then:

Perimeter = Sum of lengths of all the sides = AB + BC + CD + DA.

Area = ½ × (Sum of lengths of parallel sides) × h = ½ × (AB + CD) × h.

A trapezium is another type of quadrilateral in which it follows a single property where only one pair of opposite sides of trapezium should be parallel to each other.

Kite: A Special Quadrilateral

Kite is a quadrilateral that has the following properties.

It has two pairs of consecutive sides that are equal in size.

Diagonals intersect each other at 90°, therefore the diagonals of a kite are perpendicular to each other.

When diagonals intersect each other, only one of them will be bisected.

NCERT Solutions for Class 8 Maths - Chapterwise Solutions

Chapter 1 - Rational Numbers

Chapter 2 - Linear Equations in One Variable

Chapter 4 - Practical Geometry

Chapter 5 - Data Handling

Chapter 6 - Squares and Square Roots

Chapter 7 - Cubes and Cube Roots

Chapter 8 - Comparing Quantities

Chapter 9 - Algebraic Expressions and Identities

Chapter 10 - Visualising Solid Shapes

Chapter 11 - Mensuration

Chapter 12 - Exponents and Powers

Chapter 13 - Direct and Inverse Proportions

Chapter 14 - Factorisation

Chapter 15 - Introduction to Graphs

Chapter 16 - Playing with Numbers

We Cover all the Given Exercises of Chapter 3 Understanding Quadrilaterals:-

Benefits of ncert solutions for class 8 maths chapter 3 understanding quadrilaterals.

Our subject specialists worked hard to make the solutions to NCERT Answers for Class 8 Mathematics Chapter 3 Understanding Quadrilaterals easier to comprehend for students. We have answered all of the questions from the chapter in the NCERT textbook. Students will be able to recognise the problems and solve them correctly in the test if they refer to these solutions.

Quick Revision 

Quadrilaterals are majorly of 6 types - Squares, Rectangles, Parallelograms, Trapeziums, Rhombuses, and Kites. It is important that students learn the formulas of area and perimeter for these quadrilaterals. It is also imperative that they revise the same so that they can use these to solve sums from this chapter quickly and efficiently.

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

Perimeter of quadrilaterals.

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

In conclusion, NCERT Solutions for Class 8 Maths Chapter 3 - Understanding Quadrilaterals provide 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 begins by introducing the concept of a quadrilateral and its different types, such as parallelograms, rectangles, squares, rhombuses, and trapeziums. Each type is explained in terms of its defining properties, including sides, angles, diagonals, and symmetry.

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

1. What is the Area of a Field in the Shape of a Rectangle with Dimensions of 20 Meters and 40 Meters?

We know that the field is rectangular. Hence, we can apply the area of a rectangle to find the field area.

Length of the field = 40 Metre

Width of the field = 20 Metre

Area of the rectangular field = Length × Width = 40 × 20 = 800 Sq. Meters.

We know if the length of the rectangle is L and breadth is B then,

Area of a rectangle = Length × Breadth or L × B

Perimeter = 2 × (L + B)

So, the properties and formulas of quadrilaterals that are used in this question:

Area of the Rectangle = Length × Width

So, we used only a specific property to find the answer.

2. Find the Rest of the Angles of a Parallelogram if one Angle is 80°?

For a parallelogram ABCD, as we know the properties:

The summation of any two adjacent angles = 180 degrees.

So, the angles opposite to the provided 80° angle will likewise be 80°.

Like we know, know that the Sum of angles of any quadrilateral = 360°.

So, if ∠A = ∠C = 80° then,

Sum of ∠A, ∠B, ∠C, ∠D = 360°

Also, ∠B = ∠D

Sum of 80°, ∠B, 80°, ∠D = 360°

Or, ∠B +∠ D = 200°

Hence, ∠B = ∠D = 100°

Now, we found all the angles of the quadrilateral, which are:

3. Why are the NCERT Solutions for Class 8 Maths Chapter 3 important?

The questions included in NCERT Solutions for Chapter 3 of Class 8 Maths are important not only for the exams but also for the overall understanding of quadrilaterals. These questions have been answered by expert teachers in the subject as per the NCERT (CBSE) guidelines. As the students answer the exercises, they will grasp the topic more comfortably and in a better manner.

4. What are the main topics covered in NCERT Solutions for Class 8 Maths Chapter 3?

All the topics of the syllabus of Class 8 Maths Chapter 3 have been dealt with in detail in the NCERT Solutions by Vedantu. The chapter is Understanding Quadrilaterals and has four exercises. All the important topics in Quadrilaterals have also been carefully covered. Students can also refer to the important questions section to get a good idea about the kind of questions usually asked in the exam.

5. Do I need to practice all the questions provided in the NCERT Solutions Class 8 Maths “Understanding Quadrilaterals”?

It helps to solve as many questions as possible because Mathematics is all about practice. If you solve all the practice questions and exercises given in NCERT Solutions for Class 8 Maths, you will be able to score very well in your exams comfortably. This will also help you understand the concepts clearly and allow you to apply them logically in the questions.

6. What are the most important concepts that I need to remember in Class 8 Maths Chapter 3?

For Class 8 Maths Chapter 3, you must remember the definition, characteristics and properties of all the quadrilaterals prescribed in the syllabus, namely, parallelogram, rhombus, rectangle, square, kite, and trapezium. Also know the properties of their angles and diagonals. Regular practise will help students learn the chapter easily.

