Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 6 Triangles Ex 6.2 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 7 Maths Chapter 6 Triangles Ex 6.2

1. State, if a triangle is possible with the following angles.

Question (a).

35°, 70°, 65°

Answer:

No Reason :

Sum of three angles

= 35° + 70° + 65° = 170°

But, we know that sum of angles of a triangle is always 180°

∴ A triangle cannot have angles 35°, 70° and 90°.

Question (b).

70°, 50°, 60°

Answer:

Yes

Reason :

Sum of three angles

= 70° + 50° + 60°

= 180°

By angle sum property

∴ A triangle can have angles 70°, 50° and 60°.

Question (c).

90°, 80°, 20°

Answer:

No

Reason :

Sum of three angles

= 90° + 80° + 20°

=190°

But, we know that sum of angles of a triangle is always 180°

(Angle sum properly)

∴ A triangle cannot have angles 90°, 80° and 20°.

Question (d).

60°, 60°, 60°

Answer:

Yes

Reason :

Sum of three angles

= 60° + 60° + 60°

= 180°

by angle sum property.

∴ A triangle can have angles 60°, 60° and 60°.

Question (e).

90°, 90°, 90°

Answer:

No Reason :

Sum of three angles

= 90° + 90° + 90°

= 270°

But, we know that sum of angles of a triangle is always 180° (Angle sum properly)

∴ A triangle cannot have angles 90°, 90° and 90°.

2. Find the value of x in the following figures :

Question (i).

Answer:

By angle sum property of a triangle

x + 53° + 60° = 180°

x + 113° = 180°

x = 180° – 113°

x = 67°

Question (ii).

Answer:

By angle sum property of a triangle

90° + x + 42° = 180°

132° + x = 180°

x = 180° – 132°

x = 48°

Question (iii).

Answer:

By angle sum property of a triangle

x + x + 70° = 180°

2x + 70° = 180°

2x = 180° – 70°

2x = 110°

x = \(\frac{110^{\circ}}{2}\)

x = 55°

Question (iv).

Answer:

By angle sum property of a triangle

x + 3x + 2x = 180°

6x = 180°

x = \(\frac{180^{\circ}}{6}\)

x = 30°

Question (v).

Answer:

By angle sum property of a triangle

x + x + x = 180°

3x = 180°

x = \(\frac{180^{\circ}}{3}\)

x = 60°

Question (vi).

Answer:

By angle sum property of a triangle

x – 5° + 60° + x + 5° = 180°

2x + 60° = 180°

2x = 180° – 60°

2x = 120°

x = \(\frac{120^{\circ}}{2}\)

x = 60°

3. Find the values of x and y in the following figures :

Question (i).

Answer:

Since in ΔABC, BC is produced to D

∴ 60° + x = 110°

(By exterior angle property)

x = 110°- 60°

x = 50° ………. (1)

Now, in ΔABC

60° + x + y = 180°

(By angle sum property of triangle)

60° + 50° + y = 180° [(By using (1)]

110° + y = 180°

y = 180° – 110°

y = 70°

Hence, x = 50°,

y = 70°

Question (ii).

Answer:

In ΔPQR,

∠P + ∠Q + ∠R = 180°

60° + 40° + x = 180°

(By angle sum property of triangle)

100° + x = 180°

x = 180° – 100°

x = 80°

Now,in ΔPQR, QR is produced

∴ y = 60° + 40°

(By exterior angle property)

y = 100°

Hence, x = 80°,

y = 100°

Question (iii).

Answer:

∠ACB = ∠ECD

∴ x = 80°….(1)

(vertically opposite angles)

∠ACD + ∠ECD = 180° (Linear pair)

∴ ∠ACD + 80° = 180° [by using (1)]

∠ACD = 180°- 80°

∠ACD = 100° ….(2)

In ΔABC, BC is produced to D

∴ x + y = ∠ACD

(By exterior angle property)

80° + y = 100°

(By using (1) and (2))

y = 100° – 80°

y = 20°

Hence, x = 80° and y = 20°

Question (iv).

Answer:

By angle sum property of a triangle

∠L + ∠M + ∠N = 180°

y + 90° + y = 180°

2 y + 90° = 180°

2y = 180° – 90°

2y = 90°

y = \(\frac{90^{\circ}}{2}\)

y = 45°

Question (v).

