# Joy of Mathematics Class 6 Solutions Chapter 15

## Joy of Mathematics Class 6 Solutions Chapter 15 Triangles

Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Joy of Mathematics Class 6 Book, Chapter 15 Triangles. Here students can easily find step by step solutions of all the problems for Triangles. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Students will get chapter 15 solutions. Here students will find exercise wise solutions for chapter 15 Exercise 15.1, 15.2 and 15.3

Triangles Exercise 15.1 Solution :

Question no – (1)

Solution :

The figure of triangle. (a) The angles is ‘C’

(b) The angle is AC.

(c) Vertex is ‘C,

(d) The side is ‘BC,

Question no – (2)

Solution :

(a) 2 triangle are formed. Name the triangles :

△ ABD, △ACD

Question no – (3)

Solution :

(a) There are 5 triangle,

(b) Which triangles have

(i) △ABC and △CDE

(ii) △CDE and △DEF

Question no – (4)

Solution :

(a) Pediments of building

(b) Slice of pizza

(c) Traffic sigh board

(d) Bicycle Frame

Question no – (5)

Solution :

Acute triangle

(b) Right triangle

(c) Obtuse triangle

(d) Right triangle

Question no – (6)

Solution :

(a) Scalene triangle

(b) Isosceles triangle

(c) Equilateral triangle

(d) Scalene triangle

Question no – (7)

Solution :

(a) This statement is – False

(b) This statement is – True

(c) This statement is – False

(d) This statement is – True

(e) This statement is – True

(f) This statement is – True

Triangles Exercise 15.2 Solution :

Question no – (1)

Solution :

(a) x = 180° – (35 + 50°)

= 180° – 85°

= 95°

(b) X = 180° – (20° + 30°)

= 180° – 50°

= 130°

(c) ∠E = 180° – 60°

= 120°

∠DEC = 120°

and,

∠HEG = (180° – 120°) = 60°

x = 180° – (50° + 60°)

= 180° – 110°

= 70°

Question no – (2)

Solution :

(a) x = (180° – 140°) = 40°

and, Y  = 180° – (60° + 40°)

= 180° – 100°

= 80°

(b) Y = (180° – 130°) = 50°

∴ x = 180° – (70° + 50°)

= 180° – 120°

= 60°

(c) Here, x = (180° – 155°) = 65°

Then y = 180° – (65° + 65°)

= 180° – 130°

= 50°

Question no – (3)

Solution :

= The value is 360°

Question no – (4)

Solution :

There are 8 triangles formed.

All are equilateral side

So, ∠A + ∠B + ∠C + ∠D + ∠E + ∠F

= (6 × 60°)

= 360°

Question no – (5)

Solution :

Let, angles 2x, 3x, 5x,

2x + 3x + 5x

= 180°

= 10x = 180°

= x = 180/10

= 18°

Angles are,

= 3x = 3 × 18°

= 54°

= 2x = 2 × 18°

= 36°

= 5x = 5 × 18

= 90°

Question no – (6)

Solution :

= 2∠A = 3∠B = 6∠6

2∠A = 180°

∠A = 180°/2

= 90°

and, = 3∠B = 180°

= ∠B = 180°/3

= 60°

and, = 6∠C = 6 = 180°/6

= 30 °

Question no – (7)

Solution :

Let, Angles are, ∠A, ∠B, ∠C

∠A = ∠B + ∠C

and, ∠A + ∠B + ∠C = 180° (we know)

Now, substitute the value,

= ∠B + ∠C + ∠B + ∠C = 180°

= 2 (∠B + ∠C) = 180°

= ∠B + ∠C = 180°/2

= 90°

Question no – (8)

Solution :

the other acute angle

= 180° – (90° + 40°)

= 180° – 130°

= 50°

Question no – (9)

Solution :

∠C = ∠180° – (60 + 50)

= 180° – 110°

= 70°

∠BOC = 180° – (50/2 + 70/2)

= 180° – (25° + 35°)

= 180° – 60

= 120°

Question no – (10)

Solution :

Let, the angle 2x, 3x, 4x

= 2x + 3x + 4x = 180°

= x = 180°/9

= 20°

∴ 1st angle = (2 × 20) = 40°°

2nd angle = (3 × 20) = 60°

3rd angle = (4 × 20) = 80°

x + x + 1 + x + 2 = 180°

= 3x = 180° – 3

= 177

= x = 177/3

= 59°

(x + 1)° = (54 + 1)° = 60°

= (x + 2)° = (59 + 2) = 61°

Question no – (11)

Solution :

Let, the 3rd angle be x

= other 2 angle = (x + 15)

x + x + 15 + x + 15 = 180°

= 3x = 180° – 30°

= x = 150/3

= 50°

Angle are 50°, 65°, 65°

Question no – (12)

Solution :

(a) Given statement is – False

(b) Given statement is – False

(c) Given statement is – True

(d) Given statement is – True

(e) Given statement is – True

(f) Given statement is – True

Triangles Exercise 15.3 Solution :

Question no – (1)

Solution :

(a) From figure,

∠x = 180° – (45° + 60°)

= 180 – 105°

= 75°

Now, ∠Y = (180° – 75°)

= 105°

(b) From figure,

∠Y = (180° – 50°) = 130°

= 130°

∠x = 180° – (130° + 20°)

= 180° – 150°

= 30°

(c) From △BCD

∠3 = 180° – (50° + 30°)

= 180° – 80°

= 100°

∠x = (180° – 100°) = 80°

∠y = 180° – (80°)

= 180° – 80°

= 100°

= 100°

Question no – (2)

Solution :

(a) 2 + 3 = 5 > 4

= Yes

(b) 3 + 5 = 8 > 6

= Yes

(c) 8 + 10 = 18 > 12

= Yes

(d) 2 + 6 = 8 < 4

= No

(e) 4 + 5 = 9 < 10

= No

Question no – (4)

Solution :

(a) AB < AO + OB

(b) AC < AO + OC

(c) BC < BO + OC

(d) AB + BC > AC

(e) AB + BC + AC > OA + OB + OC

Question no – (5)

Solution :

(a) If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

(b) The sum of any two sides of triangle is greater than the third side.

(c) If a, b, and c are the three sides of a triangle, then a + b – c is always greater than ‘o’

Next Chapter Solution :

Updated: June 23, 2023 — 2:30 pm