# NCERT Solutions Class 8 Maths Chapter 3

## NCERT Solutions for Class 8 Mathematics Chapter 3: Understanding Quadrilaterals

Mathematics is an exciting subject which teaches calculations and theorems to students. It builds a foundation for the students for their higher studies. To understand the concepts in this subject, students need a deeper understanding of the theories and concepts. Class 8 is also an important stage in their academic life, as it prepares them for the entrance exams and test series.

Class 8 Mathematics Chapter 3: Understanding Quadrilaterals introduces a polygon as a simple closed curve made of only line segments. The chapter begins with the introduction of important concepts which need to be learned before moving on to studying quadrilaterals. The topics include the Classification of polygons based on sides, examining diagonals, concave polygons, and regular and irregular polygons.

NCERT Solutions for Class 8 Chapter 3 Mathematics are available on our Extramarks website. It covers all the chapter-end questions along with their comprehensive answers explained with proper illustrations.

By referring to these solutions students can clarify any doubts related to textual exercises of understanding quadrilaterals. Our subject matter experts provide appropriate solutions as per CBSE guidelines.

Extramarks is one of the trusted online learning platforms, and lakhs of students refer to our NCERT Solutions. Through regular practice, students will be able to increase their understanding of the chapter subjects and build a strong foundation on the concepts.

Students can also regularly check our website for the latest update. In addition, students can also refer to the NCERT solutions Class 10, NCERT solutions Class 11, and NCERT solutions Class 12.

## Key Topics Covered in NCERT Solutions for Class 8 Mathematics Chapter 3:

NCERT Solutions for Class 8 Mathematics Chapter 3 introduces many new concepts to the students. This chapter talks about the polygon as a simple closed curve. Understanding Quadrilaterals Class 8 solutions elaborate on squares, rectangles, parallelograms, kites, and rhombuses. It shows students how to solve problems based on these figures. Some of the key topics featured in NCERT Solutions for Class 8 Mathematics Chapter 3 are:

Topics

• Introduction
• Polygon
• Sum of the Measures of the Exterior Angles of a Polygon
• Some Special Parallelograms

### 1. Introduction

Paper is a model for a plane surface. We achieve a plane curve by joining several points without lifting a pencil from the paper. Besides, quadrilaterals are one type of polygon with four sides, four vertices, four angles, and two diagonals. There are various types of planes elaborated for NCERT Class 8 Mathematics Chapter 3 as follows:

• Simple closed curve
• A closed curve that is not simple
• A simple curve that is not closed
• Not a simple curve

### 2. Polygon

A simple closed curve that is made up of only line segments is called a polygon.

• Curves that are polygons
• Some curves are not polygons.
• Quadrilaterals are also one type of polygon.

Classification of polygons

We Classify polygons according to the number of sides they have. Therefore, the following are the different polygons explained in NCERT solutions for Class 8 Mathematics Chapter 3:

 Number of sides or vertices Classification 3 Triangle 4 Quadrilateral 5 Pentagon 6 Hexagon 7 Heptagon 8 Octagon 9 Nonagon 10 Decagon

Diagonals

A Diagonals is a line segment connecting two non-consecutive vertices of a polygon.

Convex and Concave Polygons

Convex polygons have no portions of their diagonals in their exteriors or any line segment joining any two different points in the interior of the polygon that lies wholly in its interior. In simple words, convex polygons have angles less than 180 degrees, and concave polygons have more than 180 degrees. To understand the crucial difference between concave and convex polygons, students can refer to NCERT solutions for Class 8 Mathematics Chapter 3.

Regular and Irregular Polygons

A regular polygon is both equiangular as well as equilateral. For example, a square has sides of equal length and angles of equal measure. Therefore, it is a regular polygon. On the other hand, a rectangle is equiangular but not equilateral.

Angle sum property

The sum of the measures of the three angles of a triangle is 180 degrees.

Some of the Measures of the Exterior Angles of a Polygon

A polygon is an enclosed 2-dimensional figure of three or more line segments. These line segments are known as the sides of the polygon. A vertex is a point at which two sides meet. An interior angle is an angle that forms between two sides adjacent to any of the vertices. An exterior angle is outside of the enclosure of a polygon by one of its sides and the extension on its adjacent side.

The summation of all exterior angles of a polygon equals 360 degrees.

Before we start studying different types of quadrilaterals, let’s first remember what a quadrilateral looks like. A quadrilateral can be described as a polygon with the following properties.

• Four sides and four vertices surround four angles.
• 360 degrees is the sum of all interior angles in a quadrilateral.
• The formula of polygon, i.e. (n -2) x180, where n is the number of sides of a polygon.

Quadrilaterals generally have sides with different lengths and angles for different measures. However, squares, rectangles, etc., some quadrilaterals have all sides and all angles equal. Therefore, the area for quadrilaterals is dependent on the type of quadrilateral.

It gets unique names based on the nature of the sides or angles of a quadrilateral. The types of quadrilaterals elaborated in our NCERT solutions for Class 8 Mathematics Chapter 3 are:

1. Trapezium

A trapezium convex quadrilateral has precisely one pair of opposing sides parallel to the other. The trapezium is a 2-dimensional shape that looks like a table when drawn on paper. A quadrilateral in Euclidean Geometry is a polygon that has four sides and four vertices.

Many examples of trapezium can be observed in real life. The trapezium rule is a significant application of trapezium. This involves dividing the area under the curve into several trapeziums and then evaluating the area of each trapezium.

To understand the essential formulas and formation of the trapezium, students can refer to the NCERT solutions for Class 8 Mathematics Chapter 3.

