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**Mathsight Class 7 Solutions Chapter 10 Properties of Triangles**

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students to solve MathSight Class 7 Mathematics Book, Chapter 10, Properties of Triangles. Here students can easily get all the exercise questions solution for Chapter 10, Properties of Triangles Exercise 10.1 and 10.2

**Properties of Triangles Exercise 10.1 Solution : **

**Question no – (1)**

**Solution :**

**(a)** Equilateral Triangle.

**(b)** Isosceles Triangle.

**(c)** Scalene Triangle.

**Question no – (2)**

**Solution :**

**(a)** x + 19° + 21° = 180°

Or, x + 40°** = **180°

Or, x = 180° – 40°

**∴** x = 40°

**(b)** x + h + 50 = 180°

Or, 2x = 180 – 50° = 130°

Or, x = 130/2 = 56

Or, x = 56°

**(c)** a = 180° – (90° + 40°)

0= 180° – 130° = 50°

**(d)** x + x + 90 = 180°

Or, 2x = 180° – 90° = 90°

Or, x = 90/x = 45°

**(e)** b = 180° – (40° + 40°)

= 180° – 80°

= 100°

**(f)** z = 180° – (108° – 32°)

= 180° – 140°

= 40°

**Question no – (3)**

**Solution :**

**(a)** x + 2x + 3x = 180°

Or, 6x = 180°

Or, x = 180°/6 = 30°

**(b)** 8x + 5x + 5x = 180°

Or, 18x = 180°

Or, x = 180°/18 = 10

**(c)** 3x + 2x + 5x = 180°

Or, 10x = 180°

Or, x = 180°/10 = 18

**(d)** 7x + 5x + 6x = 180°

Or, 18x = 180°

Or, x = 180°/18 = 10°

**Question no – (4)**

**Solution :**

3^{rd} angle = 180° – (45° + 55°)

= 180° – 100°

= 80°

Therefore, it is acute triangle.

