## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3

Question 1.

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

Solution:

(i) In ∆ABC,

AB = AC

∠B = ∠C (Angles opposite to equal sides)

x = 50°

(ii) In ∆ABC,

AB = AC

∠B = ∠C (Angles opposite to equal sides)

∠B = ∠C = x

and Ext. ∠BAD = ∠B + ∠C = x + x

110° = ∠B + ∠C

110° = x + x = 2x

x = 55°

(iii) In the given figure,

CA = CB

∠A = ∠ABC = x

∠EBD = 36°

∠ABC = ∠EBD = 36° (Vertically opposite angles)

x = 36°

Question 2.

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

Solution:

(i) In ∆ABC,

AC = BC, ∠C = 40°

∠A = ∠B (Angles opposite to equal sides)

∠A = ∠B = x

But A + B + C = 180° (Angles of a triangle)

⇒ x + x + 40° = 180°

⇒ 2x = 180° – 40° = 140°

⇒ x = 70°

x = 70°

(ii) In ∆ABC,

∠B = 45°

AC = BC

∠A = ∠B = 45°

But ∠A + ∠B + ∠C = 180°

⇒ 45° + 45° + x° = 180°

⇒ x + 90°= 180°

⇒ x = 180° – 90° = 90°

x = 90°

(iii) In the given figure, EF || BC

In ∆ABC, Ext. ∠DAF = 50°

AB = BC

∠A = ∠C = x

EF || BC

∠DAF = ∠ABC = 50°

Now in ∆ABC

∠BAC + ∠ABC + ∠BCA = 180°

⇒ x + 50° + x = 180°

⇒ 2x = 180° – 50° = 130°

⇒ x = 65°

x = 65°

Question 3.

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

Solution:

(i) In the given figure of ∆ABC

AB = AC

∠ABC = ∠ACB = y

But, Ext. ∠ACD + ∠ACB = 120° (Linear pair)

∠ACB = 180° – 120° = 60°

y = 60°

Now in ∆ABC,

∠A + ∠B + ∠ACB = 180° (Angles of a triangle)

⇒ x + y + y = 180°

⇒ x + 60° + 60° = 180°

⇒ x + 120° = 180°

⇒ x = 180° – 120° = 60°

Here, x = 60°, y = 60°

(ii) Here, ∠A = ∠ACB

∠ACB = x (∵ angles opposite to equal sides)

⇒ x + 115° = 180°

⇒ x = 180° – 115° = 65

∠ACB = ∠A = 65°

Now, ∠A + ∠B + ∠ACB = 180° (Sum of ∠s of a ∆)

65° + 65° + ∠y = 180°

130° + ∠y = 180°

∠y = 180° – 130° = 50°

(iii) In ∆ABC,

∠ABC = ∠ACB (∠s opposite to equal sides)

∠ABC = x = ∠ACB

Now, In ∆ABC

48° + x + x = 180° (Sum of ∠s of a ∆)

⇒ 2x = 180° – 48°

⇒ 2x = 132°

⇒ x = 66°

In ∆ACD,

∠CAD = ∠CDA (∠s opposite to equal sides)

∠CAD = y = ∠CDA

Now, x + ∠ACD = 180° (Linear pair ∠s)

66° + ∠ACD = 180°

∠ACD = 180° – 66° = 114°

Now, in ∆ACD

y + y + 114°= 180° (Sum of ∠s of a ∆)

⇒ 2y + 114° = 180°

⇒ 2y = 180° – 114°

⇒ 2y = 66°

⇒ y = 33°

Hence, x = 66° and y = 33°

Question 4.

Calculate the size of each lettered angle in the following figures:

Solution:

(i) In ∆ABC,

AC = BC (Given)

∠ABC = ∠BAC (∠s opposite to equal sides)

∠ABC = 32°

⇒ x = 32°

Now, y = 32° + 32° = 64°

(∵ Exterior angle = Sum of two opposite interior ∠s)

In ∆BCD,

BC = BD

∠BDC = ∠BCD (∠s opposite to equal sides)

∠BDC = y = ∠BCD

∠BDC = 64° = ∠BCD

Now, z + 64° + 64° = 180° (Sum of ∠s of a ∆)

⇒ z + 128°= 180°

⇒ z = 180° – 128° = 52°

Hence, x = 32°, y = 64°, z = 52°

(iii) In ∆ABC,

AC = BC (Given)

∠ABC = ∠BAC (∠s opposite to equal sides)

∠ABC = 35° = ∠BAC

Now, ∠ACD = ∠ABC + ∠BAC (Exterior angle = Sum)

= 35° + 35° = 70°.

In ∆ACD,

AC = AD (Given)

∠ADC = ∠ACD (∠s opposite to equal sides)

∠ADC = 70°

Now, ∠x = ∠ABD + ∠ADB

(Exterior angle = Sum of two opposite interior ∠s)

= 35° + 70° = 105°

Hence, ∠x = 105°

Question 5.

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

Solution:

Ratio in the angles of a triangle are 1 : 2 : 1

Sum of angles of a triangle = 180°

Let first angle = x

Then second = 2x

and third angle x

x + 2x + x = 180°

⇒ 4x = 180°

⇒ x = 45°

Angles are 45°, 45° × 2 = 90° and 45°

Two angles are equal

Their opposite sides are also equal

It is an isosceles triangle

It’s one angle is 90°

It is a right-angled triangle.

Question 6.

In an isosceles triangle, a base angle is four times its vertical angle. Find all the angles of the triangle.

Solution:

In an isosceles triangle ABC, AB = AC

Base angles are equal

Let ∠B = ∠C = x

Angles of the triangle will be 80°, 80°, 20°