1. Kite

A polygon is a plane figure bounded by finite line segments to create a closed figure. A quadrilateral is a 4-sided polygon. 360 degrees is the sum of all interior angles in a quadrilateral—the number of diagonals in an n-sided polygon. Thus, a quadrilateral has two diagonals. For any n-sided polygon, the sum of all exterior angles is always 360 degrees. Therefore, a kite is a quadrilateral in which two adjacent sides are equal in length, and the diagonals intersect at right angles.

1. Parallelogram

Parallelograms are closed, four-sided, two-dimensional figures with opposite sides equal in length and are parallel to each other. The opposite angles can also be equal. It is helpful to learn the properties of a parallelogram to find the sides and angles of a parallelogram. These are the four most important properties of a parallelogram:

1. Angles of a parallelogram
• Parallelograms have opposite sides that are equal in size and are parallel to one another.
• Parallelograms can be viewed from opposite sides.
• 360 degrees is the sum of all interior angles in a parallelogram.
• Adjacent angles of Parallelograms are supplementary (180deg).
1. Diagonals of a parallelogram

Parallelogram is a Greek word that means “bounded by parallel lines” in Greek. A parallelogram is a quadrilateral that has parallel lines on the opposite side. Therefore, parallelograms will have equal and parallel sides. The line segments that connect the opposite vertices of the parallelogram are called the diagonals.

Some Special Parallelograms

1. Rhombus

A rhombus can be described as a particular type of parallelogram where opposite sides are parallel, and opposite angles are equal. These sides are all equal in length. The diagonals also intersect at right angles. A rhombus can also be called a diamond or rhombus diamond. Rhombi is the plural form for a rhombus.

1. Rectangle

A rectangle is a quadrilateral with parallel sides and all four vertices equal to 90 degrees. It is also known as an equiangular trilateral. It can also be called a parallelogram because the opposite sides of a rectangle are equal and parallel. A rectangle is a 2-dimensional flat shape. An XY plane can be used to represent a rectangle. The arms of the x-axis or y-axis indicate the rectangle’s length and width.

1. Square

A Square is a regular quadrilateral that has all sides having equal length. All four angles are equal. The angles of the square are right-angles or equal to 90 degrees. The diagonals of the square are also equal and intersect at 90 degrees. A rectangle can also be described as a square if the lengths of its opposite sides are equal.

Special parallelograms have unique formulas and properties. Thus students can refer to NCERT solutions for Class 8 Mathematics Chapter 3.

## NCERT Solutions for Class 8 Mathematics Chapter 3: Exercise & Solutions

NCERT Solutions for Class 8 Mathematics Chapter 3 provides detailed explanations that help students better understand the concepts. Students are able to develop higher-order thinking abilities and analytical skills, In addition, they can quickly and accurately memorize extensive information with the help of study material by Extramarks

This helps students deal with the pressure of the lengthy board exam syllabus. As a result, they can study in a structured manner and outperform their peers. Kindly register on our website to get access to our material

Class 8-Chapter 3 Mathematics has many questions like filling in the blanks and matching the pair. In addition, students will see objective and descriptive-type questions at the end of each solution. Click the links below for specific questions and solutions.

 Chapter 3 – Understanding Quadrilaterals Exercises Exercise 3.1 Classification of polygons Questions & Solutions Exercise 3.2 Angle Sum Property Questions & Solutions Exercise 3.3 Parallelogram Questions & Solutions Exercise 3.4 Diagonals of a parallelogram Questions & Solutions

## NCERT Exemplar for Class 8 Mathematics:

Mathematics is fascinating and requires an in-depth understanding of the concepts. Answering a range of questions can help students to understand these concepts. This helps to remove confusion and gives them a better understanding.

Students can access the NCERT exemplar on our website at any time. They are created by the best professors who are mathematics experts and have good knowledge of the subject. The questions asked in CBSE annual exams are based on NCERT books. Exemplar consists of various types of questions such as fill in the blanks, match the pair, and true or false, etc.. It will help students crack the annual exam and clear the toughest competitive exams.

The NCERT Solutions for Class 8 Mathematics Chapter 3 also includes questions from the NCERT exemplar. In addition, students will be able to examine different sub-topics, such as polygons’ types per angles, sides, and planes.

### Key Features of NCERT Solutions for Class 8 Mathematics Chapter 3:

Extramarks has a dedicated group of teachers who provide appropriate solutions for every topic in the chapter. To fully understand the topics and concepts, students can refer to our NCERT solutions.

These are the characteristics we included in our NCERT Solutions for Class 8 Mathematics Chapter 3 solutions:

• These solutions answer all questions of Chapter 3 of Class 8 Mathematics.
• NCERT Solutions for Class 8 Mathematics Chapter 3 has detailed answers to all the textual exercises to help students learn how to convey their answers in a precise way.
• This way they can comprehend the topics, and learn and retain them easily .
• These solutions include a step-by-step description of every question and concept.

Q.1 Given here are some figures. Classify each of them on the basis of the following.
(a) Simple curve
(b) Simple closed curve
(c)Polygon
(d) Convex polygon
(e) Concave polygon

Ans.

(a) Simple curve: 1, 2, 5, 6, 7
(b) Simple closed curve: 1, 2, 5, 6, 7
(c) Polygon: 1, 2
(d) Convex polygon: 2
(e) Concave polygon: 1

Q.2 How many diagonals does each of the following have?
(b) A regular hexagon
(c) A triangle

Ans.

(a) A convex quadrilateral has 2 diagonals.

(b) A regular hexagon has 9 diagonals.

(c) A triangle does not have any diagonal.

