## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 10 Lines and Angles Check Your Progress

Question 1.

Find the supplementary angle of each of the following angles:

(i) \(\frac { 1 }{ 2 }\) of 90°

(ii) \(\frac { 3 }{ 7 }\) of 280°

Solution:

(i) \(\frac { 1 }{ 2 }\) of 90° = 45°

Supplementary angle of 45° = 180° – 45° = 135°

Hence, supplementary angle of \(\frac { 1 }{ 2 }\) of 95° is 135°.

(ii) \(\frac { 3 }{ 7 }\) of 280° = 120°

Supplementary of 120° is 180° – 120° = 60°

Question 2.

How many degrees are there in an angle which is one-fifth of its complement?

Solution:

Let one angle = x

But it is \(\frac { 1 }{ 5 }\) th of its complementary angle

Complementary angle = 5x

x + 5x = 90°

⇒ 6x = 90°

⇒ x = 15°

Required angle = 15°

Question 3.

If two angles are supplementary and one angle is 5° more than four-times the other, find the angles.

Solution:

Let one angle of supplementary angles = x

The second = 4x + 5°

x + 4x + 5° = 180°

⇒ 5x = 180° – 5 = 175°

⇒ x = 35°

One angle = 35°

and other angle = 180° – 35° = 145°

Question 4.

The given diagram shows two intersecting straight lines. Find the values of x, y and z.

Solution:

In the given figure,

2x + 15 = 3x – 10° (Vertically opposite angles)

⇒ 3x – 2x = 15° + 10°

⇒ x = 25°

Now, y + 2x + 15° = 180° (Linear pair)

y + 2 × 25° + 15° = 180°

⇒ y + 50° + 15° = 180°

⇒ y + 65° = 180°

⇒ y = 180° – 65° = 115°

and z = y = 115° (Vertically opposite angles)

x = 25°, y = 115°, z = 115°

Question 5.

In the given diagram, lines AB and CD intersect at O. If ∠1 + ∠3 = 78°, find the size of ∠2.

Solution:

In the given figure,

AB and CD intersect each other at O

∠1 + ∠3 = 78°

But ∠1 = ∠3 (Vertically opposite angles)

∠1 = ∠3 = \(\frac { 78 }{ 2 }\) = 39°

But ∠1 + ∠2 = 180° (Linear pair)

39° + ∠2 = 180°

∠2 = 180° – 39° = 141°

∠2 = 141°

Question 6.

(a) In the figure (i) given below, l || m. If ∠5 = 65°, find all other angles.

(b) In the figure (ii) given below, l || m. Find the values of x, y and z.

(c) In the figure (iii) given below, l || m and p || q. Find the values of x, y and z.

Solution:

In the given figure, l || m

∠5 = 65°

∠5 = ∠3 (Alternate angle)

∠3 = 65°

∠5 = ∠1

∠1 = 65°

∠4 + ∠5 = 180° (Sum of co-interior angle)

∠4 + 65° = 180°

∠4 = 180° – 65° = 115°

∠2 = ∠4 = 115° (Vertically opposite angles)

∠6 = ∠4 = 115° (Alternate angles)

∠6 = ∠8 = 115° (Vertically opposite angles)

∠7 = ∠5 = 65° (Vertically opposite angles)

∠1 = 65°, ∠2 = 115°, ∠3 = 65°, ∠4 = 115° ∠6 = 115°,

∠7 = 65°, ∠8 = 115°

(ii) In the given figure, l || m

Given ∠s are 40° and 105°

x = 40° (Alternate angles)

z + 105° = 180° (Sum of co-interior angles)

⇒ z = 180°- 105° = 75°

Now, 40° + y + z = 180°

(Angles on one side of a straight line)

⇒ 40° + y + 75° = 180°

⇒ y + 115° = 180°

⇒ y – 180° – 115° = 65°

x = 40°, y = 65°, z = 75°

(iii) In the given figure,

Given angle is 105°

x + 105° = 180°

x = 180° – 105° = 75°

But x + y = 180° (Sum of co-interior anlges)

⇒ 75° + y = 180° (Sum of co-interior anlges)

⇒ y = 180° – 75° = 105°

Similarly, y + z = 180° (Sum of co-interior anlges)

⇒ 105° + z = 180°

⇒ z = 180° – 105° = 75°

Hence, x = 75°, y = 105°, z = 75°