NCERT Solutions for Exercise 6.1 Class 9 Maths Chapter 6

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

Edited By Safeer PP | Updated on Jul 18, 2022 02:16 PM IST

A line with two endpoints is referred to as a line segment, and a portion of a line with one endpoint is referred to as a ray.

Collinear points are those that have three or more points on the same line; otherwise, they are non-collinear points. When two rays originate from the same endpoint, an angle is formed.

The rays that form an angle are referred to as the angle's arms, and the endpoint is referred to as the vertex of the angle.

NCERT Solutions Class 9 Maths exercise 6.1 – In NCERT solutions for Class 9 Maths chapter 6 exercise 6.1 we get to know that an acute angle has a range of 0° to 90°, whereas a right angle has a value of 90°. An obtuse angle is one that is greater than 90° but less than 180°.

Also, keep in mind that a straight angle equals 180°. A reflex angle is an angle that is greater than 180° but less than 360°. Furthermore, two angles with a sum of 90° are referred to as complementary angles, while two angles with a sum of 180° are referred to as supplementary angles.

Along with Class 9 Maths, chapter 8 exercise 6.1 the following exercises are also present.

Lines and Angles Class 9 Maths Chapter 6 Exercise: 6.1

Q1 In Fig. 6.13, lines AB and CD intersect at O. If $\angle$ AOC + $\angle$ BOE = 70° and $\angle$ BOD = 40°, find $\angle$ BOE and reflex $\angle$ COE.

1640149435860

Answer:

1595609937996
Given that,
AB is a straight line. Lines AB and CD intersect at O. $\angle AOC + \angle BOE = 70^0$ and $\angle$ BOD = $40^0$
Since AB is a straight line
$\therefore$ $\angle$ AOC + $\angle$ COE + $\angle$ EOB = $180^0$
$\Rightarrow \angle COE = 180^0-70^0=110^0$ [since $\angle AOC + \angle BOE = 70^0$ ]

So, reflex $\angle$ COE = $360^0-110^0 = 250^0$
It is given that AB and CD intersect at O
Therefore, $\angle$ AOC = $\angle$ BOD [vertically opposite angle]
$\Rightarrow \angle COA = 40^0$ [ GIven $\angle$ BOD = $40^0$ ]
Also, $\angle AOC + \angle BOE = 70^0$
So, $\angle$ BOE = $30^0$

Q2 In Fig. 6.14, lines XY and MN intersect at O. If $\angle POY = 90^o$ and a : b = 2 : 3, find c.

1640149490726

Answer:

1595610210463
Given that,
Line XY and MN intersect at O and $\angle$ POY = $90^0$ also $a:b = 2:3 \Rightarrow b = \frac{3a}{2}$ ..............(i)

Since XY is a straight line
Therefore, $\\a+b+\angle POY = 180^0\\ a+b = 180^0-90^0 = 90^0$ ...........(ii)
Thus, from eq (i) and eq (ii), we get
$\\\Rightarrow \frac{3a}{2}+a = 90^0\\$
$\\\Rightarrow a = 36^0\\$
So, $b = 54^0\\$
Since $\angle$ MOY = $\angle$ c [vertically opposite angles]
$\angle$ a + $\angle$ POY = c
$126^0 =c$

Q3 In Fig. 6.15, $\angle$ PQR = $\angle$ PRQ, then prove that $\angle$ PQS = $\angle$ PRT.

1640149622015

Answer:

1595610294843
Given that,
ABC is a triangle such that $\angle$ PQR = $\angle$ PRQ and ST is a straight line.
Now, $\angle$ PQR + $\angle$ PQS = $180^0$ {Linear pair}............(i)
Similarly, $\angle$ PRQ + $\angle$ PRT = $180^0$ ..................(ii)

equating the eq (i) and eq (ii), we get

$\angle PQR +\angle PQS =\angle PRT + \angle PRQ$ {but $\angle$ PQR = $\angle$ PRQ }
Therefore, $\angle$ PQS = $\angle$ PRT
Hence proved.

