## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.2

Question 1.

Find the value of the unknown exterior angle x in each of the following diagrams:

Solution:

We know that the exterior angle of a triangle

is equal to the sum of its interior opposite angles.

Therefore,

(i) Ext. ∠x = 45° + 65° = 110°

(ii) Ext. ∠x = 55° + 40° = 95°

(iii) Ext. ∠x = 50° + 50° = 100°

Question 2.

Find the value of the unknown interior angle x in each of the following diagrams:

Solution:

We know that the exterior angle of a triangle

is equal to the sum of its interior opposite angles.

Therefore,

(i) In the given triangle,

Ext. ∠115° = x + 50°

⇒ x = 115° – 50° = 65°

⇒ x = 65°

(ii) In given triangle,

Ext. ∠80° = 30° + x

⇒ x = 80° – 30° = 50°

⇒ x = 50°

(iii) In given triangle,

Ext. ∠70° = x + 36°

⇒ x = 70° – 36° = 34°

⇒ x = 34°

Question 3.

Find the value of x in each of the following diagrams:

Solution:

We know that the exterior angle of a triangle

is equal to the sum of its interior opposite angles.

Therefore,

(i) In a given triangle,

Ext. ∠105° = 2x + x

⇒ 3x = 105°

⇒ x = 35°

x = 35°

(ii) In given triangle,

Ext. ∠125° = 2x + 3x

⇒ 5x = 125°

⇒ x = 25°

x = 25°

(iii) In given triangle,

∠A + ∠B + ∠C = 180°

(Sum of angles of a triangle)

⇒ x + 60° + 50° = 180°

⇒ x + 110° = 180°

⇒ x = 180° – 110° = 70°

x = 70°

Question 4.

Find the value of unknown x in each of the following:

Solution:

(i) In given triangle = Let ∆ABC

∠A + ∠B + ∠C = 180°

⇒ 50° + x + x = 180°

⇒ 2x = 180° – 50° = 130°

Hence, x = 65°

(ii) In the given figure,

Let the name of ∆ be ABC

∠ABC + ∠ABE = 180°

⇒ ∠ABC + 110° = 180°

⇒ ∠ABC = 180° – 110° = 70°

Similarly,

∠ACB + ∠ACD = 180°

⇒ ∠ACB + 120° = 180°

⇒ ∠ACB = 180° – 120° = 60°

Now in ∆ABC

∠BAC + ∠ABC + ∠ACB = 180°

⇒ x + 70° + 60° = 180°

⇒ x + 130° = 180°

⇒ x= 180° – 130° = 50°

⇒ x = 50°

(iii) Let the given triangle be named as ∆ABC,

where ∠C = 90°

In ∆ABC,

∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

⇒ 2x + x + 90° = 180°

⇒ 3x = 180° – 90° = 90°

⇒ x = 30°

Question 5.

Find the values of x and y in each of the following diagrams:

Solution:

We know that an exterior angle of a triangle

is equal to the sum of its interior opposite angle.

Therefore,

(i) Let the ∆’s name = ∆ABC

In ∆ABC

Ext. ∠ACD = ∠A + ∠B

120° = x + 50°

⇒ x = 120° – 50° = 70°

But ∠ACD + ∠ABC = 180° (Linear pair)

120° + y = 180°

⇒ y = 180° – 120° = 60°

x = 70°, y = 60°

(ii) In the given figure,

∠ACB = ∠DCE (Vertically opposite angles)

x = 60°

But ∠A + ∠B + ∠ACB = 180°

(Sum of angles of a triangle)

⇒ y + 40° + x = 180°

⇒ y + 40° + 60° = 180°

⇒ y + 100° = 180

⇒ y = 180° – 100° = 80°

Hence, x = 60°, y = 80°

(iii) In the given figure,

∠BAC = ∠EAF (Vertically opposite angles)

y = 90°

In ∆ABC,

∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

⇒ y + x + x = 180°

⇒ 90° + 2x = 180°

⇒ 2x = 180° – 90° = 90°

⇒ x = 45°

Hence, x = 45°

Question 6.

