# NCERT Solutions For Class 9 Math Chapter – 6 Exercise – 6.1 NCERT Solutions For Class 9 Math Chapter – 6 Exercise – 6.1

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Q1.  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°

So, reflex ∠COE = 360o – 110o = 250o

1. 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 linear pair are 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°

1. 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)

Q4.  In fig 6.16 If x + y = w + z then prove that AOB is a line Solution:-

X + y  = 1800

X + y + w + z = 3600

X+y = w + z

So , (x+y) + (x+y) = 3600

2(x+y) = 3600

(x+y) = 1800 (Hence proved)

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 ROS = 1/2(QOS – POS). Solution:-

POQ = 1800

POS + ROS + ROQ = 1800 – 900 (Since POR = ROQ = 900)

POS + ROS = 900

QOS = ROQ + ROS

QOS = 900 + ROS

QOS – ROS = 900

POS + ROS = 900 and  QOS – ROS = 900 we get

POS + ROS = QOS – ROS

2ROS + POS = QOS

Or , ROS = 1/2 (QOS – POS) (Hence proved).

Q6. It is given that XYZ = 640 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 = 1800

Putting the value of XYZ =  640 we get ,

640 + ZYP = 1800

ZYP = 1160

YQ bisects ZYP ,

ZYQ = QYP

ZYP = 2ZYQ

ZYQ = QYP = 580

XYQ = XYZ + ZYQ

By putting the value of XYZ = 640 and ZYQ = 580 we get

XYQ = 640 + 580

XYQ = 1220

QYP = 1800 + XYQ

XYQ = 1220

QYP = 1800 + 1220

QYP = 3020