**Question no – (5)**

**Solution :**

Base angles = 180° – 90° = 90°

**∴** no of angles = 90°/2 = 45°

**∴** base angles each = 45°, 45°

**Question no – (6)**

**Solution :**

Let, CF = x

We know that, 2x + 3x + 5x = 180°

Or, 10x = 180°

Or, x = 180/10 = 18°

**∴** 2x = 2 × 18 = 36°

3x = 3 × 18 = 54°

5x = 5 × 18 = 90°

**∴** It is right angle triangle.

**Question no – (7)**

**Solution :**

x + x + x = 90°

Or, 3x = 90°

Or, x = 90/3

**∴** x = 30°

**Question no – (8)**

**Solution :**

**(a)** ∠PRS = ∠RPS + ∠PQR

**(b)** ∠QPU = ∠PQR + ∠QRP

**(c)** ∠RQT = ∠QPR + ∠PRQ

**Question no – (9)**

**Solution :**

**(a)** x = 55° + 60° = 115°

**(b)** x = 120° – 60° = 60°

**(c)** y = 180° – 75° = 105°

**(d)** x = 120° – 70° = 50°

Now, x + x + 105° = 180°

Or, 2x = 180° – 105° = 75°

Or, x = 75°/2

**∴** x = 37.5°

**Question no – (10)**

**Solution :**

∠PRQ = 180° – 110°

= 70°

∠P = 120° – 70°

**∴** ∠P = 50°

**Question no – (11)**

**Solution :**

**(a)** Equilateral triangle.

**(b)** Right angled triangle.

**Question no – (12)**

**Solution :**

**(a)** False

**(b)** False

**(c)** True

**Properties of Triangles Exercise 10.2 Solution : **

**Question no – (1)**

**Solution :**

**(a)** 7 + 7 > 10

= 14 > 10

**∴** Triangle possible

**(b)** 3 + 3.5 < 8

= 6.5 < 8

**∴** Not possible

**(c)** 8 + 9 > 11

= 17 > 11

∴ Possible

**(d)** 2 + 2 < 4.5

4 < 4.5

**∴** Not possible

**Question no – (2)**

**Solution :**

**(a)** OA + OB > AB = T

**(b)** OB + OC = F

**(c)** F

**Question no – (3)**

**Solution :**

∠ABD = 90° – 64° = 26°

We know that,

90° + ∠ACB + ∠ABC = 180°

Or, 90° + 2 ∠ABC + ∠ABC = 180°

Or, 3 + ∠ABC = 180° – 90° = 90°

Or, ∠ABC = 90°/3 = 30°

**∴** ∠ABC = 30°

∠ACB = 2 ∠ABC

= 2 × 30°

= 60°

**Question no – (4)**

**Solution :**

**(a)** 4^{2} + 6^{2}

= 16 + 36 = 52

8^{2} = 64

**∴** 4^{2} + 6^{2} ≠ 8^{2}

**∴** not right angle triangle

**(b)** 1.5^{2 }+ 2^{2} = 2.7 + 2

= 4.7

(2.5)^{2} = 6.25

(1.5)^{2} + 2^{2} ≠ (2.5)^{2}

**∴** not right angled triangle.

**(c)** 9^{2 }+ 40^{2}

= 81 + 1600

= 1681

(41)^{2} = 1681

∴ 9^{2} + 40^{2}

= (41)^{2}

**∴** It is right angled triangle.

**(d)** 6^{2} + 8^{2}

= 36 + 64

= 100

(10)^{2} = 100

**∴** 6^{2} + 8^{2} = 10^{2}

**∴** It is right angled triangle.

**Question no – (5)**

**Solution :**

**(a)** (12)^{2} + (16)^{2} = 20^{2}

Or, 144 + 256 = 400

**∴ **It is Pythagoras Triplet.

**(b)** 5^{2} + 12^{2} = 13^{2}

Or, 25 + 144 = 169

**∴** It Pythagoras triplet.

**(c)** 8^{2} + 15^{2} = 17^{2}

64 + 225 = 289

**∴** It is Pythagoras triplet.

**(d)** 7^{2} + 8^{2} ≠ 9^{2}

Or, 49 + 64 ≠ 81

∴ It is not Pythagoras triplet.

**Question no – (6)**

**Solution :**

**(a)** x^{2} = 15^{2} – 9^{2}

Or, x^{2} = 225 – 81

Or, x^{2} = 144

Or, x^{2} = 12^{2}

Or, x = 12 cm

**(b)** x^{2} = 8^{2} + 6^{2}

Or, x^{2} = 64 + 36

Or, x^{2} = 100

Or, x^{2} = (10)^{2}

Or, x = 10 cm

**(c)** x^{2} = 4^{2} + 3^{2}

Or, x^{2} = 16 + 9 = 25

Or, x = 5 cm

**(d)** x^{2} = (41)^{2} – (40)^{2}

Or, x^{2} = 1681 – 1600

Or, x^{2} = 81

Or, x = 9 cm

**Question no – (7)**

**Solution :**

**∵** Isosceles right angle triangle,

PQ = QR

∠PQR = 90°

**∴** ∠P + ∠R = 180° – 90° = 90°

Or, ∠P + ∠P = 90° [∵ PQ = QR]

Or, 2 ∠P = 90

Or, ∠P = 90/2 = 45°

**∴** ∠P = ∠R = 45°

**∵** (PR)^{2} = (PQ)^{2} + (QR)^{2}

= (PQ)^{2} + (PQ)^{2}

Or, PR^{2} = 2 (PQ)^{2} ……(Proved)

**Question no – (8)**

**Solution :**

(6)^{2} + (4.5)^{2} = (7.5)^{2}

Or, 36 + 20.25

= 56.25

∴ Yes it is right angled triangle.

Therefore, the Length of hypotenuse 7.5 cm.

**Question no – (9)**

**Solution :**

= (diagonal’s length)^{2}

= 8^{2} + 6^{2}

= 64 + 36

= 100

Or, diagonal’s length = 10 cm.

**Question no – (10)**

**Solution :**

(16)^{2} + (12)^{2}

= 256 + 144

= 400 = 10x.

∴ x = 400/100

∴ x = 40

Thus, the value of x will be 40

**Question no – (11)**

**Solution :**

Pole height = h

∴ h^{2} = 13^{2} – 12^{2}

= 169 – 144

= 25

= 5^{2}

Or, h = 5 cm.

Hence, the height of the pole will ne 5 cm.

**Question no – (12)**

**Solution :**

∴ Distance = d

∴ d^{2} = (24)^{2} + 7^{2}

= 576 + 49

= 625 = (25)^{2}

Or, d = 25 cm.

Therefore, the shortest distance will be 25 cm.

**Previous Chapter Solution : **