Q.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. Let ABCD be a convex quadrilateral. Draw a diagonal BD. Now, this quadrilateral is divided into two triangles ABD and BC

$\begin{array}{l}\text{We know that, the sum of interior angles of a triangle is 18}{0}^{\circ }.\\ \therefore \text{The sum of interior angles of a quadrilateral}=\text{18}{0}^{\circ }\text{+18}{0}^{\circ }\\ \text{}={360}^{\circ }\\ \text{Hence,the sum of the measures of the angles of a convex}\\ \text{quadrilateral is}\text{\hspace{0.17em}}{360}^{\circ }.\end{array}$

If the quadrilateral is not convex, then also this property holds. This quadrilateral can also be divided into two triangles ABD and ACD.

$\begin{array}{l}\text{We know that, the sum of interior angles of a triangle is 18}{0}^{\circ }.\\ \therefore \text{The sum of interior angles of a quadrilateral}=\text{18}{0}^{\circ }\text{+18}{0}^{\circ }\\ \text{}={360}^{\circ }\\ \text{Hence,the sum of the measures of the angles of a concave}\\ \text{quadrilateral is}\text{\hspace{0.17em}}{360}^{\circ }.\end{array}$

Q.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?

(a) 7 (b) 8 (c) 10 (d) n

Ans.

From the table it can be observed that the angle sum of a convex polygon of n sides is (n – 2)×180°.

Therefore, for given number of sides, the sum of the angles are:

$\begin{array}{l}\left(\mathrm{a}\right)\text{}7:\left(7-2\right)×{180}^{\circ }={900}^{\circ }\\ \left(\mathrm{b}\right)\text{}8:\left(8-2\right)×{180}^{\circ }={1080}^{\circ }\\ \left(\mathrm{c}\right)\text{}10:\left(10-2\right)×{180}^{\circ }={1440}^{\circ }\\ \left(\mathrm{d}\right)\text{}\mathrm{n}:\left(\mathrm{n}-2\right)×{180}^{\circ }\end{array}$

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

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

Ans.

A polygon which is “equilateral” and “equiangular” is called a regular polygon.

(i) 3 sides: Equilateral triangle (ii) 4 sides: Square (iii) 6 sides: Regular Hexagon Q.6 Find the angle measure x in the following figures.    Ans. $\begin{array}{l}\text{We know that the sum of interior angles of a}\\ \text{quadrilateral is}{360}^{\circ }.\\ \\ \therefore \text{From the given quadrilateral ,we have}\\ {50}^{\circ }+{130}^{\circ }+{120}^{\circ }+\mathrm{x}={360}^{\circ }\\ ⇒300\mathrm{°}+\mathrm{x}={360}^{\circ }\\ ⇒\mathrm{x}={360}^{\circ }-300\mathrm{°}\\ ⇒\mathrm{x}=60\mathrm{°}\end{array}$ $\begin{array}{l}\mathrm{From}\text{}\mathrm{the}\text{}\mathrm{figure},\\ \text{}90\mathrm{°}+\mathrm{a}=180\mathrm{°}\text{}\left(\mathrm{Linear}\text{}\mathrm{pair}\right)\\ ⇒\mathrm{a}=180\mathrm{°}-90\mathrm{°}=90\mathrm{°}\\ \\ \mathrm{Now},\mathrm{the}\text{}\mathrm{sum}\text{}\mathrm{of}\text{}\mathrm{all}\text{}\mathrm{the}\text{}\mathrm{interior}\text{}\mathrm{angles}\text{}\mathrm{of}\\ \mathrm{a}\text{}\mathrm{quadrilateral}\text{}\mathrm{is}\text{}360\mathrm{°}.\text{}\\ \\ \therefore \mathrm{in}\text{}\mathrm{the}\text{}\mathrm{given}\text{}\mathrm{quadrilateral},\\ 60\mathrm{°}+70\mathrm{°}+\mathrm{x}+90\mathrm{°}=360\mathrm{°}\\ ⇒220\mathrm{°}+\mathrm{x}=360\mathrm{°}\\ ⇒\mathrm{x}=140\mathrm{°}\end{array}$ From the figure, we have
70+a =180° (Linear Pair)
a = 110°
Also, 60°+b = 180° (Linear Pair)
b= 120°
Since the pentagon has 5 sides, so the sum of all the interior
angles of a pentagon is 540°.

Therefore, we have
120°+110°+30°+x+x = 540°
260° + 2x = 540°
2x= 280°
x = 140° $\begin{array}{l}\text{We know that the sum of all the interior angles of a}\\ \text{pentagon is 54}{0}^{\circ }.\\ \text{Since, all the sides are equal,therefore, all the angles are equal.}\\ \text{Hence, it is a regular pentagon.}\\ \therefore \text{5}\mathrm{x}=\text{54}0\mathrm{°}\\ ⇒\mathrm{x}=\text{1}0\text{8}\mathrm{°}\end{array}$

Q.7  (a) Find x + y + z
(b) Find x + y + z + w

Ans.