Answer:

∠ABC = ∠HBI

(Vertically opposite angles)

∴ y = x …(1)

∠BAC = ∠GAF

(Vertically opposite angles)

∴ ∠BAC = x ….(2)

∠ACB = ∠EFD

(Vertically opp. angles)

∠ACB = x …(3)

Now, in ΔABC

∠BAC + ∠ABC + ∠ACB = 180°

(By angle sum property of triangle)

x + x + x = 180°

[by using (1), (2) and (3)]

3x = 180°

x = \(\frac{180^{\circ}}{3}\)

x = 60°

y = x

= 60° (by using (1) and (4))

Hence, x = 60°, y = 60°

Question (vi).

Answer:

In ΔPQR, QR is produced to S,

∴ 2x – 5° = 50° + x + 5°

(By exterior angle property.)

2x – 5°= 55° + x

2x – x = 55° + 5°

x = 60° ….(i)

Now, by angle sum property of a ΔPQR

50° + x + 5° + y = 180°

55° + 60° + y = 180°

115°+ y = 180°

y = 180° – 115°

y = 65°

Hence, x = 60° and y = 65°

4. The angles of a triangle are in the ratio 5:6:7. Find the measure of each of the angles.

Solution:

Let the measure of the given angles be

(5x)°, (6x)°, (7x)°

By angle sum property of a triangle

(5x)° + (6x)° + (7x)° = 180°

(18x)° = 180°

x = \(\frac{180^{\circ}}{18}\)

x = 10

Required angles

= (5 × 10)°, (6 × 10)°, (7 × 10)°

= 50°, 60°, 70°

5. One angle of a triangle is 60°. The other two angles are in the ratio 4 : 8. Find the angles.

Solution:

One angle of triangle = 60°

Let the other two angles be (4x)° and (8x)°

By angle sum property of a triangle

60° + (4x)° + (8x)° = 180°

60° + (12x)° = 180°

(12x)° = 180° – 60°

(12x)° = 120°

x = \(\frac{120^{\circ}}{12}\)

x = 10

Required angles = (4x)°, (8x)°

(4 × 10)°, (8 × 10)°

= 40°, 80°

6. In a triangle ABC, ZB = 50°, ∠C = 62°. Find ∠A.

Solution:

In a ΔABC, ∠B = 50°, ∠C = 62°

By angle sum property of a triangle

∠A + ∠B + ∠C = 180°

∠A + 50° + 62° = 180°

∠A + 112° = 180°

∠A = 180° – 112°

∠A = 68°

7. In a right angled triangle two acute angles are in the ratio 2 : 3. Find the angles.

Solution:

In a right angle triangle one angle = 90°

Let the other two angles be (2x)°, (3x)°

By angle sum property of a triangle.

90° + (2x)° + (3x)° = 180°

90° + (5x)° = 180°

(5x)° = 180° – 90°

(5x)° = 90°

x = \(\frac{90^{\circ}}{5}\)

x = 18

Required angles = (2x)°, (3x)°

= (2 × 18)°, (3 × 18)°

= 36°, 54°

8. Three angles of a triangle are (2x + 20)°, (x + 30)° and (2x – 10)°. Find the angles.

Solution:

Since, we know that the sum of angles of a triangle is always 180°

∴ (2x + 20)° + (x + 30)° + (2x – 10)° = 180°

(5x + 40)° = 180°

(5x)° = 180° – 40°

(5x)° = 140°

x = \(\frac{140^{\circ}}{5}\)

x = 28

Required angles

= (2x + 20)°, (2x + 30)° and (2x – 10)°

= (2 × 28 + 20)°, (28 + 30)° and (2 × 28 -10)°

= (56 + 20)°, (58)° and (56 – 10)°

= 76°, 58° and 46°

9. Multiple choice questions :

Question (i).

A triangle can have two …………….

(a) Acute angles

(b) Obtuse angles

(c) Right angles

(d) None of these.

Answer:

(a) Acute angles

Question (ii).

A triangle is possible with measure of angles

(a) 30°, 40°, 100°

(b) 60°, 60°, 70°

(c) 60°, 50°, 70°

(d) 90°, 89°, 92°

Answer:

(c) 60°, 50°, 70°

Question (iii).

One of the equal angles of an isosceles triangle is 45° then its third angle is

(a) 45°

(b) 60°

(c) 100°

(d) 90°

Answer:

(d) 90°

Question (iv).

The number of obtuse angles that a triangle can have

(a) 2

(b) 1

(c) 3

(d) 4.

Answer:

(b) 1