7. Is Class 8 Maths Chapter 3 Easy?

Class 8 chapter 3 of Maths is a really interesting but critical topic. It's important not only for the Class 8 exams but also for understanding future concepts in higher classes. So, to stay focused and get a good grip of all concepts, it is advisable to download the NCERT Solutions for Class 8 Maths from the Vedantu website or from the Vedantu app at free of cost. This will help the students to clear out any doubts and allow them to excel in the exams. 

NCERT Solutions for Class 8 Maths

Ncert solutions for class 8.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

Ncert solutions for class 8 maths chapter 3 understanding quadrilaterals| pdf download.

• Exercise 3.1 Chapter 3 Class 8 Maths NCERT Solutions
• Exercise 3.2 Chapter 3 Class 8 Maths NCERT Solutions
• Exercise 3.3 Chapter 3 Class 8 Maths NCERT Solutions
• Exercise 3.4 Chapter 3 Class 8 Maths NCERT Solutions

NCERT Solutions for Class 8 Maths Chapters:

How many exercises in chapter 3 understanding quadrilaterals, what is equilateral triangle, in a quadrilateral abcd, the angles a, b, c and d are in the ratio 1 : 2 : 3 : 4. find the measure of each angle of the quadrilateral., the interior angle of a regular is 108°. find the number of sides of the polygon., contact form.

Class 9 Maths Case Study Questions of Chapter 8 Quadrilaterals PDF Download

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Class 9 Maths Case Study Questions Chapter 8  are very important to solve for your exam. Class 9 Maths Chapter 8 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Maths Chapter 8 Quadrilaterals

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

These case study questions challenge students to apply their knowledge of quadrilaterals in practical scenarios, enhancing their problem-solving abilities. This article provides the Class 9 Maths Case Study Questions of Chapter 8: Quadrilaterals, enabling students to practice and excel in their examinations.

Quadrilaterals Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Maths Chapter 8 Quadrilaterals

Case Study/Passage-Based Questions

Case Study 1: Laveena’s class teacher gave students some colorful papers in the shape of quadrilaterals. She asked students to make a parallelogram from it using paper folding. Laveena made the following parallelogram.

How can a parallelogram be formed by using paper folding? (a) Joining the sides of a quadrilateral (b) Joining the mid-points of sides of a quadrilateral (c) Joining the various quadrilaterals (d) None of these

Answer: (b) Joining the mid-points of sides of quadrilateral

Which of the following is true? (a) PQ = BD (b) PQ = 1/2 BD (c) 3PQ = BD (d) PQ = 2BD

Answer: (b) PQ = 1/2 BD

Which of the following is correct combination? (a) 2RS = BD (b) RS = 1/3 BD (c) RS = BD (d) RS = 2BD

Answer: (a) 2RS = BD

Which of the following is correct? (a) SR = 2PQ (b) PQ = SR (c) SR = 3PQ (d) SR = 4PQ

Answer: (b) PQ = SR

Case Study/Passage Based Questions

Case Study 2: Anjali and Meena were trying to prove mid-point theorem. They draw a triangle ABC, where D and E are found to be the midpoints of AB and AC respectively. DE was joined and extended to F such that DE = EF and FC is also joined.

▲ADE and ▲CFE are congruent by which criterion? (a) SSS (b) SAS (c) RHS (d) ASA

Answer: (b) SAS

∠EFC is equal to which angle? (a) ∠DAE (b) ∠EDA (c) ∠AED (d) ∠DBC

Answer: (b)∠EDA

∠ECF is equal to which angle? (a) ∠EAD (b) ∠ADE (c) ∠AED (d) ∠B

Answer: (a) ∠EAD

CF is equal to (a) EC (b) BE (c) BC (d) AD

Answer: (d) AD

CF is parallel to (a) AE (b) CE (c) BD (d) AC

Answer: (c) BD

Case Study 3. A group of students is exploring different types of quadrilaterals. They encountered the following scenario:

Four friends, Aryan, Bhavana, Chetan, and Divya, participated in a geometry project. They constructed a figure with four sides and made the following observations:

• The opposite sides of the figure are parallel.
• The opposite angles of the figure are congruent.
• The figure has two pairs of congruent adjacent sides.
• The sum of the measures of the interior angles of the figure is 360 degrees.

Based on this information, the students were asked to analyze the properties of the quadrilateral they constructed. Let’s see if you can answer the questions correctly:

MCQ Questions:

Q1. The type of quadrilateral formed by their figure is: (a) Parallelogram (b) Rhombus (c) Rectangle (d) Square

Answer: (a) Parallelogram

Q2. The measure of each angle in the figure is: (a) 90 degrees (b) 120 degrees (c) 135 degrees (d) 180 degrees

Answer: (d) 180 degrees

Q3. The figure is an example of a quadrilateral that satisfies the: (a) Opposite sides are equal condition (b) Opposite angles are congruent condition (c) Diagonals bisect each other condition (d) None of the above

Answer: (b) Opposite angles are congruent condition

Q4. The sum of the measures of the exterior angles of the figure is: (a) 90 degrees (b) 180 degrees (c) 270 degrees (d) 360 degrees

Answer: (d) 360 degrees

Q5. The figure has rotational symmetry of: (a) Order 1 (b) Order 2 (c) Order 3 (d) Order 4

Answer: (a) Order 1

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Mathematics Chapter 8 Quadrilaterals with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Maths Quadrilaterals Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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