$\begin{array}{l}\left(\text{a}\right)\\ \mathrm{x}+\text{9}0\mathrm{°}=\text{18}0\mathrm{°}\text{}\left(\text{Linear pair}\right)\\ ⇒\mathrm{x}=\text{9}0\mathrm{°}\\ \\ \mathrm{z}+\text{3}0\mathrm{°}\text{}=\text{18}0\mathrm{°}\text{}\left(\text{Linear pair}\right)\\ ⇒\mathrm{z}=\text{15}0\mathrm{°}\\ \\ \text{We know that, the exterior angle is equal to the}\\ \text{sum of interior opposite angles.}\\ \therefore \mathrm{y}=\text{9}0\mathrm{°}\text{}+\text{3}0\mathrm{°}\text{}\\ ⇒\mathrm{y}=\text{12}0\mathrm{°}\\ \\ \mathrm{Hence},\\ \mathrm{x}\text{}+\text{}\mathrm{y}\text{}+\text{}\mathrm{z}=\text{9}0\mathrm{°}+\text{12}0\mathrm{°}+\text{15}0\mathrm{°}\text{}\\ \text{}=\text{36}0\mathrm{°}\end{array}$ $\begin{array}{l}\text{(b)}\\ \text{We know that, the sum of the measures of all interior}\\ \text{angles of a quadrilateral is 3}{60}^{\circ }.\text{}\\ \\ \text{Therefore,}\\ a+\text{6}0\mathrm{°}+\text{8}0\mathrm{°}+\text{12}0\mathrm{°}=\text{36}0\mathrm{°}\\ ⇒a+\text{26}0\mathrm{°}=\text{36}0\mathrm{°}\\ ⇒a=\text{1}00\mathrm{°}\end{array}$ $\begin{array}{l}\mathrm{Also},\\ \mathrm{x}+\text{12}0\mathrm{°}=\text{18}0\mathrm{°}\text{}\left(\text{Linear pair}\right)\\ ⇒\mathrm{x}=\text{6}0\mathrm{°}\\ \\ \mathrm{y}+\text{8}0\mathrm{°}=\text{18}0\mathrm{°}\text{}\left(\text{Linear pair}\right)\\ ⇒\mathrm{y}=\text{1}00\mathrm{°}\\ \\ \mathrm{z}+\text{6}0\mathrm{°}=\text{18}0\mathrm{°}\text{}\left(\text{Linear pair}\right)\\ ⇒\mathrm{z}=\text{12}0\mathrm{°}\\ \\ \mathrm{w}+\text{1}00\mathrm{°}=\text{18}0\mathrm{°}\text{}\left(\text{Linear pair}\right)\\ \mathrm{w}=\text{8}0\mathrm{°}\\ \\ \text{Hence,}\\ \mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{w}=\text{6}0\mathrm{°}+\text{1}00\mathrm{°}+\text{12}0\mathrm{°}+\text{8}0\mathrm{°}\\ \text{}=\text{36}0\mathrm{°}\end{array}$

Q.8 Find x in the following figures.  Ans.

$\begin{array}{l}\text{(a)}\\ \text{We know that the sum of all exterior angles of any}\\ \text{polygon is 36}{0}^{\circ }.\\ \therefore \text{125}\mathrm{°}+\text{125}\mathrm{°}+\mathrm{x}=\text{36}0\mathrm{°}\\ ⇒\text{25}0\mathrm{°}+\mathrm{x}=\text{36}0\mathrm{°}\\ ⇒\mathrm{x}=\text{11}0\mathrm{°}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{We know that the sum of all exterior angles of any}\\ \text{polygon is 36}{0}^{\circ }.\\ \therefore \text{6}0°+\text{9}0°+\text{7}0°+x+\text{9}0°=\text{36}0°\\ ⇒\text{31}0°+x=\text{36}0°\\ ⇒x=\text{5}0°\end{array}$

Q.9 Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides.

Ans.

$\begin{array}{l}\text{(i)}\\ \text{We know that, the}\mathrm{sum}\mathrm{of}\mathrm{}\mathrm{all}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angles}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\\ \mathrm{given}\text{}\mathrm{polygon}={360}^{\circ }\\ \\ \mathrm{Thus},\mathrm{}\mathrm{measure}\mathrm{}\mathrm{of}\mathrm{}\mathrm{each}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angle}\mathrm{}\mathrm{of}\mathrm{}\mathrm{a}\mathrm{}\mathrm{regular}\mathrm{}\\ \mathrm{polygon}\text{}\mathrm{of}\text{}9\text{}\mathrm{sides}=\frac{{360}^{\circ }}{9}={40}^{\circ }\\ \end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\\ \mathrm{We}\mathrm{know}\mathrm{that},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{}\mathrm{all}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angles}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\\ \mathrm{given}\text{}\mathrm{polygon}={360}^{\circ }\\ \\ \mathrm{Thus},\mathrm{}\mathrm{measure}\mathrm{}\mathrm{of}\mathrm{}\mathrm{each}\mathrm{}\mathrm{exterior}\mathrm{}\mathrm{angle}\mathrm{}\mathrm{of}\mathrm{}\mathrm{a}\mathrm{}\mathrm{regular}\mathrm{}\\ \mathrm{polygon}\mathrm{}\mathrm{of}\mathrm{}15\mathrm{}\mathrm{sides}=\frac{{360}^{\circ }}{15}={24}^{\circ }\end{array}$

Q.10 How many sides does a regular polygon have if the measure of an exterior angle is 24°?

Ans.

$\begin{array}{l}\text{We know that, the sum of all exterior angles of the given}\\ \text{polygon}=\text{36}0°\\ \\ \text{It is given that the measure of each exterior angle}=\text{24}°\\ \\ \text{Thus},\text{number of sides of the regular polygon=}\frac{{360}^{\circ }}{24°}=15\end{array}$

Q.11 How many sides does a regular polygon have if each of its interior angles is 165°?

Ans.

$\begin{array}{l}\text{Given,}\\ \text{Measure of each interior angle}=\text{165}°\\ \\ \therefore \text{Measure of each exterior angle}=\text{18}0°-\text{165}°=\text{15}°\\ \\ \text{Also, the sum of all exterior angles of any polygon is 36}0º.\\ \\ \mathrm{Therefore},\text{number of sides of the polygon =}\frac{{360}^{\circ }}{{15}^{\circ }}=24\end{array}$

Q.12 (a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?
(b) Can it be an interior angle of a regular polygon? Why?

