Haryana State Board HBSE 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.1 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 6 Lines and Angles Exercise 6.1

Question 1.

In figure 6.26, lines AB and CD intersect at O. If âˆ AOC + âˆ BOE = 70Â° and âˆ BOD = 40Â°, find âˆ BOE and reflex âˆ COE.

Solution:

âˆ AOC = âˆ BOD, (Vertically opposite angles)

â‡’ âˆ AOC = 40Â°,

âˆµ âˆ BOD = 40Â°, given …(i)

Now, âˆ AOC + âˆ BOE = 70Â°, (Given)

â‡’ 40Â° + âˆ BOE = 70Â°

â‡’ âˆ BOE = 70Â° – 40Â° = 30Â° …(ii)

Ray OC stands on line AB.

âˆ´ âˆ AOC + âˆ COB = 180Â°,

(Linear pair axiom)

â‡’ âˆ AOC + âˆ COE + âˆ BOE = 180Â°,

âˆµ âˆ COB = âˆ COE + âˆ BOE

â‡’ 40Â° + âˆ COE +30Â° = 180Â°, (From (i) and (ii), we have âˆ AOC = 40Â°, âˆ BOE = 30Â°)

â‡’ âˆ COE + 70Â° = 180Â°

â‡’ âˆ COE = 180Â° – 70Â° = 110Â°

So, Reflex

âˆ COE = 360Â° – 110Â° = 250Â°

Hence, âˆ BOE = 30Â°

and reflex âˆ COE = 250Â°.

Question 2.

In figure 6-27, lines XY and MN intersect at O. If âˆ POY = 90Â° and a : b = 2 : 3, find c.

Solution:

Since ray OP stands on line XY. Therefore âˆ POY + âˆ POX = 180Â°,

(Linear pair axiom)

â‡’ 90Â° + âˆ POX = 180Â°,

[âˆµ âˆ POY = 90Â°, given]

â‡’ âˆ POX = 180Â° – 90Â° = 90Â°

â‡’ a + b = 90Â°

But a : b= 2 : 3 (given)

Sum of ratios = 2 + 3 = 5

âˆ´ a = \(\frac {2}{5}\) Ã— 90Â° = 36Â°

and b = \(\frac {3}{5}\) Ã— 90Â° = 54Â°

Now, MN is a line.

Since, ray OX stands on line MN.

Therefore, âˆ MOX + âˆ XON = 180Â°,

(Linear pair axiom)

â‡’ b + c = 180Â°

â‡’ 54Â° + c = 180Â°

â‡’ c = 180Â° – 54Â° = 126Â°

Hence, c = 126Â°.

Question 3.

In figure 6.28, âˆ PQR = âˆ PRQ, then prove that âˆ PQS = âˆ PRT.

Solution :

Since, ray QP stands on line SQT. Therefore, âˆ PQS + âˆ PQR = 180Â°,

(Linear pair axiom) …(i)

Since, ray RP stands on line SQT. Therefore, âˆ PRT + âˆ PRQ = 180Â°,

(Linear pair axiom) ..(ii)

From (i) and (ii), we get

âˆ PQS + âˆ PQR = âˆ PRT + âˆ PRQ

â‡’ âˆ PQS + âˆ PQR = âˆ PRT + âˆ PQR,

[âˆµ âˆ PRQ = âˆ PQR, given)

â‡’ âˆ PQS = âˆ PRT, [Subtracting âˆ PQR from both sides]

Hence proved

Question 4.

In figure 6.29, if x + y = w + z, then prove that AOB is a line.

Solution:

Since, sum of all the angles round a point is equal to 360Â°. Therefore,

âˆ AOC + âˆ BOC + âˆ BOD + âˆ AOD = 360Â°

â‡’ y + x + w + z = 360Â°

â‡’ x + y + x + y = 360Â° [Given w + z = x + y]

â‡’ 2x + 2y = 360Â°

â‡’ 2(x + y) = 360Â°

â‡’ x + y = \(\frac {360Â°}{2}\) = 180Â°

According to Axiom 6.2, if the sum of two adjacent angles is 180Â°, then non-common arms of the angles form a line.

Hence, AOB is a line. Hence Proved

Question 5.

In figure 6.30, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that:

âˆ ROS = \(\frac {1}{2}\)(âˆ QOS – âˆ POS).

Solution:

Since, ray OR âŠ¥ line PQ.

âˆ ROQ + âˆ ROP = 180Â°,

(Linear pair axiom)

â‡’ 90Â° + âˆ ROP = 180Â° [âˆµâˆ ROQ = 90Â°)

â‡’ âˆ ROP = 180Â° – 90Â° = 90Â°

â‡’ âˆ ROS + âˆ POS = 90Â°,

âˆµ [âˆ ROP = âˆ ROS + âˆ POS)

â‡’ âˆ POS = 90Â° – âˆ ROS …(i)

âˆ QOS =âˆ ROS + âˆ ROQ

â‡’ âˆ QOS = âˆ ROS + 90Â° …(ii)

Subtracting (i) from (ii), we get

âˆ QOS – âˆ POS = âˆ ROS + 90Â° – (90Â° – âˆ ROS)

â‡’ âˆ QOS – âˆ POS = âˆ ROS + 90Â° – 90Â° + âˆ ROS

â‡’ âˆ QOS – âˆ POS = 2âˆ ROS

â‡’ \(\frac {1}{2}\)(âˆ QOS – âˆ POS) = âˆ ROS

â‡’ âˆ ROS = \(\frac {1}{2}\)(âˆ QOS – âˆ POS).

Hence proved

Question 6.

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

Solution:

It is given that XY produced to P. So, XP is a straight line.

Now, ray YZ stands on line XP.

âˆ´ âˆ XYZ + âˆ ZYP = 180Â°

(Linear pair axiom)

â‡’ 64Â° + âˆ ZYP = 180Â°, [âˆµâˆ XYZ = 64Â°, given]

â‡’ âˆ ZYP = 180Â° – 64Â° = 116Â°

Since, ray YQ bisects âˆ ZYP.

âˆ ZYQ = âˆ QYP = \(\frac {116Â°}{2}\) = 58Â°

Now, âˆ XYQ = âˆ XYZ + âˆ ZYQ

= 64Â° + 58Â°

= 122Â°

Reflex âˆ QYP = 360Â° – âˆ QYP

= 360Â° – 58Â° = 302Â°

Hence, âˆ XYQ = 122Â° and Reflex âˆ QYP = 302Â°.