Find the values of x and y in each of the following diagrams:

Solution:

(i) In the given figure,

In ∆ABC,

x = y (Vertically opposite angles)

Similarly,

∠BAC = y, ∠ABC = y, ∠BCA = y

But ∠BAC + ∠ABC + ∠BCA = 180°

(Angles of a triangle)

⇒ y + y + y = 180°

⇒ 3y = 180°

⇒ y = 60°

x = 60°, y = 60°

(ii) In ∆ABC,

∠ABC + ∠ABD = 180°

⇒ x + 125° = 180°

⇒ x = 180°- 125° = 55°

and Ext. ∠ABD = x + y

⇒ 125° = 55° + y

⇒ y = 125° – 55° = 70°

x = 55°, y = 70°

(iii) In ∆ABC,

Ext. ∠ABD = ∠A + ∠B = 50° + 70° = 120°

But ∠ABC + ∠ABD = 180° (Linear pair)

⇒ x + ∠ABD = 180°

⇒ x + 120° = 180°

⇒ x = 180° – 120° = 60°

But in ∆ABD

Ext. ∠ABC = ∠D + ∠DAB

⇒ x = y + 30°

⇒ 60° = y + 30°

⇒ y = 60° – 30° = 30°

x = 60°, y = 30°

Question 7.

In the adjoining figure, find the size of each lettefed angle.

Solution:

In the given figure,

In ∆ABC

∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

⇒ x + 80° + 52° = 180°

⇒ x + 132° = 180°

⇒ x = 180° – 132° = 48°

∠DEC + ∠DEA = 180° (Linear pair)

⇒ 143° + z = 180°

⇒ z = 180° – 143° = 37°

Now in ∆ADE

Ext. ∠DEC = ∠A + ∠ADE

⇒ 143° = x + y

⇒ 143° = 48° + y

⇒ y = 143° – 48° = 95°

x = 48°, y = 95°, z = 37°

Question 8.

One of the angles of a triangle measures 80° and the other two angles are equal. Find the measure of each of the equal angles.

Solution:

One angle of an ∆ABC = 80°

Let ∠A = 80° and the other two angles are equal

Let ∠B = ∠C = x

In ∆ABC,

∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

⇒ 80° + x + x = 180°

⇒ 2x = 180° – 80° = 100°

⇒ x = 50°

∠A = 80°, ∠B = 50°, ∠C = 50°

Question 9.

If one angle of a triangle is 60° and the other two angles are in the ratio 2 : 3, find these angles.

Solution:

One angle of a triangle = 60°

Other two angles are in the ratio 2 : 3

Sum of other two angles = 180° – 60° = 120°

Let one of other two angles = 2x

Then third angle = 3x

2x + 3x = 120°

⇒ 5x = 120°

⇒ x = 24

Other two angles are 2x = 2 × 24 = 48°

and 3x = 3 × 24 = 72°

Other two angles of the triangle are 48°, 72°

Question 10.

If the angles of a triangle are in the ratio 1 : 2 : 3, find the angles. Classify the triangle in two different ways.

Solution:

Sum of angles of a triangle = 180°

Ratio in the angles of a triangle = 1 : 2 : 3

Let first angle = x

Second angle = 2x

Third angle = 3x

x + 2x + 3x = 180°

⇒ 6x = 180°

⇒ x = 30°

∴ First angle = 30°

Second angle = 30° × 2 = 60°

and third angle = 30° × 3 = 90°

∵ One angles is 90°

∴ It is a right angled triangle

∵ Sides an different

∴ It is a scalene triangle.

Question 11.

Can a triangle have three angles whose measures are

(i) 65°, 74°, 39°?

(ii) \(\frac { 1 }{ 3 }\) right angle, 1 right angle, 60°?

Solution:

We know that sum of angles of a triangle = 180°

(i) Angles are 65°, 74°, 39°

Sum of angles = 65° + 74° + 39° = 178°

178° ≠ 180°

There three angles can not be of triangle

(ii) \(\frac { 1 }{ 3 }\) right angle = \(\frac { 1 }{ 2 }\) × 90° = 30°

1 right angle = 90°

Third angle = 60°

Sum of angles = 30° + 90° + 60° = 180°

These angles are of a triangle.