# NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1

## Introduction :

In this chapter you come to know about polygons. A simple curved made up of only line segments is called polygon. Polygons are classified according to the number of sides or vertices they have.

A diagonal is a line segment connecting two non- consecutive vertices of a polygon. There are two types of polygons named convex and concave polygons.

NCERT Class 8 Maths Chapter 3 Understanding Quadrilaterals :

- Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1
- Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.2
- Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3
- Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.4

# Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1

**Q1. **Given here are some figures.

Classify each of the above figure on the basis of the following:

(a) Simple curve

(b) Simple closed curve

(c) Polygon

(d) Convex polygon

(e) Concave polygon

**Solution:** 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

**Q2. **How many diagonals does each of the following have?

(a) A convex quadrilateral

(b) A regular hexagon

(c) A triangle

**Solution: **

(a) A convex quadrilateral

from the above figure, the diagonals formed in a convex quadrilateral is 2( AC and BD).

(b) A regular hexagon

from the above figure, the diagonals formed in a regular hexagon is 9( AC, AD, AE, BF, BE, BD, CF,CE and DF).

(c) A triangle

from the above figure, the diagonal formed in a triangle is 0.

**Q3. **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 verify)

**Solution: **First we draw a convex quadrilateral

In quadrilateral ABCD,

∠A+∠B+∠C+∠D

∠1+∠2+∠3+∠4+∠5+∠6

arranging according to ΔABD and ΔBCD

(∠1+∠2+∠6) + (∠3+∠4+∠5)

∠1+∠2+∠6 = 180°( By angle sum property of Δ)

∠3+∠4+∠5 = 180°( By angle sum property of Δ)

s0, 180°+180° = 360°

In case of Non Convex Quadrilateral

First we draw a Non Convex Quadrilateral

considering ΔABD and ΔBCD

∠1+∠2+∠3 = 180°( By angle sum property of Δ)

∠4+∠5+ ∠6 = 180°( By angle sum property of Δ)

So,180°+180° = 360°

From the above observation, this property hold if the quadrilateral is not convex.

**Q4.**Examine the table. (Each figure is divided into triangles and the sum of the angles reduced 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

**Solution: **From the table, the angle sum of a convex polygon of n side is = (n-2) × 180°

(a) 7 = (n-2) × 180°

= (7-2) × 180°

=5× 180°

= 900°

(b) 8 = (n-2) × 180°

= (8-2) × 180°

= 6× 180°

= 1080°

(c) 10 = (n-2) × 180°

= (10-2) × 180°

= 8× 180°

=1440°

(d) n= (n-2) × 180°

**Q5. **What is a regular polygon? State the name of a regular polygon of

(i) 3 sides

(ii) 4 sides

(iii) 6 sides

**Solution: **A polygon with equal side and equal angle is called regular polygon.

i) a polygon with 3 side is called equilateral triangle.

ii) a polygon with 4 side is called square.

iii) a polygon with 6 side is called hexagon.

**Q6. **Find the angle measure x in the following figures:

**Solution:**

a) Angle sum of Quadrilateral = 360°

50°+130°+120°+x = 360°

180°+ 120°+x =360°

300°+ x = 360°

x = 360° - 300°

x = 60°

b) Angle sum of Quadrilateral = 360°

70°+ 60°+ 90° +x = 360°

220°+x = 360°

x = 360° - 220°

x = 140°

c) Angle sum of pentagon = 540°

x + x+ 30 +110°+ 120° = 540° ( 180° - 70° = 110° - by linear pair) ( 180°- 60° = 120° - by linear pair)

2x + 260° = 540°

2x = 540°- 260°

2x = 280°

x = \(\displaystyle \frac{{280°}}{2}\)

x = 140°

d) Angle sum of regular pentagon = 540°

x + x +x +x +x = 540°

5x = 540°

x = \(\displaystyle \frac{{540°}}{5}\)

x = 108°

**Q7. **(a) Find x + y + z

b) Find x + y + z + w

**Solution: **

**a)**

∠1+90°+30° = 180° (angle sum property of triangle)

∠1+120° = 180°

∠1 = 180° - 120° = 60°

Using linear pair

∠x + 90° = 180°

∠x = 180° - 90°

∠x = 90°

∠1+∠y = 180°

60°+∠y= 180°

∠y = 180°- 60°

∠y = 120°

∠z+30° = 180°

∠z = 180° - 30°

∠z = 150°

now,

x + y + z

90° + 120° + 150°

= 360°

so, x + y + z = 360°

**b)**

** **∠1+120°+80° + 60° = 360° (angle sum property of quadrilateral)

∠1+260° = 360°

∠1 = 360° - 260°

∠1 = 100°

Using linear pair

∠x + 120° = 180°

∠x = 180° - 120°

∠x = 60°

∠y + 80° = 180°

∠y = 180° - 80°

∠y = 100°

∠z + 60° = 180°

∠z = 180° - 60°

∠z = 120°

∠w+∠1 = 180°

∠w + 100° = 180°

∠w = 180° - 100°

∠w = 80°

so, x + y + z +w

= 60° + 100° + 120° + 80°

= 360°

thus, x + y + z +w = 360°

NCERT Class 8 Maths Chapter 3 Understanding Quadrilaterals :

- Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1
- Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.2
- Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3
- Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.4