Q4 In Fig. 6.16, if x + y = w + z , then prove that AOB is a line.

1640149658149

Answer:

1595610388882
Given that,
$x+y = z+w$ ..............(i)
It is known that, the sum of all the angles at a point = $360^0$
$\therefore$ $x+y+z+w=360^0$ ..............(ii)

From eq (i) and eq (ii), we get

$\\2(x+y)=360^0\\ x+y = 180^0$

Hence proved AOB is a line.

Q5 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 $\angle \textup{ROS} = \frac{1}{2}(\angle \textup{QOS} - \angle \textup{POS})$

1640149689300

Answer:

1595610457579
Given that,
POQ is a line, OR $\perp$ PQ and $\angle$ ROQ is a right angle.
Now, $\angle$ POS + $\angle$ ROS + $\angle$ ROQ = $180^0$ [since POQ is a straight line]
$\\\Rightarrow \angle POS + \angle ROS = 90^0\\ \Rightarrow \angle ROS = 90^0-\angle POS$ .............(i)
and, $\angle$ ROS + $\angle$ ROQ = $\angle$ QOS
$\angle ROS = \angle QOS -90^0$ ..............(ii)

Add the eq (i ) and eq (ii), we get

$\angle \textup{ROS} = \frac{1}{2}(\angle \textup{QOS} - \angle \textup{POS})$

hence proved.

Q6 It is given that $\angle$ XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects $\angle$ ZYP, find $\angle$ XYQ and reflex $\angle$ QYP.

Answer:

1595611505816
Given that,
$\angle$ XYZ = $64^0$ and XY produced to point P and Ray YQ bisects $\angle$ ZYP $\Rightarrow \angle QYP = \angle ZYQ$
Now, XYP is a straight line
So, $\angle$ XYZ + $\angle$ ZYQ + $\angle$ QYP = $180^0$
$\Rightarrow 2(\angle QYP )=180^0 - 64^0 = 116^0$
$\Rightarrow (\angle QYP )= 58^0$

Thus reflex of $\angle$ QYP = $360^0- 58^0 = 302^0$

Since $\angle$ XYQ = $\angle$ XYZ + $\angle$ ZYQ [ $\because$ $\angle QYP = \angle ZYQ$

$\angle XYQ$ = $64^0+58^0 = 122^0$

Benefits of NCERT Solutions for Class 9 Maths Exercise 6.1

Exercise 6.1 Class 9 Maths, is based on the concepts of lines and angles
From Class 9 Maths chapter 6 exercise 6.1 we learn about parallel lines and pairs of angles
Understanding the concepts from Class 9 Maths chapter 6 exercise 6.1 will allow us to understand the concepts related to angle sum properties of LINES AND ANGLES.

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This Story also Contains

Lines and Angles Class 9 Maths Chapter 6 Exercise: 6.1
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Also see

NCERT Solutions of Class 9 Subject Wise

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Frequently Asked Questions (FAQs)

1. What is the main concept of NCERT solutions for Class 9 Maths exercise 6.1?

This exercise is about lines and angles, types of angles like vertically opposite angles and axioms and concepts like parallel lines and pair of angles.

2. What is an angle?

An angle is a figure which is formed by two rays, called the sides of the angle, both of which shares a common endpoint, called the vertex of the angle.

3. What are collinear points?

If we have three or more points that lie on the same line then they are called collinear points.

4. Name the different types of angles?

There are various types of angles are:

Acute angle
Right angle
Obtuse angle
Straight angle
Reflex angle

5. What do you understand by reflex angles?

A reflex angle is an angle greater than 180° and less than 360°.

6. What are intersecting lines?

When we have two or more lines meeting or crossing each other in a plane, they are termed as intersecting lines.

7. What is pair of angles?

When we have two lines sharing a common endpoint, called Vertex then an angle that is formed between these two lines is known as the pair of angles.

8. What do you understand by vertical opposite angles?

When we have two lines intersecting each other, the opposite angles are equal. These angles are known as the vertically opposite angles because they are equal and opposite to each other at the vertex.

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