Mathsight Class 7 Solutions Chapter 10

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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) 4² + 6²

= 16 + 36 = 52

8² = 64

∴ 4² + 6² ≠ 8²

∴ not right angle triangle

(b) 1.5² + 2² = 2.7 + 2

= 4.7

(2.5)² = 6.25

(1.5)² + 2² ≠ (2.5)²

∴ not right angled triangle.

(c) 9² + 40²

= 81 + 1600

= 1681

(41)² = 1681

∴ 9² + 40²

= (41)²

∴ It is right angled triangle.

(d) 6² + 8²

= 36 + 64

= 100

(10)² = 100

∴ 6² + 8² = 10²

∴ It is right angled triangle.

Question no – (5)

Solution :

(a) (12)² + (16)² = 20²

Or, 144 + 256 = 400

∴ It is Pythagoras Triplet.

(b) 5² + 12² = 13²

Or, 25 + 144 = 169

∴ It Pythagoras triplet.

(c) 8² + 15² = 17²

64 + 225 = 289

∴ It is Pythagoras triplet.

(d) 7² + 8² ≠ 9²

Or, 49 + 64 ≠ 81

∴ It is not Pythagoras triplet.

Question no – (6)

Solution :

(a) x² = 15² – 9²

Or, x² = 225 – 81

Or, x² = 144

Or, x² = 12²

Or, x = 12 cm

(b) x² = 8² + 6²

Or, x² = 64 + 36

Or, x² = 100

Or, x² = (10)²

Or, x = 10 cm

(c) x² = 4² + 3²

Or, x² = 16 + 9 = 25

Or, x = 5 cm

(d) x² = (41)² – (40)²

Or, x² = 1681 – 1600

Or, x² = 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, PR² = 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

= 8² + 6²

= 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

∴ h² = 13² – 12²

= 169 – 144

= 25

= 5²

Or, h = 5 cm.

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

Question no – (12)

Solution :

∴ Distance = d

∴ d² = (24)² + 7²

= 576 + 49

= 625 = (25)²

Or, d = 25 cm.

Therefore, the shortest distance will be 25 cm.

Previous Chapter Solution : 

👉 Chapter 9