# Lines and Angles Chapter 6 Class 9 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 9 Maths Chapter 6‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 9 Maths Chapter 6 Exercise Solution’ using the download button.

## 6.1 Introduction

In Chapter 5, you have studied that a minimum of two points are required to draw a line.

You have also studied some axioms and, with the help of these axioms, you proved some other statements.

In this chapter, you will study the properties of the angles formed when two lines intersect each other, and also the properties of the angles formed when a line intersects two or more parallel lines at distinct points.

Further, you will use these properties to prove some statements using deductive reasoning (see Appendix 1). You have already verified these statements through some activities in the earlier classes.

In your daily life, you see different types of angles formed between the edges of plane surfaces. For making a similar kind of model using the plane surfaces, you need to have a thorough knowledge of angles.

For instance, suppose you want to make a model of a hut kept in the school exhibition using bamboo sticks.

Imagine how you would make it? You would keep some of the sticks parallel to each other, and some sticks would be kept slanted.

Whenever an architect has to draw a plan for a multistoried building, she has to draw intersecting lines and parallel lines at different angles.

Without the knowledge of the properties of these lines and angles, do you think she can draw the layout of the building?

### NCERT Solutions Class 9 Maths Chapter 6 Lines and Angles

1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC +∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:

From the diagram, we have

(∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) forms a straight line.

So, ∠AOC+∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180°

Now, by putting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get

∠COE = 110° and ∠BOE = 30°

So, reflex ∠COE = 360o – 110o = 250o

2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a: b = 2 : 3, find c.

Solution:

We know that the sum of linear pair is always equal to 180°

So,

∠POY +a +b = 180°

Putting the value of ∠POY = 90° (as given in the question) we get,

a+b = 90°

Now, it is given that a : b = 2 : 3 so,

Let a be 2x and b be 3x

∴ 2x+3x = 90°

Solving this we get

5x = 90°

So, x = 18°

∴ a = 2×18° = 36°

Similarly, b can be calculated and the value will be

b = 3×18° = 54°

From the diagram, b+c also forms a straight angle so,

b+c = 180°

c+54° = 180°

∴ c = 126°

3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution:

Since ST is a straight line so,

PQS+PQR = 180° (linear pair) and

PRT+PRQ = 180° (linear pair)

Now, PQS + PQR = PRT+PRQ = 180°

Since PQR =PRQ (as given in the question)

PQS = PRT. (Hence proved).