Ans.

$\begin{array}{l}\text{(a) Given:}\\ \text{Exterior angle}={22}^{\circ }\\ \\ \text{We know that,}\\ \text{the sum of all the exterior angles of a polygon is 3}{60}^{\circ }.\end{array}$

$\begin{array}{l}\mathrm{Also},\text{all the exterior angles are of equal measure.That}\\ {\text{means 22}}^{\circ }\text{should be exact divisor of 3}{60}^{\circ }.\\ \\ \mathrm{Since},{\text{\hspace{0.17em}22}}^{\circ }\text{\hspace{0.17em}is not an exact divisor of 3}{60}^{\circ },\mathrm{therefore},\text{such}\\ \text{a polygon is not possible.}\\ \end{array}$

$\begin{array}{l}\text{(b) Given:}\\ \text{Interior angle}=\text{22}°\\ \\ \therefore \text{Exterior angle}=\text{18}0°-\text{22}°=\text{158}°\\ \\ \text{As 36}0°\text{is not a perfect multiple of 158}°,so\text{such a polygon}\\ \text{is not possible}\text{.}\end{array}$

Q.13 (a) What is the minimum interior angle possible for a regular polygon? Why?

(b) What is the maximum exterior angle possible for a regular polygon?

Ans. For a regular polygon, as the number of sides decreases, interior angle also decreases and exterior angle increases.

Therefore, polygon with maximum exterior angle and minimum interior angle is an equilateral triangle as it has the minimum number of sides possible for a polygon.

Consider an equilateral triangle. The exterior angle of this triangle will be the maximum exterior angle possible for any regular polygon.

Therefore,

$\mathrm{Exterior}\text{}\mathrm{angle}\text{}\mathrm{of}\text{}\mathrm{an}\text{}\mathrm{equilateral}\text{}\mathrm{triangle}=\frac{{360}^{\circ }}{3}=120\mathrm{°}$

$\begin{array}{l}\mathrm{Hence},\text{}\mathrm{maximum}\text{}\mathrm{possible}\text{}\mathrm{measure}\text{}\mathrm{of}\text{}\mathrm{exterior}\text{}\mathrm{angle}\text{}\mathrm{for}\text{}\mathrm{any}\\ \mathrm{polygon}\text{}\mathrm{is}\text{}120\mathrm{°}.\text{}\\ \\ \mathrm{Also},\text{}\mathrm{we}\text{}\mathrm{know}\text{}\mathrm{that}\text{}\mathrm{an}\text{}\mathrm{exterior}\text{}\mathrm{angle}\text{}\mathrm{and}\text{}\mathrm{an}\text{}\mathrm{interior}\text{}\mathrm{angle}\\ \mathrm{are}\text{}\mathrm{always}\text{}\mathrm{in}\text{}\mathrm{a}\text{}\mathrm{linear}\text{}\mathrm{pair}.\\ \mathrm{Hence},\text{}\mathrm{minimum}\text{}\mathrm{interior}\text{}\mathrm{angle}=180\mathrm{°}-120\mathrm{°}=60\mathrm{°}\end{array}$

Q.14 Given a parallelogram ABCD. Complete each statement along with the definition or property used. 2. ∠ DCB = ……
3. OC = ……
4. m ∠DAB + m ∠CDA = …..

Ans.

(Opposite sides are of equal length in a parallelogram.)

2. ∠ DCB = ∠ DAB

(Opposite angles are equal in measure.)

3. OC = OA

(In a parallelogram, diagonals bisect each other)

4. m ∠DAB + m ∠CDA = 180°

(In a parallelogram, adjacent angles are supplementary to each other.)

Q.15 Consider the following parallelograms. Find the values of the unknowns x, y, z. Ans.

(i) $\begin{array}{l}\mathrm{x}+\text{1}00\mathrm{°}=\text{18}0\mathrm{°}\left(\begin{array}{l}\text{Adjacent}\mathrm{}\text{angles of a parallelogram}\\ \text{are supplementary}\end{array}\right)\\ ⇒\mathrm{x}=\text{8}0\mathrm{°}\\ \\ \text{Since opposite angles of a parallelogram are equal}\\ \therefore \angle \mathrm{A}=\angle \mathrm{C}\\ ⇒\mathrm{z}=\mathrm{x}=\text{8}0\mathrm{º}\\ \\ \mathrm{Also},\angle \mathrm{B}=\angle \mathrm{D}\\ \mathrm{Hence},\mathrm{y}=\text{1}00\mathrm{°}\text{}\\ \\ \therefore \mathrm{x}=\text{8}0\mathrm{°},\mathrm{y}=100\mathrm{°}\text{\hspace{0.17em}}\mathrm{and}\text{}\mathrm{z}=\text{8}0\mathrm{°}\end{array}$

(ii) $\begin{array}{l}\mathrm{Since}\text{}\mathrm{adjacent}\text{}\mathrm{angles}\text{}\mathrm{of}\text{}\mathrm{a}\text{}\mathrm{parallelogram}\text{}\mathrm{are}\\ \mathrm{supplementary},\text{}\mathrm{therefore},\\ 50\mathrm{°}\text{}+\mathrm{y}=\text{}180\mathrm{°}\\ ⇒\mathrm{y}=\text{}130\mathrm{°}\\ \mathrm{Also},\text{opposite angles of a parallelogram are equal}\\ \therefore \mathrm{x}=\mathrm{y}=130\mathrm{°}\\ \\ \text{Now,’z’ and ‘x’ are the corresponding angles.}\\ \therefore \mathrm{z}=\mathrm{x}=130\mathrm{°}\\ \\ \therefore \mathrm{x}=130\mathrm{°},\mathrm{y}=130\mathrm{°}\text{\hspace{0.17em}}\mathrm{and}\text{}\mathrm{z}=\text{13}0\mathrm{°}\end{array}$

