Most of the time students don’t ask their doubts in class due to shyness and fear of teacher. Due to this, they don’t get to know how the questions are being solved by the teacher. Most often this happens in Maths class. In other subjects, students can still find the answer if they re-read the chapter. But for mathematics, if they miss the concepts, then they would not be able to solve it. So, to help students here, we have provided the detailed NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles Exercise 6.1.

Students can download the NCERT Solutions of Class 9 Maths Chapter 6 Exercise 6.1 in PDF format by clicking on the link below. These solutions are prepared by subject experts and are described in detail. By going through NCERT Solutions PDF, students can easily understand the method of solving the questions.

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### Access other Exercise Solutions of Class 9 Maths Chapter 6 – Lines and Angles

The NCERT Class 9 Chapter 6 Lines and Angles has two more exercises. Students can access the solutions of these exercises by visiting the link below.

Exercise 6.2 Solutions 6 Questions (3 Short Answer Questions, 3 Long Answer Question)

Exercise 6.3 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

## NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles Exercise 6.1

**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°

**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 a 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).

**4. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.**

**Solution:**

For proving AOB is a straight line, we will have to prove x+y is a linear pair

i.e. x+y = 180°

We know that the angles around a point are 360° so,

x+y+w+z = 360°

In the question, it is given that,

x+y = w+z

So, (x+y)+(x+y) = 360°

2(x+y) = 360°

∴ (x+y) = 180° (Hence proved).

**5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ROS = ½ (QOS – POS).**

**Solution:**

In the question, it is given that (OR ⊥ PQ) and POQ = 180°

So, POS+ROS+ROQ = 180°

Now, POS+ROS = 180°- 90° (Since POR = ROQ = 90°)

∴ POS + ROS = 90°

Now, QOS = ROQ+ROS

It is given that ROQ = 90°,

∴ QOS = 90° +ROS

Or, QOS – ROS = 90°

As POS + ROS = 90° and QOS – ROS = 90°, we get

POS + ROS = QOS – ROS

2 ROS + POS = QOS

Or, ROS = ½ (QOS – POS) (Hence proved).

**6. It is given that XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ZYP, find XYQ and reflex QYP.**

**Solution:**

Here, XP is a straight line

So, XYZ +ZYP = 180°

Putting the value of XYZ = 64° we get,

64° +ZYP = 180°

∴ ZYP = 116°

From the diagram, we also know that ZYP = ZYQ + QYP

Now, as YQ bisects ZYP,

ZYQ = QYP

Or, ZYP = 2ZYQ

∴ ZYQ = QYP = 58°

Again, XYQ = XYZ + ZYQ

By putting the value of XYZ = 64° and ZYQ = 58° we get.

XYQ = 64°+58°

Or, XYQ = 122°

Now, reflex QYP = 180°+XYQ

We computed that the value of XYQ = 122°.

So,

QYP = 180°+122°

∴ QYP = 302°

This exercise will recall students about the types of angles (acute, obtuse, right, straight, reflex, complementary and supplementary) that they have studied in earlier classes. The new concepts in the exercise will start from the Intersecting and non-intersecting lines, and then Pairs of Angles are explained in detail along with a theorem proof and some examples in the NCERT textbook.

The questions in the exercise are based on finding the value of angles from the given diagram and proving the statements. In the NCERT Class 9 Maths Solutions, we have provided the step by step solution along with the diagram for students convenience. Also, the questions are solved by the easiest method, so can student can easily understand it.

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