NCERT solution class 9 chapter 6 Lines and Angles exercise 6.1 mathematics

EXERCISE 6.1


Page No 96:

Question 1:

In the given figure, lines AB and CD intersect at O. If and  find ∠BOE and reflex ∠COE.

Answer:


Page No 97:

Question 2:

In the given figure, lines XY and MN intersect at O. If ∠POY =  and a:b = 2 : 3, find c.

Answer:

Let the common ratio between a and b be x.

∴ a = 2x, and b = 3x

XY is a straight line, rays OM and OP stand on it.

∴ ∠XOM + ∠MOP + ∠POY = 180º

b + a + ∠POY = 180º

3x + 2x + 90º = 180º

5x = 90º

x = 18º

a = 2x = 2 × 18 = 36º

b = 3x= 3 ×18 = 54º

MN is a straight line. Ray OX stands on it.

∴ b + c = 180º (Linear Pair)

54º + c = 180º

c = 180º − 54º = 126º

∴ c = 126º

Question 3:

In the given figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Answer:

In the given figure, ST is a straight line and ray QP stands on it.

∴ ∠PQS + ∠PQR = 180º (Linear Pair)

∠PQR = 180º − ∠PQS (1)

∠PRT + ∠PRQ = 180º (Linear Pair)

∠PRQ = 180º − ∠PRT (2)

It is given that ∠PQR = ∠PRQ.

Equating equations (1) and (2), we obtain

180º − ∠PQS = 180 − ∠PRT

∠PQS = ∠PRT


Page No 96:

Question 1:

In the given figure, lines AB and CD intersect at O. If and  find ∠BOE and reflex ∠COE.

Answer:

Video Solution

Page No 97:

Question 2:

In the given figure, lines XY and MN intersect at O. If ∠POY =  and a:b = 2 : 3, find c.

Answer:

Let the common ratio between a and b be x.

∴ a = 2x, and b = 3x

XY is a straight line, rays OM and OP stand on it.

∴ ∠XOM + ∠MOP + ∠POY = 180º

b + a + ∠POY = 180º

3x + 2x + 90º = 180º

5x = 90º

x = 18º

a = 2x = 2 × 18 = 36º

b = 3x= 3 ×18 = 54º

MN is a straight line. Ray OX stands on it.

∴ b + c = 180º (Linear Pair)

54º + c = 180º

c = 180º − 54º = 126º

∴ c = 126º

Question 3:

In the given figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Answer:

In the given figure, ST is a straight line and ray QP stands on it.

∴ ∠PQS + ∠PQR = 180º (Linear Pair)

∠PQR = 180º − ∠PQS (1)

∠PRT + ∠PRQ = 180º (Linear Pair)

∠PRQ = 180º − ∠PRT (2)

It is given that ∠PQR = ∠PRQ.

Equating equations (1) and (2), we obtain

180º − ∠PQS = 180 − ∠PRT

∠PQS = ∠PRT

Video Solution

Question 4:

In the given figure, if then prove that AOB is a line.

Answer:

It can be observed that,

x + y + z + w = 360º (Complete angle)

It is given that,

x + y = z + w

∴ x + y + x + y = 360º

2(x + y) = 360º

x + y = 180º

Since x and y form a linear pair, AOB is a line.

Question 5:

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

Answer:

It is given that OR ⊥ PQ

∴ ∠POR = 90º

⇒ ∠POS + ∠SOR = 90º

∠ROS = 90º − ∠POS … (1)

∠QOR = 90º (As OR ⊥ PQ)

∠QOS − ∠ROS = 90º

∠ROS = ∠QOS − 90º … (2)

On adding equations (1) and (2), we obtain

2 ∠ROS = ∠QOS − ∠POS

∠ROS = (∠QOS − ∠POS)


Question 6:

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

Answer:

It is given that line YQ bisects ∠PYZ.

Hence, ∠QYP = ∠ZYQ

It can be observed that PX is a line. Rays YQ and YZ stand on it.

∴ ∠XYZ + ∠ZYQ + ∠QYP = 180º

⇒ 64º + 2∠QYP = 180º

⇒ 2∠QYP = 180º − 64º = 116º

⇒ ∠QYP = 58º

Also, ∠ZYQ = ∠QYP = 58º

Reflex ∠QYP = 360º − 58º = 302º

∠XYQ = ∠XYZ + ∠ZYQ

= 64º + 58º = 122º


 

Leave a Reply

Your email address will not be published. Required fields are marked *