(iii) $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{x}=\text{9}0\mathrm{°}\text{}\left(\text{Vertically opposite angles}\right)\\ \\ \mathrm{Also},\\ \mathrm{x}+\mathrm{y}+\text{3}0\mathrm{°}=\text{18}0\mathrm{°}\text{}\left(\text{By angle sum property of triangles}\right)\\ ⇒\text{12}0\mathrm{°}+\mathrm{y}=\text{18}0\mathrm{°}\\ ⇒\mathrm{y}=\text{6}0\mathrm{°}\\ \\ \mathrm{Now},\mathrm{z}=\mathrm{y}=\text{6}0\mathrm{°}\text{}\left(\text{Alternate interior angles}\right)\\ \\ \therefore \mathrm{x}=90\mathrm{°},\mathrm{y}=60\mathrm{°}\text{\hspace{0.17em}}\mathrm{and}\text{}\mathrm{z}=60\mathrm{°}\end{array}$

(iv) $\begin{array}{l}\text{In the given figure},\\ \mathrm{z}=\text{8}0\mathrm{°}\text{}\left(\text{Corresponding angles}\right)\\ \\ \mathrm{y}=\text{8}0\mathrm{°}\left(\text{Opposite angles are equal in a parallelogram}\right)\\ \mathrm{Also},\\ \mathrm{x}+\mathrm{y}=\text{18}0\mathrm{°}\text{}\left(\text{Adjacent angles are supplementary}\right)\\ ⇒\mathrm{x}=\text{18}0\mathrm{°}-\text{8}0\mathrm{°}\\ \text{}=\text{1}00\mathrm{°}\\ \\ \therefore \mathrm{x}=100\mathrm{°},\mathrm{y}=80\mathrm{°}\text{\hspace{0.17em}}\mathrm{and}\text{}\mathrm{z}=80\mathrm{°}\end{array}$

(v)

$\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{y}=\text{112}\mathrm{°}\text{}\left(\text{Opposite angles are equal}\right)\\ \\ \mathrm{In}\text{}\mathrm{a}\text{​}\mathrm{triangle},\\ \mathrm{x}+\text{}\mathrm{y}+\text{4}0\mathrm{°}=\text{18}0\mathrm{°}\text{}\left(\text{By angle sum property}\right)\\ ⇒\mathrm{x}+\text{112}\mathrm{°}+\text{4}0\mathrm{°}=\text{18}0\mathrm{°}\\ ⇒\mathrm{x}+\text{152}\mathrm{°}\text{}=\text{18}0\mathrm{°}\\ ⇒\mathrm{x}=\text{28}\mathrm{°}\\ \\ \mathrm{Also},\\ \mathrm{z}=\mathrm{x}=\text{28}\mathrm{°}\text{}\left(\text{Alternate interior angles}\right)\\ \\ \therefore \mathrm{x}=\text{28}\mathrm{°},\mathrm{y}=\text{112}\mathrm{°}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\mathrm{z}=\text{28}\mathrm{°}\end{array}$

Q.16 Can a quadrilateral ABCD be a parallelogram if

(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?

Ans. A parallelogram has opposite sides equal and parallel. The opposite angles are also equal. The adjacent angles are supplementary.

1. So, if ∠D + ∠B = 180°, a quadrilateral ABCD may or may not be a parallelogram as all the other conditions of the parallelogram should also be fulfilled by a quadrilateral ABCD.
2. In a parallelogram the opposite sides should be of equal length. Here, AB is equal to CD but AD is not equal to BC. Hence, ABCD is not a parallelogram.
3. In a parallelogram the opposite angles are of equal length. Here, ∠A is not equal to ∠C. Hence, ABCD is not a parallelogram.

Q.17 Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

Ans. Here, ABCD is a quadrilateral that is not a parallelogram but ∠B=∠D.

Q.18 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. $\begin{array}{l}\text{The two angles are in the ratio 3 : 2.}\\ \text{Let}\angle \text{A = 3x and}\angle \text{B = 2x}\\ \text{We know that , the adjacent angles are supplementary in a parallelogram.}\end{array}$ $\begin{array}{l}\therefore \angle \mathrm{A}\text{}+\angle \mathrm{B}\text{}=\text{}180\mathrm{°}\\ ⇒3\mathrm{x}+\text{}2\mathrm{x}=\text{}180\mathrm{°}\\ ⇒5\mathrm{x}=\text{}180\mathrm{°}\\ ⇒\mathrm{x}=\frac{180\mathrm{°}}{5}=36\mathrm{°}\\ \mathrm{Hence},\\ \angle \mathrm{A}=3\mathrm{x}=3×36\mathrm{°}=108\mathrm{°}\text{}\\ \angle \mathrm{B}=2\mathrm{x}=2×36\mathrm{°}=72\mathrm{°}\\ \\ \mathrm{Since},\text{opposite angles of a paralle}\mathrm{log}\text{ra}\mathrm{m}\text{}\mathrm{are}\\ \mathrm{equal},\mathrm{therefore},\\ \text{}\angle \text{A=}\angle \text{C and}\angle \text{B=}\angle \text{D}\\ \therefore \angle \mathrm{A}=\angle \mathrm{C}\text{}=\text{}108\mathrm{°}\text{\hspace{0.17em}}\mathrm{and}\text{}\angle \mathrm{B}=\angle \mathrm{D}=72\mathrm{°}\\ \text{}\end{array}$

Thus, the measures of the angles of the parallelogram are 108°, 72°, 108°, and 72°.

