**Collins Maths Solutions Class 7 Chapter 14 Properties of Triangles**

Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Collins Maths Class 7 Mathematics, Chapter 14, Properties of Triangles. Here students can easily find chapter 7 solutions with exercise wise explanation. Students will find proper solutions for Exercise 14.1 and 14.2 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily. Here in this post all the solutions are based on ICSE latest Syllabus.

**Properties of Triangles Exercise 14.1 Solution :**

**Question no – (1) **

**Solutions :**

**(a)** Yes, it form a triangle.

**(b)** Yes, it form a triangle.

**(c)** No, it not form triangle.

**(d)** Yes, it form a triangle.

**(e)** No, it not form a triangle.

**(f)** Yes, it form a triangle.

**Question no – (2) **

**Solutions : **

**(a) (25° + 75° + 80°)**

= 180°

**∴** yes, it form a triangle.

**(b) (80° + 80° + 30°)**

= 190°

**∴** No, it is not form a triangle.

**(c) (20° + 60° + 100°)**

= 180°

**∴** Yes, it form a triangle.

**(d) (40° + 110° + 30°)**

= 180°

**∴** Yes, it form a triangle.

**(e) (140° + 50° + 10°)**

= 180°

**∴** Yes, it form a triangle.

**(f) (80° + 70° + 30°)**

= 180°

**∴** Yes, it form a triangle.

**(g) (55° + 35° + 90°)**

= 180°

**∴** Yes, it form a triangle.

**(h) (62° + 38° + 80°)**

= 180°

**∴** Yes, it form a triangle.

**Question no – (3) **

**Solution : **

∠C = 180° – (75° + 55°)

= 180° – 130°

= 50°

Therefore, the measure of ∠C will be 50°

**Question no – (4) **

**Solution : **

ΔRST,

∠R = 80°

**∴** (180° – 80°) = 100°/2

= 50°

So, ∠S = ∠T = 50°

**Question no – (5) **

**Solution : **

Let, angles 4x, 5x, 6x

**∴** 4x + 5x + 6x = 180°

= 15x = 180 °

= x = 180/15

= x = 12°

**∴** 4x = (4 × 12°) = 48°

**∴** 5x = (5 × 12°) = 60°

**∴** 6x = (6 × 12°) = 72°

Hence, The measures of each angle will be 48°, 60° and 72 °

**Question no – (6) **

**Solution : **

Let, other angle 3x and 2x

**∴** 80 + 3x + 2x = 180°

= 65x = 180° – 80°

= x = 100°/5

= x = 20°

**∴** (3 × 20°) = 60°

**∴** (2 × 20°) = 40°

Thus, the angles will be 60° and 40°.

**Question no – (7) **

**Solution : **

Each angle be 50°, 50°

And, (180° – (50 + 50)°

= 180° – 100°

= 80°

Therefore, the measures of each angle will be 80°.

**Question no – (8) **

**Solution : **

Let, angle 3x and 6x

**∴** 3x + 6x + 90° = 180°

= 9x = 180° – 90°

= x = 90/9

= x = 10°

**∴** Angles be (3 × 10) = 30°

And, (6 × 10) = 60°

Therefore, the angles will be 30° and 60°

**Question no – (9) **

**Solution : **

Isosceles right angle triangle,

= 90° and other x = x

**∴** x + x + 90° = 180°

= 2x = 180° – 90°

= x = 90/2

= x = 45°

**∴ **45°, 45°, 90°

Therefore, the measures of the angles will be 45°, 45°, 90°

**Question no – (10) **

**Solution : **

As we know that,

Equilateral triangle = all side are equal,

Let, angle be = x

**∴** x + x + x = 180°

= 3x = 180°

= x = 180°/3

= x = 60°

Therefore, the measures of an equilateral triangle will be 60°

**Properties of Triangles Exercise 14.2 Solution :**

**Question no – (1) **

**Solutions : **

**(a) 9 cm, 40 cm, 41 cm**

**∴** 9^{2} + 40^{2}

= 81 + 1600

= 1681

And, 41^{2} = 1681

**∴** 41^{2} = 9^{2} + 40^{2}

So, yes, it form a right angle triangle.

**(b) 6ft, 8ft, 10ft**

**∴** 6^{2} + 8^{2}

= 36 + 64

= 100

And, 10^{2} = 100

**∴ **6^{2 }+ 8^{2} = 10^{2}

So, it form a right angled triangle

**(c) 14 m, 40, 50m**

**∴** 14^{2} = 196

And, 40^{2} + 50^{2}

= 1600 + 2500

= √4100

**∴** This is not equal

So, it will not form a right angled triangle.

**Question no – (2) **

**Solutions : **

**(a) Third side,**

= √13^{2} – 5^{2}

= √169 – 25

= √ 144

= 12 cm

**(b) Hypotenuse,**

= √24^{2} + 10^{2}

= √5676 + 100

= √676

= 26 cm

**(c) Another side,**

= √50^{2} – 48^{2}

= √2500 – 2304

= √196

= 14 cm

**(d) Hypotenuse,**

= √18^{2 }+ 80^{2}

= √324 + 6400

= √6724

= 82 cm.

**Question no – (3) **

**Solution : **

Distance, = √12^{2} + 9^{2}

= √144 + 81

= √ 255

= 15 km

Therefore, he is 15 km far from his starting point.

**Question no – (4) **

**Solution : **

Distance ground does the ladder touch building.

= √29^{2} – 21^{2}

= √841 – 441

= √ 400

= 20ft.

Therefore, 20ft above, the ladder touch the building.

**Question no – (5) **

**Solution : **

Width, = √20^{2} – 12^{2}

= √400 -144

= √256

= 16 ft

Therefore, the width of the carpet will be 16ft.

**Question no – (6) **

**Solution : **

Height, = √4^{2 }– 2^{2}

= √16 – 4

= √12 cm

Therefore, the height of the equilateral triangle will be 12 cm.

**Question no – 7**

**Solution : **

Length of diagonal,

= √15^{2} + 8^{2}

= √289

= 17 m

Therefore, the length of the diagonal of the house will be 17 m.

**Next Chapter Solution : **

👉 Chapter 15 👈