ML Aggarwal Solutions for Class 7 Maths Chapter 10 - Lines and Angles

ML Aggarwal Solutions for Class 7 Maths Chapter 10 – Lines and Angles are provided here in a step-by-step manner to help students grasp the concepts easily and score well in their exams. This chapter mainly deals with problems based on lines and angles. For a better understanding of the concepts, students can solve the problems by referring to the ML Aggarwal Solutions, which are designed by the expert faculty team at BYJU’S with utmost care. Students aspiring to secure high marks in their annual examinations are advised to practise the solutions on a regular basis.

Chapter 10, Lines and Angles, provides solutions to questions related to each and every topic covered in this chapter. Students can easily download ML Aggarwal Class 7 Solutions in PDFs for free from the links given below and can use them for future reference, as well.

ML Aggarwal Solutions for Class 7 Maths Chapter 10 – Lines and Angles

Download PDF

carouselExampleControls112

Previous Next

Access Answers to ML Aggarwal Solutions for Class 7 Maths Chapter 10 – Lines and Angles

1. (i) Can two right angles be complementary?

(ii) Can two right angles be supplementary?

(iii) Can two adjacent angles be complementary?

(iv) Can two adjacent angles be supplementary?

(v) Can two obtuse angles be adjacent?

(vi) Can an acute angle be adjacent to an obtuse angle?

(vii) Can two right angles form a linear pair?

Solution:

(i) No, as the sum of two complementary angles is 90^o.

(ii) Yes

(iii) Yes

(iv) Yes

(v) Yes

(vi) Yes

(vii) Yes

2. Find the complement of each of the following angles:

Solution:

(i) Complement of 25^o is 90^o – 25^o = 65^o.

(ii) Complement of 63^o is 90^o – 63^o = 27^o.

(iii) Complement of 57^o is 90^o – 57^o = 33^o.

3. Find the supplement of each of the following angles:

Solution:

(i) Supplement of 105^o is 180^o – 105^o = 75^o.

(ii) Supplement of 87^o is 180^o – 87^o = 93^o.

(iii) Supplement of 142^o is 180^o – 142^o = 38^o.

4. Identify which of the following pairs of angles are complementary and which are supplementary:
(i) 55°, 125°
(ii) 34°, 56°
(iii) 137°, 43°
(iv) 112°, 68°
(v) 45°, 45°
(vi) 72°, 18°
Solution:

(i) 55°, 125° are supplementary angles.

(ii) 34°, 56° are complementary angles.

(iii) 137°, 43° are supplementary angles.

(iv) 112°, 68° are supplementary angles.

(v) 45°, 45° are complementary angles.

(vi) 72°, 18° are complementary angles.

5. (i) Find the angle which is equal to its complement.
(ii) Find the angle which is equal to its supplement.
Solution:

(i) The angle which is equal to its complement is

90/2 = 45^o.

So, 45^o is a complement to 45^o.

(ii) The angle which is equal to its supplement is

180/2 = 90^o.

So, 90^o is a supplement to 90^o.

6. Two complementary angles are (x + 4)° and (2x – 7)°, find the value of x.
Solution:

Given:

Two complementary angles are (x + 4)° and (2x – 7)°

x + 4 + 2x – 7 = 90^o

3x – 3 = 90^o

3x = 90 + 3

3x = 93

x = 93/3

x = 31^o

∴ The value of x is 31^o.

7. Two supplementary angles are in the ratio of 2: 7; find the angles.
Solution:

Given:

Two supplementary angles are in the ratio of 2: 7

We know the sum of the angles is 180^o.

So, first angle = [180^o/(2+7)] × 2

= (180/9) × 2

= 40^o

Second angle = [180^o/(2+7)] × 7

= (180/9) × 7

= 140^o

∴ The angles are 40^o and 140^o.