Gujarat Board GSEB Solutions Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 6 Lines and Angles Ex 6.1

Question 1.

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

Solution:

It is given that

âˆ AOC +âˆ BOE = 70Â° ………..(1)

âˆ AOC = âˆ BOD

â‡’ âˆ AOC = 40Â°

Putting in eqn. (1),

40Â° + âˆ BOE = 70Â°

âˆ BOE = 70Â° – 40Â°

â‡’ âˆ BOE = 30Â°

âˆ AOE + âˆ BOE = 180Â° (Linear pair)

âˆ AOC + âˆ COE + âˆ BOE = 180Â°

â‡’ âˆ AOC + âˆ BOE + âˆ COE = 1800

â‡’ 70Â° +âˆ COE = 180Â°

â‡’ âˆ COE = 180Â° – 70Â°

â‡’ âˆ COE = 110Â°

Now, Reflex âˆ COE = 360Â° – 110Â° = 250Â°

Question 2.

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

Solution:

Let a = 2 x and b = 3x

âˆ POX + âˆ POY = 180Â° (Linear pair)

âˆ POX + 90Â° = 180Â°

âˆ POX = 180Â° – 90Â° = 90Â°

â‡’ a + b = 90Â°

â‡’ 2x + 3x = 90Â°

â‡’ 5x = 90Â°

â‡’ x = \(\frac {90Â°}{5}\) = 18Â°

âˆ´ a = 2x = 2 x 18Â°= 36Â°

â‡’ b = 3x = 3 x 18Â°= 54Â°

Now, âˆ XON + âˆ YON = 180Â° (Linear pair)

â‡’ c + 54Â° = 180Â° (âˆ XOM = b and âˆ XON = c)

â‡’ c = 180Â° – 54Â°

â‡’ c = 126Â°

Question 3.

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

Solution:

Given: âˆ PQR = âˆ PRQ

To Prove: âˆ PQS = âˆ PRT

Proof: âˆ PQS + âˆ PQR = 180Â° ………. (1) (Linear pair)

âˆ PRQ +âˆ PRT = 180Â° ………(2) (Linear pair)

But âˆ PQR =âˆ PRQ

Putting value âˆ PQR from eqn. (3) in eqn. (1)

âˆ PQS + âˆ PRQ = 180Â°

Now from eqn. (2) and (4), we have

âˆ PQS + âˆ PRQ = âˆ PRQ + âˆ PRT

âˆ PQS = âˆ PRT

Question 4.

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

Solution:

Given: x + y = w + z

To Prove: AOB is a line.

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

(Sum of all angles round a point is equal to 360Â°)

x + y + w + z = 360Â° ……..(1)

But x + y = w + z ……..(2)

From eqn. (1) and (2), we have

w + z + w + z = 360Â°

â‡’ 2w + 2z = 360Â°

â‡’2(w + z) = 360Â°

â‡’ w + z = \(\frac{360Â°}{2}\)

â‡’ w + z = 180Â°

w + z = 180Â° (Linear pair)

Hence,

âˆ BOD + âˆ AOD = 180Â°

Therefore AOB is a line.

Question 5.

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

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

Solution:

Given: Ray OR âŠ¥ PQ and ray OS is another ray lying between OP and OR.

To Prove: âˆ ROS = (âˆ QOS – âˆ POS)

Proof: Ray OR âŠ¥ PQ

âˆ´ âˆ POR = âˆ ROQ = 90Â°

Now, âˆ POS = âˆ POR – âˆ ROS

âˆ POS = 90Â° – âˆ ROS ………..(1)

âˆ QOS = âˆ QOR + âˆ ROS

âˆ QOS = 90Â° + âˆ ROS ……..(2)

Subtracting eqn (2) from eqn (1),

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

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

âˆ QOS – âˆ POS = 2âˆ ROS

2 âˆ ROS = âˆ QOS – âˆ POS

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

Question 6.

It is given that âˆ XYZ = 640 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:

âˆ XYZ = 64Â°

Ray YZ stands on PX.

âˆ XYZ + âˆ PYZ = 180Â° (Linear pair)

64Â° + âˆ PYZ = 180Â°

â‡’ âˆ PYZ = 180Â°- 64Â°

â‡’ âˆ PYZ = 116Â°

â‡’ âˆ QYZ = \(\frac{âˆ PYZ}{2}\)

(YQ is the bisector of âˆ PYZ)

âˆ QYZ = \(\frac{116Â°}{2}\)

âˆ QYZ = 58Â°

âˆ XYQ = âˆ XYZ + âˆ ZYQ = 64Â° + 58Â°

âˆ XYQ = 122Â°

âˆ QYP = âˆ QYZ

(YQ is the bisector of âˆ PYZ)

âˆ QYP = 58Â°

Reflex âˆ QYP = 360Â° – 58Â°

Reflex âˆ QYP = 302Â°