Lines and Angles Class 7 Questions

Lines and angles Class 7 questions and solutions are given here in an easily understandable way. As we know, lines and angles are one of the important concepts of Class 7 maths, where you can learn the relationship between different angles and lines. In this article, you will learn how to solve lines and angles Class 7 questions using simple techniques.

What are Lines and Angles?

In geometry, we get a line when we extend the two endpoints in either direction endlessly. Angles are formed when lines or line segments meet each other at a common point.

Also, check: Lines and Angles Class 7 Notes

Lines and Angles Class 7 Questions and Answers

1. Find the angle which is complementary to the following.

Solution:

Two angles are said to be complementary if the sum of their measures is 90°.

The given angle is 45°.

Let x° be the measure of its complementary angle.

Then, x + 20° = 90°

x = 90° – 45°

x = 45°

Hence, the complement of the given angle measures 45°.

2. What measure of an angle is supplementary to the given angle?

Solution:

Two angles are said to be supplementary if the sum of their measures is 180°.

The given angle is 105°.

Let x° be the measure of its complementary angle.

Then, x + 105° = 180°

x = 180° – 105°

x = 75°

Hence, the supplementary of the given angle measures 75°.

3. An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°?

Solution:

Let x and y be the complementary angles.

We know that the sum of measures of the pair of complementary angles is 90°.

That means, x + y = 90°

It is given in the question that x > 45°.

Adding y on both the sides, we get;

x + y > 45° + y

90° > 45° + y

90° – 45° > y

y < 45°

4. In the given figure, identify:

(i) Five pairs of adjacent angles.

(ii) Three linear pairs.

(iii) Two pairs of vertically opposite angles.

Solution:

From the given figure,

(i) Five pairs of adjacent angles are:

∠AOE and ∠EOC

∠EOC and ∠COB

∠AOC and ∠COB

∠COB and ∠BOD

∠EOB and ∠BOD

(ii) Linear pairs are:

∠AOE and ∠EOB

∠AOC and ∠COB

∠COB and ∠BOD

(iii) Vertically opposite angles are: ∠COB and ∠AOD

∠AOC and ∠BOD

5. Find the values of the angles x, y, and z in the following figure.

Solution:

In the given figure,

55° and x are vertically opposite angles.

So, x = 55°

x and y are the linear pair of angles.

That means x + y = 180°

55° + y = 180°

y = 180° – 55° = 125°

Also, y and z are vertically opposite angles.

Thus, y = z = 125°

Therefore, x = 55°, y = 125° and z = 125°.

6. In the figure, find the value of z.

Solution:

If a transversal cuts two parallel lines, then each pair of interior angles on the same

side of the transversal is supplementary.

So, z + 60° = 180°

z = 180° – 60° = 120°

Hence, the value of z is 120°.

7. In the below figure, p || q. Find the values of unknown angles.

Solution:

In the given figure,

d and 125° are corresponding angles.

d = ∠125°

We know that, the linear pair, i.e., the sum of adjacent angles is 180°.

That means e + 125° = 180°

e = 180° – 125°

e = 55°

e and f are vertically opposite angles.

So, e = f = 55°

Similarly, b = d = 125°

From the property of corresponding angles,

c = f = 55°

a = e = 55°

8. From the given figure, find the value of x.

Solution:

As we know, the sum of all angles about a point is equal to 360°.

Then, 100° + 46° + 64° + x = 360°

210° + x = 360°

x = 360° – 210°

x = 150°

Therefore, the value of x is 150°.

9. In the figure, lines PQ and ST intersect at O. If ∠POR = 90° and x : y = 3 : 2, then find the value of z.

Solution:

The sum of all angles about a straight line given in the figure, i.e., PQ is equal to 180°.

Then, ∠POR + ∠ROT + ∠TOQ = 180°

Given, x : y = 3 : 2

Let x = 3a, y = 2a

90° + 3a + 2a = 180°

90° + 5a = 180°

5a = 180° – 90°

5a = 90°

a = 90°/5

a = 18°

So, x = 3a = 3 × 18° = 54°

y = 2a = 2 × 18° = 36°

From the figure SOT is a straight line,

Then, z + y = 180°

z + 36° = 180°

z = 180° – 36°

z = 144°

10. In the figure, PQ||ST. Find the value of x + 2y.

Solution:

In the given figure, PO is a straight line.

As we know, the sum of angles on the straight line is equal to 180°.

Then,

y + ∠ PQR = 180°

y + 130° = 180°

y = 50°

Then,

∠QOS = ∠TSO [Co-interior angles]

x = 85°

Therefore, x + 2y = 85° + 2(50°) = 85° + 100° = 185°

Practice Questions on Lines and Angles Class 7

1. In the figure, ∠ROS is a right angle and ∠POR and ∠QOS are in the ratio 1 : 5. Find the measure of ∠QOS.
   
   
2. Find the angle which makes a linear pair with an angle of 61°.
3. Can two angles be supplementary if both of them are:

(i) acute? (ii) obtuse? (iii) right?

4. Find the measure of an angle whose measure is complementary to 27°
5. In the below figure, if QP || SR, find the value of a.