NCERT Class 8 Mathematics Chapter 3 ‘Understanding Quadrilaterals’ Exercise 3.3

Introduction:

Welcome to Online Tutorial Classes! We are excited to present a valuable resource for CBSE students studying Class 8 Mathematics. In this post, we will delve into the NCERT class 8 Mathematics Chapter 3 ‘Understanding Quadrilaterals’ Exercise 3.3, from the CBSE curriculum and provide you with a comprehensive notes and worksheets to help you master this important topic.

NCERT Class 8 Mathematics Chapter 3, Exercise 3.3 ‘Understanding Quadrilaterals’, introduces students to various types of quadrilaterals, their properties, and the relationships between their angles and sides. It covers concepts from basic polygons to special quadrilaterals like parallelograms, rectangles, and squares, enhancing students’ geometric understanding and reasoning skills.

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NCERT class 8 Mathematics Chapter 3 ‘Understanding Quadrilaterals’ Overview

NCERT Class 8 Mathematics Chapter 3, ‘Understanding Quadrilaterals’, Exercise 3.3, provides a comprehensive introduction to various types of quadrilaterals and their properties. The chapter begins with an introduction to the fundamental concepts of convex and concave polygons, followed by discussions on regular and irregular polygons. Students explore various types of quadrilaterals such as trapeziums, kites, parallelograms, rhombuses, rectangles, and squares, each defined by specific characteristics like side lengths and angle measures. Through visual aids, geometric reasoning, and practical activities, students develop a deep understanding of the properties and classifications of these quadrilaterals, along with insights into their elements like sides, angles, and diagonals.

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NCERT class 8 Maths Chapter 3 ‘Understanding Quadrilaterals’ Exercise 3.3 Solutions

Question 1

Solution:

(i) AD = BC (Opposite sides of a parallelogram are equal)

(ii) ∠ DCB = ∠DAB (Opposite angles of a parallelogram are equal)

(iii) OC = OA (Diagonals of a parallelogram bisect each other)

(IV) m ∠DAB + m ∠CDA = 180° (In a parallelogram, adjacent angles are supplementary)

Question 2

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

Solution:

(i)

y = 100° (opposite angles of a parallelogram)

x + 100° = 180° (adjacent angles of a parallelogram)

⇒ x = 180° – 100° = 80°

x = z = 80° (opposite angles of a parallelogram)

∴, x = 80°, y = 100° and z = 80°

(ii)

50° + x = 180°

⇒ x = 180° – 50° = 130° (adjacent angles of a parallelogram)

x = y = 130° (opposite angles of a parallelogram)

x = z = 130° (corresponding angle)

(iii)

x = 90° (vertical opposite angles)

x + y + 30° = 180° (sum of the angles of a triangle)

⇒ 90° + y + 30° = 180°

⇒ y = 180° – 120° = 60°

y = z = 60° (alternate angles)

(iv)

⇒ x + 80° = 180° (adjacent angles are supplementary)

⇒ x = 180° – 80° = 100°

y = 80° (opposite angles of a parallelogram are of equal)

z = y = 80° = (Alternate angles are equal)

(V)

y = 112° (opposite angles of a parallelogram are of equal)

x + y + 40° = 180° (angles of a triangle)

⇒ x = 180° – 40° – y

x = 180° – 40° – 112° = 28°

Also z = x = 28° (alternate angles)

Question 3

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°?

Solution:

(i) ∠D + ∠B = 180°

Yes, Opposite angle and hence can be a parallelogram

(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm

No, it cannot make a parallelogram, since AD ≠ BC. In a parallelogram opposite sides are equal.

(iii) ∠A = 70° and ∠C = 65°

No. Opposite angles of a parallelogram are equal.

Question 4

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

Solution:

A kite has only one pair of equal angles.

Question 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.

Solution:

Let the measures of two adjacent angles ∠A and ∠B be 3x and 2x,

∠A + ∠B = 180°

⇒ 3x + 2x = 180°

⇒ 5x = 180°

⇒ x = 36°

∴ ∠A = 3x = 3 x 36° = 108°

and ∠B = 2x = 2 × 36° = 72°

Also opposite sides of a parallelogram are equal.

∴ ∠C = ∠A = 108°

and ∠D = ∠B = 72°

Question 6

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

Solution:

Let ABCD be a parallelogram.

Sum of adjacent angles of a parallelogram = 180°

∠A + ∠B = 180°

⇒ 2∠A = 180° (∠A = ∠B)

⇒ ∠A = 90°

⇒ ∠B = 90° (∠A = ∠B)

⇒ ∠C = ∠A = 90° (Opposite angles)

⇒ ∠B = ∠D = 90° (Opposite angles)

Question 7

Solution:

y = 40° (alternate interior angle)

∠P = 70° (alternate interior angle)

∠P = ∠H = 70° (opposite angles of a parallelogram)

z = ∠H – 40°= 70° – 40° = 30°

∠H + x = 180°

⇒ x = 180° – ∠H

⇒ x = 180° – 70° = 110°

Question 8

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

Solution:

(i)

SG = NU (Opposite sides of a parallelogram)

∴ 3x = 18 cm

⇒ x = 18/3 = 6 cm

GU = SN = 26 Cm

⇒ 3y – 1 = 26

⇒ 3y = 26 + 1 = 27

⇒ y = 27/3 = 9 cm

(ii)

x + y = 16 and y + 7 = 20 (diagonals of a parallelogram bisect each other)

Solving the two equations:

y + 7 = 20

⇒ y = 20 – 7 = 13 cm

Now x + y = 16

⇒ x + 13 = 16

⇒ x = 16 – 13 = 3 cm

Question 9

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

Solution:

∠UEC = ∠L = 70° (Opposite angles of a parallelogram)

Also ∠KSI = 180 – 120 = 60° (Opposite angles of a parallelogram)

∠SEC + ∠ESI + x = 180°

⇒ x = 180° – ∠SEC – ∠ESI

⇒ x = 180° – 70° – 60° = 50°

Question 10

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

Solution:

A trapezium is a quadrilateral with a pair of parallel sides.

in the given figure:

∠M + ∠L = 100° + 80° = 180°

⇒ MN || LK

Since the quadrilateral KLMN has one pair of parallel lines, it is a trapezium.

Question 11

Find m∠C in Fig 3.27 if AB || DC.

Solution:

m∠C + m∠B = 180° (angles on the same side of transversal)

⇒ m∠C + 120° = 180°

⇒ m∠C = 180°- 120° = 60°

Question 12

Find the measure of ∠P and ∠S if SP || RQ in Fig 3.28.
(If you find m∠R, is there more than one method to find m∠P?)

Solution:

SP || RQ (given)

∴ ∠P + ∠Q = 180° (angles on the same side of transversal)

⇒ ∠P + 130° = 180°

⇒ ∠P = 180° – 130° = 50°

Also ∠R = 90°

⇒ SP ⟂ RS

∴ ∠S = 90°

Therefore ∠P = 50° and ∠S = 90°.

Another method for finding m∠P

PQRS is a quadrilateral

∴ ∠P + ∠Q + ∠R + ∠S = 360°

⇒ ∠P = 360° – ∠Q – ∠R – ∠S

⇒ ∠P = 360° – 130° – 90° – 90°

⇒ ∠P = 360° – 130° – 90° – 90° = 50°

Thus we can say that ∠P = 50°

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NCERT class 8 Mathematics Notes

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