## NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

- Class 8 Maths Understanding Quadrilaterals Exercise 3.1
- Class 8 Maths Understanding Quadrilaterals Exercise 3.2
- Class 8 Maths Understanding Quadrilaterals Exercise 3.3
- Class 8 Maths Understanding Quadrilaterals Exercise 3.4
- Understanding Quadrilaterals Class 8 Extra Questions

**NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1**

Ex 3.1 Class 8 Maths Question 1.

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) and (7)

(b) Simple closed curve: (1), (2), (5), (6) and (7)

(c) Polygon: (1) and (2)

(d) Convex polygon: (2)

(e) Concave polygon: (1)

Ex 3.1 Class 8 Maths Question 2.

How many diagonals does each of the following have?

(a) A convex quadrilateral

(b) A regular hexagon

(c) A triangle

Solution:

(a) In Fig. (i) ABCD is a convex quadrilateral which has two diagonals AC and BD.

(b) In Fig. (ii) ABCDEF is a regular hexagon which has nine diagonals AE, AD, AC, BF, BE, BD, CF, CE and DF.

(c) In Fig. (iii) ABC is a triangle which has no diagonal.

Ex 3.1 Class 8 Maths Question 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 verify)

Solution:

In the given figure, we have a quadrilateral ABCD. Join AC diagonal which divides the quadrilateral into two triangles ABC and ADC.

In ∆ABC, ∠3 + ∠4 + ∠6 = 180°…(i) (angle sum property)

In ∆ADC, ∠1 + ∠2 + ∠5 = 180° …(ii) (angle sum property)

Adding, (i) and (ii)

∠1 + ∠3 + ∠2 + ∠4 + ∠5 + ∠6 = 180° + 180°

⇒ ∠A + ∠C + ∠D + ∠B = 360°

Hence, the sum of all the angles of a convex quadrilateral = 360°.

Let us draw a non-convex quadrilateral.

Yes, this property also holds true for a non-convex quadrilateral.

Ex 3.1 Class 8 Maths Question 4.

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 above table, we conclude that the sum of all the angles of a polygon of side ‘n’

= (n – 2) × 180°

(a) Number of sides = 7

Angles sum = (7 – 2) × 180° = 5 × 180° = 900°

(b) Number of sides = 8

Angle sum = (8 – 2) × 180° = 6 × 180° = 1080°

(c) Number of sides = 10 Angle sum = (10 – 2) × 180° = 8 × 180° = 1440°

(d) Number of sides = n

Angle sum = (n – 2) × 180°

Ex 3.1 Class 8 Maths Question 5.

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 sides and equal angles is called a regular polygon.

(i) Equilateral triangle

(ii) Square

(iii) Regular Hexagon

Ex 3.1 Class 8 Maths Question 6.

Find the angle measure x in the following figures:

Solution:

(a) Angle sum of a quadrilateral = 360°

⇒ 50° + 130° + 120° + x = 360°

⇒ 300° + x = 360°

⇒ x = 360° – 300° = 60°

(b) Angle sum of a quadrilateral = 360°

⇒ x + 70° + 60° + 90° = 360° [∵ 180° – 90° = 90°]

⇒ x + 220° = 360°

⇒ x = 360° – 220° = 140°

(c) Angle sum of a pentagon = 540°

⇒ 30° + x + 110° + 120° + x = 540° [∵ 180° – 70° = 110°; 180° – 60° = 120°]

⇒ 2x + 260° = 540°

⇒ 2x = 540° – 260°

⇒ 2x = 280°

⇒ x = 140°

(d) Angle sum of a regular pentagon = 540°

⇒ x + x + x + x + x = 540° [All angles of a regular pentagon are equal]

⇒ 5x = 540°

⇒ x = 108°

Ex 3.1 Class 8 Maths Question 7.

(a) Find x + y + z

(b) Find x + y + z + w

Solution:

(a) ∠a + 30° + 90° = 180° [Angle sum property]

⇒ ∠a + 120° = 180°

⇒ ∠a = 180° – 120° = 60°

Now, y = 180° – a (Linear pair)

⇒ y = 180° – 60°

⇒ y = 120°

and, z + 30° = 180° [Linear pair]

⇒ z = 180° – 30° = 150°

also, x + 90° = 180° [Linear pair]

⇒ x = 180° – 90° = 90°

Thus x + y + z = 90° + 120° + 150° = 360°

(b) ∠r + 120° + 80° + 60° = 360° [Angle sum property of a quadrilateral]

∠r + 260° = 360°

∠r = 360° – 260° = 100°

Now x + 120° = 180° (Linear pair)

x = 180° – 120° = 60°

y + 80° = 180° (Linear pair)

⇒ y = 180° – 80° = 100°

z + 60° = 180° (Linear pair)

⇒ z = 180° – 60° = 120°

w = 180° – ∠r = 180° – 100° = 80° (Linear pair)

x + y + z + w = 60° + 100° + 120° + 80° = 360°.

#### More CBSE Class 8 Study Material

- NCERT Solutions for Class 8 Maths
- NCERT Solutions for Class 8 Science
- NCERT Solutions for Class 8 Social Science
- NCERT Solutions for Class 8 English
- NCERT Solutions for Class 8 English Honeydew
- NCERT Solutions for Class 8 English It So Happened
- NCERT Solutions for Class 8 Hindi
- NCERT Solutions for Class 8 Sanskrit
- NCERT Solutions