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

NCERT Solutions for Class
8 Maths Chapter 3
Understanding Quadrilaterals Ex 3.1 are the
part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT
Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1**.**

**Ex 3.1 Class 8 Maths Question 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

**Solution:****
**(a)
Simple curve: (1), (2), (5), (6) and (7) are simple curves.

(b) Simple closed curve: (1), (2), (5), (6) and (7) are simple curves.

(c) Polygon: (1) and (2) are polygons.

(d) Convex polygon: (2) is convex polygon.

(e) Concave polygon: (1) is concave polygon.

**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 the following figure, ABCD is a convex
quadrilateral which has two diagonals AC and BD.

(c) In the following figure, 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 following figure, we have a quadrilateral
ABCD. Join the diagonal AC, which divides the quadrilateral into two triangles
ABC and ACD.

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

Adding equations (i) and (ii), we get

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

⇒ ∠A + ∠C + ∠D + ∠B = 360°

Hence, the sum of the measures of the angles of a convex quadrilateral is 360°.

Let us draw a non-convex quadrilateral.

Join the diagonal BD and prove as above in triangles
BCD and ABD.

Thus, 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 deduced
from that).(a) 7 (b) 8 (c) 10 (d) n

**Solution:****
**From
the given table, we can conclude that the sum of the angles of a convex polygon
with side ‘n’ = (n – 2) × 180°

(a) If the number of sides = 7

Then the angle sum = (7 – 2) × 180° = 5 × 180° = 900°

(b) If the number of sides = 8

Then the angle sum = (8 – 2) × 180° = 6 × 180° = 1080°

(c) If the number of sides = 10

Then
the angle sum = (10 – 2) × 180° = 8 × 180° = 1440°

(d) If the number of sides = n

Then the 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, whose all sides and all angles are equal,
is called a regular polygon.

(i) A polygon with 3 sides is called an

**equilateral triangle**.

(ii) A polygon with 4 sides is called a

**square**.

(iii) A polygon with 6 sides is called a

**regular hexagon**.

**Ex 3.1 Class 8 Maths Question 6.**

Find the angle measure x in the following figures:**Solution:
**(a) Sum of the angles of a quadrilateral = 360°

⇒ 50° + 130° + 120° + x = 360°

⇒ 300° + x = 360°

⇒ x = 360° – 300°

⇒ x = 60°

(b) Sum of the angles of a quadrilateral = 360°

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

⇒ x + 220° = 360°

⇒ x = 360° – 220°

⇒ x = 140°

(c) Sum of the angles 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) Sum of the angles of a regular pentagon = 540°

⇒ x + x + x + x + x = 540°

[All the angles of a regular pentagon are equal]

⇒ 5x = 540°

⇒ x = 108°

**Ex 3.1 Class 8 Maths Question 7.**

(a)
Find x + y + z**Solution:****
**(a) ∠a + 30° + 90° =
180° [Angle sum property of a
triangle]

⇒ ∠a + 120° = 180°

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°

and y + 80° = 180° (Linear pair)

⇒ y = 180° – 80° = 100°

again, z + 60° = 180° (Linear pair)

⇒ z = 180° – 60° = 120°

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

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

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