RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 8 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 8. Students can also read RBSE Class 8 Maths Important Questions for exam preparation. Students can also go through RBSE Class 8 Maths Notes to understand and remember the concepts easily. Practicing the class 8 maths chapter 6 try these solutions will help students analyse their level of preparation.

RBSE Class 8 Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1

Question 1.
Given here are some figures.
RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 1
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
Answer:
The classification of the given figures is given below :
(a) Simple curve: (i), (ii), (v), (vi) and (vii)
(b) Simple closed curve: (i), (ii), (v), (vi) and (vii)
(c) Polygon: (i), (ii) and (iv)
(d) Convex Polygon: (ii)
(e) Concave Polygon: (i) and (iv)

RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1

Question 2.
How many diagonals does each of the following have?
(a) A convex quadrilateral
(b) A regular hexagon (C) A triangle
Answer:
(a) A convex quadrilateral has two diagonals.
(b) A regular hexagon has nine diagonals.
(c) A triangle has no diagonal.

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 try!)
Answer:
We know that the sum of measures of the angles of a convex quadrilateral is 360°.
Yes, this property holds in case of the quadrilateral is not convex.
RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 2
In given quadrilateral ABCD,
= ∠A + ∠B + ∠C + ∠D
= 110° + 120° + 50° + 80°
= 360°

Question 4.
Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)
RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 3
What can you say about the angle sum of a convex polygon with number of sides?
(a) 7
(b) 8
(c) 10
(d) n
Answer:
From the given table, it is clear that the sum of angles (interior angles) of a polygon with n sides = (n - 2) × 180°
(a) Here n = 7
∴ The sum of the angles of a polygon of 7 sides = (7 - 2) × 180°
= 5 × 180° = 900°

(b) Here n = 8
∴ The sum of the angles of a polygon of 8 sides = (8 - 2) × 180°
= 6 × 180°
= 1080°

(c) Here n = 10
∴ The sum of the angles of a polygon of 10 sides = (10 - 2) × 180°
= 8 ×180°
= 1440°

(d) It is clear- from the given table that the number of triangles is two less than the number of sides in the polygon. Therefore if the polygon has n sides, the number of triangles formed will be (n - 2).
The sum of angles of a triangle = 180°
∴ The sum of angles of a polygon of n sides = (n - 2) × 180°

RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1

Question 5.
What is a regular polygon? State the name of a regular polygon of
(i) 3 sides
(ii) 4 sides
(iii) 6 sides
Answer:
A polygon is said to be a regular polygon, if all its,
(i) interior angles are equal;
(ii) sides are equal; and
(iii) exterior angles are equal.

The name of a regular polygon of
(i) 3 sides is equlateral triangle.
(ii) 4 sides is square
(iii) 6 sides is regular hexagon.

Question 6.
Find the angle measure x in the following figures:
(a)
RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 4
Answer:
We know that the sum of the angles (interior) of a quadrilateral is 360°.
∴ x + 120° + 130° + 50° = 360°
or x + 300° = 360°
or x = 360° - 300°
= 60°

(b)
RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 5
Answer:
We know that the sum of the angles (interior) of a quadrilateral is 360°.
∴ x + 70° + 60° + 90° = 360°
or x + 220° = 360°
or x = 360° - 220°
= 140°

RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1

(c)
RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 6
Answer:
According to question, the given figure has = (n - 2) × 180°
= (5 - 2) × 180°
= 3 × 180°
= 540°
RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 7
It is clear from the figure that
m∠1 + 60°= 180° [Linear pair]
or m∠1 = 180° - 60° = 120°
and m∠2 + 70°= 180° [Linear pair]
or m∠2 = 180° - 70° = 110°
∴ m∠1 + m∠2 + x + x + 30° = 540°
or 120° + 110° + 2x + 30° = 540°
or 2x + 260° = 540°
or 2x = 540° - 260°
= 280°
or x = \(\frac{540^{\circ}}{5}\) = 140°.

(d)
RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 8
Answer:
According to question, the given figure has 5 sides, i.e. n = 5
∴ The sum of the angles of this figure = (n - 2) × 180°
= (5 - 2) × 180°
= 3 × 180° = 540°
∴ x + x + x + x + x = 540°
or 5x = 540°
or x = \(\frac{540^{\circ}}{5}\)
= 108°

Question 7.
RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 9
(a) Find x + y + z
(b) Find x + y + z + w
Answer:
(a) We know that the sum of the angles of a triangle is 180°,
∴ m∠ 1 + 30°,+ 90°= 180°
or m∠1 + 120°= 180°
RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 10
or m∠1 = 180° - 120°
= 60°
and x + 90° = 180° [Linear pair]
or x = 180° - 90°
= 90°
y + m∠1 = 180° [Linear pair]
or y + 60° = 180° [∵ m∠1 = 60°]
or y = 180° - 60° = 120° .......... (2)
and z + 30° = 180° [Linear pair]
or z = 180° - 30° = 150° ........... (3)
Adding (1), (2) and (3), we get
x + y + z = 90° + 120° + 150°
= 360°

RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1

(b) We know that the sum of the angles of a quadrilateral is 360°, therefore
m∠1 + 120° + 80° + 60°= 360°
or m∠1 + 260°= 360°
or m∠1 = 360° - 260° = 100°
RBSE Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1 11
and x + 120°= 180° [Linear pair]
x = 180° - 120°
= 60° .......... (1)
y + 80°= 180° [Linear pair]
y= 180° - 80°
= 100° ........... (2)
z + 60°= 180° [Linear pair]
z = 180° - 60°
= 120° ............. (3)
m∠1 + w = 180° [Linear pair]
w = 180° - m∠1
= 180° - 100
= 80° ............ (4)
Adding (1), (2), (3) and (4)
x + y + z + w = 360°
60° + 100° + 120° + 80° = 360°

Bhagya
Last Updated on May 17, 2022, 11:46 a.m.
Published May 17, 2022