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**Frank Learning Maths Class 8 Solutions Chapter 12 Properties of Quadrilaterals**

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 Book, Chapter 12 Properties of Quadrilaterals. Here students can easily find step by step solutions of all the problems for Properties of Quadrilaterals. Here students will find solutions for Exercise 12.1 and 12.2. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.

**Properties of Quadrilaterals Exercise 12.1 Solution**

**Question no – (2)**

**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°

**Question no – (3) **

**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°

**Properties of Quadrilaterals Exercise 12.2 Solution**

**Question no – (1)**

**Solution : **

**(a)** A square can be considered as a special type of a parallelogram,

Because**,** Its diagonals bisect each other and also the vertex angle, Like a parallelogram.

**(b)** A square can be considered as a special type of a rectangle,

Reason** :** A square can be considered as a special type of rectangle in which all sides are equal.

**(c)** A square can be considered as a special type of a rhombus,

Reason **:** A square can be also considered as a special type of rhombus in which all angles are right angles.

**Question no – (2)**

**Solution : **

PQ = QO

**∴** 5x – 6 = 3x + 4

= 2x = 10

= x = 5

Therefore, the value of x will be 5.

**Question no – (4)**

**Solution : **

∠A + ∠B = 180°

= 5x + 5 + 4x – 5 = 180°

= 9x = 180°

= x = 20°

**∴** ∠A = (100 + 5)°

= 105°

∠B = (80 – 5)°

= 75°

**∴** ∠B + ∠C = 180°

= 75 + ∠C = 180°

∠C = 105°

**∴ **∠A + ∠D = 180°

= 105 + ∠D = 180°

∠D = 75°

**Question no – (5)**

**Solution : **

∠P + ∠Q = 180°

= 2x + 10 + 3x – 25 = 180°

= 5x – 15 = 180

= 5x = 180 + 15°

= x = 195/5

= 39°

**∴** ∠P = (78 + 10) = 88°

∠S = 180° – ∠P

= 180°- 88°

= 92

**∴** ∠CP = (117 – 25) = 92°

∠R = 180° – ∠Q

= 180° – 92°

= 88°

**Question no – (6)**

**Solution : **

(3x° + 2x°) = 180°

54° = 180°

x = 36°

**∴** 1st = 3 × 36° = 108°

**∴** 2nd = 2 × 36° = 72°

**∴** 3rd = 180 – 72 = 108°

**∴** 4th = 180° – 108° = 72°

Therefore, the the measure of each of the angles are 108°, 72°, 108°, 72°

**Question no – (7)**

**Solution : **

Let, the equal angles = x

= (x + x) = 180°

= 2x = 180°

= x = 90°

**∴** 1st = 90°

**∴** 2nd = 90°

**∴** 3rd = 180° – 90° = 90°

**∴** 4th = 180° 90° = 90°

**Question no – (8)**

**Solution : **

Let, ABCD be a quadrilateral,

∠A = 2x,

∠B = 3x,

∠C = 5x,

∠D = 8x

**∴** 2x + 3x + 5x + 8x = 360°

= 180 = 360°

= x = 20°

Therefore the Angles are 40°, 60°, 100°, 160°

**Question no – (9)**

**Solution : **

Let, ∠P = 3x,

∠Q = 4x,

∠R = 5x,

∠S = 6x

**∴** 3x + 4x + 5x + 6x = 360°

= 180 = 360°

= x = 20°

**∴** Angle are = 60°, 80°, 100°, 120°

**∴** ∠P + ∠S = 180°

**∴** PQRS is a parallelogram.

**Question no – (10)**

**Solution : **

= x + (x – x) + (2x – 10) + (2x – 5) + (2x + 15) = 540°

= 8x – 4 = 540

= 8x = 540 + 4

= x = 544/8

= x = 680°

**∴** Angle are,

= 680° (68 – 4)°, (p36 – 10) (136 – 3), (136 + 15)

= 68°, 64°, 126°, 131°, 15°

**Question no – (12)**

**Solution : **

**Figure – (a)** ∠x = 180 – 105 = 75°

∠y = 180 – 75° = 105°

∠z = 180° – 105° = 75°

**Figure – (b)** ∠y = 180° – 40 = 140°

∠z = 180° – 40° = 140°

3rd angle = 180° – 140° = 40°

∠x = 180° – 40° = 140°

**Figure – (c)** x + 30 + 110° = 180°

∠x = 180° – 140° = 40°

∠y = 180° – 70° = 140°

∠z = 40°

**Figure** **– (d)** 90 + ∠y + 60° = 180°

∠y = 180° – 150° = 30°

∠z = 30°

∠x = 90°

**Question no – (13)**

**Solution : **

**Figure – (a)**

x + 5 = 20

x = 15 cm

y + 4 = 25

y = 21 cm

**Figure – (b)**

3y + 1 = 25

3y = 24

= y = 8 cm

5x = 15

= x = 3

**Question no – (17)**

**Solution : **

(5x – 5) + (10x + 35)° = 180°

= 15x + 30 = 180°

= 15x = 150

= x = 10°

**∴** Angles are = 45, 135

**∴** 45 : 135

= 9 : 27

= 1 : 3

Therefore, the ratio of these angles will be 1 : 3

**Previous Chapter Solution : **