Q.19 Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

Ans. In a parallelogram the adjacent angles are supplementary

$\begin{array}{l}\therefore \angle \text{A}+\angle \text{B}=\text{18}0\mathrm{°}\\ \text{Since}\angle \text{A}=\angle \text{B},\\ ⇒\text{2}\angle \text{A}=\text{18}0\mathrm{°}\text{}\\ ⇒\angle \text{A}=\text{9}0\mathrm{°}\\ ⇒\angle \text{B}=\text{9}0\mathrm{°}\\ \\ \mathrm{Now},\angle \text{C}=\angle \text{A}\left(\text{Opposite angles}\right)\\ ⇒\angle \text{C}=\text{9}0\mathrm{°}\\ \angle \text{D}=\angle \text{B}\left(\text{Opposite angles}\right)\\ ⇒\angle \text{D}=\text{9}0\mathrm{°}\\ \text{}\\ \text{Hence},\text{all the angles of the parallelogram are 9}0\mathrm{º}.\end{array}$

Q.20 The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them. Ans. $\begin{array}{l}\text{7}0\mathrm{°}=\mathrm{z}\text{}+\text{4}0\mathrm{°}\text{}\left(\text{Corresponding angles of a parallelogram}\right)\\ ⇒\text{7}0\mathrm{°}-\text{4}0\mathrm{°}=\mathrm{z}\\ ⇒\text{z}=\text{3}0\mathrm{°}\\ \\ \mathrm{Also},\\ \mathrm{y}=\text{4}0\mathrm{°}\text{}\left(\text{Alternate interior angles}\right)\\ \\ \mathrm{Now},\angle \mathrm{E}+\angle \mathrm{H}=180\mathrm{°}\left(\text{Adjacent angles are supplementary}\right)\\ ⇒\mathrm{x}+\left(\mathrm{z}\text{}+\text{4}0\mathrm{º}\right)=\text{18}0\mathrm{°}\text{}\\ ⇒\mathrm{x}+\left(30\mathrm{°}\text{}+\text{4}0\mathrm{º}\right)=\text{18}0\mathrm{°}\text{}\\ ⇒\mathrm{x}+\text{7}0\mathrm{°}=\text{18}0\mathrm{°}\\ ⇒\mathrm{x}=\text{11}0\mathrm{°}\\ \\ \therefore \mathrm{x}=\text{11}0\mathrm{°},\mathrm{y}=40\mathrm{°}\text{}\mathrm{and}\text{\hspace{0.17em}}\mathrm{z}=30\mathrm{°}\end{array}$

Q.21 The following figures GUNS and RUNS are parallelograms . Find x and y. (Lengths are in cm) Ans.

(i) We know that, the lengths of opposite sides of a parallelogram are equal to each other.

Therefore, GU = SN and SG =NU

$\begin{array}{l}⇒\text{3}\mathrm{y}-\text{1}=\text{26}\\ ⇒\text{3}\mathrm{y}=\text{27}\\ ⇒\mathrm{y}=\text{9}\\ \\ \text{Also, SG}=\text{NU}\\ ⇒\text{3}\mathrm{x}=\text{18}\\ ⇒\mathrm{x}=\text{6}\end{array}$

Hence, the measures of x and y are 6 cm and 9 cm.

(ii) We know that the diagonals of a parallelogram bisect each other.

$\begin{array}{l}\mathrm{y}+\text{7}=\text{2}0\text{}\mathrm{and}\text{\hspace{0.17em}}\mathrm{x}+\mathrm{y}=\text{16}\\ \\ \mathrm{Now},\mathrm{y}+\text{7}=\text{2}0\text{}\\ ⇒\mathrm{y}=\text{13}\\ \\ \mathrm{Also},\mathrm{x}+\mathrm{y}=\text{16}\\ ⇒\mathrm{x}+\text{13}=\text{16}\\ ⇒\mathrm{x}=\text{3}\end{array}$

Hence, the measures of x and y are 3 cm and 13 cm.

Q.22 In the above figure both RISK and CLUE are parallelograms. Find the value of x.

Ans.

$\begin{array}{l}\text{In parallelogram RISK},\\ \angle \text{RKS}+\angle \text{ISK}=\text{18}0°\begin{array}{l}\text{}\left[\text{Adjacent angles of a parallelogram}\\ \text{are supplementary.}\right]\end{array}\\ ⇒\text{12}0°+\angle \text{ISK}=\text{18}0°\\ ⇒\angle \text{ISK}=\text{6}0°\\ \\ \\ \text{In parallelogram CLUE},\\ \angle \text{ULC}=\angle \text{CEU}=\text{7}0°\left(\begin{array}{l}\text{opposite angles of}\\ \text{a parallelogram are equal}.\end{array}\right)\\ \\ \text{Now, in a triangle,}\\ x+\text{6}0°+\text{7}0°=\text{18}0°\text{}\left(\text{By angle sum property}\right)\\ ⇒x=\text{5}0°\end{array}$

Q.23 Explain how this figure is a trapezium. Which of its two sides are parallel?

Ans. Here, ∠NML + ∠MLK = 180°

Hence, NM||LK and ML is a transversal.(If a transversal intersects the two given lines such that the sum of the angles on the same side of transversal is 180º, then the given two lines will be parallel to each other.)

