Frank Learning Maths Class 8 Chapter 12 Properties of Quadrilaterals Exercise 12.1 Solutions
This post is created to help all the CBSE Class 8 students for the Solutions of Frank Learning Maths Class 8 Mathematics Book, Chapter 12 Properties of Quadrilaterals Exercise 12.1. Here students can easily find step by step solutions of all the problems for Properties of Quadrilaterals Exercise 12.1. Step by Step proper solutions for every problems. All the problem are solved with easily understandable methods with proper guidance so that all the students can understand easily.
Chapter 12 – Properties of Quadrilaterals
Properties of Quadrilaterals – Exercise 12.1 all Questions Solution
(2) What is the sum of the interior angles of a polygon having
Solution :
Required rule : (2n – 4)/n × 90°
(a) ∴ n = 5 (2 × 5 – 4) × 90°
= 6 × 90
= 540°
Therefore, the sum is 540°
(b) ∴ n = 8, (2 × 9 – 4) × 90°
= 12 × 90
= 1080°
Hence, the sum of the interior angles is 1080°
(c) ∴ n = 10, (2 × 10 – 4) × 90°
= 10 × 90
= 1450°
Thus, the sum of the interior angles is 1450°
(d) ∴ n = 9, (2 × 9 – 4) × 90°
= 14 × 90
= 1260°
Therefore, the sum of the interior angles is 1260°
(3) What is the measure of an interior angle of a regular polygon having
Solution :
Regular polygon rule : = 2n – 4/n × 90°
(a) ∴ n = 5, 2 × 5 – 4/5 × 90
= 6 × 18
= 108°
Hence, the measure of interior angle is 108°
(b) ∴ n = 9 18 – 4/9 × 90
= 14 × 10
= 140°
Thus, the measure of interior angle is 140°
(c) ∴ n = 10,
= 20 – 4/10
= 16 × 9
= 144°
Hence, the measure of interior angle is 144°
(d) ∴ n = 12,
= 24 – 4/12 × 90
= 205/4 × 30
= 150°
Therefore, the measure of interior angle is 150°
Question no – (4)
Solution :
Required Rule : 2n – 4/n × 90°
∴ (2n – 4/n) × 90° = 156°
= (2 – 4/n) = 156/90
= 4/n = 2 – 156/90
= 4/n = 180 – 156/90
= n = 90 °4/24
= n = 15
Thus, the number of sides are 15
Question no – (5)
Solution :
Given, x = 25°
No, its not a exterior angle of polygon.
Question no – (6)
Solution :
No, as 60° is the least value of measure of the interior angle of a regular polygon, and the sum of interior and exterior angle is 180°
Question no – (7)
Solution :
Given, 1 : 3 : 5 : 4 : 5
The sum of the angles of a pentagonal = 540°
∴ 1/18 × 540 = 30°
= 3/16 × 30 = 90°
= 5 × 30 = 150°
= 4 × 30 = 120°
= 5 × 30 = 150°
Therefore, the angles of the pentagon is 150°, 120° and 150°
Question no – (9)
Solution :
(a) (72 + 69 + 105 + 109)
= 355 ≠ 360° …[No]
(b) (14 + 76 + 98 + 42)
= 360° …[yes]
(c) (108 + 87 + 60 + 99)
= 360° …[Yes]
Question no – (10)
Solution :
Figure – (a) x + 40 + x + (180 – 70) + (180 – 50) = 540
= 2x + 40 + 190 + 130 = 540
= 2x + 280 = 540
= 2x + 540 = 280
= x = 260/2
= 130°
Figure – (b) It is regular,
x = 5
∴ 2x – 4/x × 90
= 10-4/5 × 90
= 6 × 18
= 108°
Figure – (c) x + 60° + (180° – 110°) + 90° = 360°
x = 150 + 70 = 360°
= x = 360 – 220°
= 140°
Question no – (11)
Solution :
Figure – (a) x = (180 – 50) = 130°
y = (180 – 70) = 110°
z = (180 – 60) = 120°
∴ x + y + z
= 130 + 110 + 120
= 360°
Figure – (b) Y = (180 – 35) = 155°
z = 90°
Now, 3rd angle = (180 – 90° – 35°) = 55°
∴ x = (180 – 55) = 125°
∴ x + y + z
= 125 + 145 + 90
= 360°
Figure – (c) x = 180° – 52° = 128°
y = 180° – 49 = 131°
Now, 3rd angle = (180 – 49 – 52°) = 79°
∴ y = 180° – 79 = 104°
∴ x + y + z
= 128° + 101° + 131°
= 360°
Question no – (12)
Solution :
(a) a = 180° – 60° = 120°
c = 180° – 105° = 75°
d = 180 – 75 = 105°
Now, 4th angle = (360 – 105 – 75 – 60) = 120°
∴ b = 180° – 120° = 60°
∴ a + b + c + d
= 120° + 60 + 75 + 105°
= 360°
(b) a = 180° – 70° = 110°
b = 180° – 95° = 85°
c = 180° – 110° = 70°
Now, 4th angle = (360 – 110 – 95 – 70) = 85°
∴ d = 180° – 85° = 95°
∴ a + b + c + d
= 110° + 85° + 70° + 95°
= 360°