Therefore, KLMN is a trapezium as it has one pair of parallel lines.

Q.24

$\mathrm{Find}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{m}\angle \mathrm{C}\mathrm{in}\mathrm{Fig}.\mathrm{if}\text{\hspace{0.17em}}\overline{\mathrm{AB}}||\overline{\mathrm{DC}}.$ Ans.

$\begin{array}{l}\mathrm{}\text{Given}:\mathrm{AB}||\mathrm{BC}\\ ⇒\angle \text{B}+\angle \text{C}=\text{18}0\mathrm{°}\begin{array}{l}\left[\text{Angles on the same side of transversal}\\ \text{are supplementary}.\right]\text{}\end{array}\\ ⇒\text{12}0\mathrm{°}+\angle \text{C}=\text{18}0\mathrm{°}\\ ⇒\angle \text{C}=\text{6}0\mathrm{°}\end{array}$

Q.25

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{measure}\mathrm{of}\angle \mathrm{P}\mathrm{and}\angle \mathrm{S}\mathrm{if}\text{\hspace{0.17em}}\mathrm{SP}||\mathrm{RQ}\mathrm{in}\mathrm{Fig}.\left(\mathrm{If}\mathrm{you}\mathrm{find}\text{\hspace{0.17em}}\\ \mathrm{m}\angle \mathrm{R},\mathrm{is}\mathrm{there}\mathrm{more}\mathrm{than}\mathrm{one}\mathrm{method}\mathrm{to}\mathrm{find}\text{\hspace{0.17em}}\mathrm{m}\angle \mathrm{P}?\right)\end{array}$ Ans.

$\begin{array}{l}\mathrm{Given}:\mathrm{SP}||\mathrm{RQ}\\ \therefore \angle \text{P}+\angle \text{Q}=\text{18}0\mathrm{°}\text{}\left(\text{Angles on the same side of transversal}\right)\\ ⇒\angle \text{P}+\text{13}0\mathrm{°}=\text{18}0\mathrm{°}\\ ⇒\angle \text{P}=\text{5}0\mathrm{°}\\ \\ \mathrm{Also},\\ \angle \text{R}+\angle \text{S =18}0\mathrm{°}\text{}\left(\text{Angles on the same side of transversal}\right)\\ ⇒\text{9}0\mathrm{°}+\angle \text{R}=\text{18}0\mathrm{°}\\ ⇒\angle \text{S}=\text{9}0\mathrm{°}\\ \end{array}$

$\begin{array}{l}\text{Yes,}\text{there is one more method to find the measure of}\angle \text{P}.\\ \text{We can apply the angle sum property of a quadrilateral}\\ \text{to find the}\mathrm{m}\angle \text{P}.\end{array}$

Q.26 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.

Ans.

(a) False
(b) True
(c) True
(d) False
(e) False
(f) True
(g) True
(h) True

Q.27 Identify all the quadrilaterals that have.

(a) four sides of equal length

(b) four right angles

Ans.

(a) four sides of equal length: Rhombus and Square

(b) four right angles: Square and rectangle

Q.28 Explain how a square is.
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle

Ans.

$\begin{array}{l}\text{(i) Quadrilateral: A square has four sides}.\mathrm{Therefore},\text{it is a}\\ \text{quadrilateral}\text{.}\\ \\ \left(\text{ii}\right)\text{Parallelogram:The opposite sides of a square are parallel}\\ \text{to each other and diagonals bisect each other}\text{.}\mathrm{Hence},\text{it is a}\\ \text{parallelogram}\text{.}\\ \\ \left(\text{iii}\right)\text{Rhombus:All the four sides of a square are of the same}\\ \text{length and diagonals bisect each other at}{90}^{\circ }.\mathrm{Hence},\text{it is a}\\ \text{rhombus}\text{.}\\ \\ \left(\text{iv}\right)\text{Rectangle:A square is a rectangle since each interior}\\ \text{angle measures 9}0°\text{and diagonals are of equal length}\text{.}\\ \mathrm{Hence},\text{it is a rectangle}\text{.}\end{array}$

Q.29 Name the quadrilaterals whose diagonals
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal

Ans.

(i) The quadrilaterals whose diagonals bisect each other are: parallelogram, square, rhombus and rectangle.
(ii) The quadrilaterals whose diagonals are perpendicular bisectors of each other are: square and rhombus.
(iii) The quadrilaterals having equal diagonals are: square and rectangle.

Q.30 Explain why a rectangle is a convex quadrilateral.

Ans. A convex quadrilateral is a quadrilateral whose diagonals lie in the interior of it and its all interior angles are less than 180°.

We know that, a rectangle has two diagonals. Both the diagonals lie in the interior of the rectangle. Also, each interior angle of a rectangle measures 90°.

Hence, it is a convex quadrilateral.

Q.31 ABC is a right-angled triangle and O is the midpoint 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). Ans.

Given: ABC is a right angled triangle and O is the midpoint of AC.

Draw lines AD and DC such that AD||BC, AB||DC and join OD.

Therefore, AD = BC, AB = DC.

Since opposite sides are equal and parallel to each other and all the interior angles are of 90º, therefore, ABCD is a rectangle.

Also, diagonals bisect each other in a rectangle.

Hence, AO= OC = BO = OD

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

## 1. Which are the key topics covered in NCERT solutions for Class 8 Mathematics Chapter 3?

Each topic has a separate exercise,., Each exercise is covered in the NCERT solutions for Class 8 Mathematics Chapter 3. The following key topics are covered in the solutions:

• Introduction
• Polygon
• Some of the Measures of the Exterior Angles of a Polygon