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# RD Chapter 8- Lines and Angles Ex-8.2 Interview Questions Answers

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Question 1 :

In the below Fig.OA and OB are opposite rays:

(i) If x = 250,what is the value of y?

(ii) If y = 350,what is the value of x?

(i) Given:x = 25

From figure: AOC andBOCform a linear pair

Which implies, AOC + BOC = 1800

From the figure, AOC = 2y + 5 and BOC = 3x

AOC + BOC = 1800

(2y + 5) + 3x = 180

(2y + 5) + 3 (25) = 180

2y + 5 + 75 = 180

2y + 80 = 180

2y = 100

y = 100/2 = 50

Therefore, y = 500

(ii) Given:y = 350

From figure: AOC + BOC = 180° (Linear pair angles)

(2y + 5) + 3x = 180

(2(35) + 5) + 3x = 180

75 + 3x = 180

3x = 105

x = 35

Therefore, x = 350

Question 2 :

In the belowfigure, write all pairs of adjacent angles and all the linear pairs.

From figure, pairs of adjacent angles are :

(AOC, COB) ; (AOD, BOD) ;(AOD, COD) ; (BOC, COD)

And Linear pair of angles are (AOD, BOD)and (AOC, BOC).

[As AOD + BOD = 1800 and AOC+ BOC = 1800.]

Question 3 : In the given figure, find x. Further find BOC , COD and AOD.

From figure, AOD andBODform a linear pair,

Therefore, AOD+ BOD = 1800

Also, AOD + BOC + COD = 1800

Given: AOD =(x+10) 0 , COD = x0 and BOC = (x + 20) 0

( x + 10 ) + x + ( x + 20 ) = 180

3x + 30 = 180

3x = 180 – 30

x = 150/3

x = 500

Now,

AOD=(x+10)=50 + 10 = 60

COD = x= 50

BOC =(x+20) = 50 + 20 = 70

Hence, AOD=600,COD=500 andBOC=700

Question 4 :

In figure, rays OA,OB, OC, OD and OE have the common end point 0. Show that AOB+BOC+COD+DOE+EOA=360°.

Given: Rays OA, OB, OC, OD and OE have the common endpoint O.

Draw an opposite ray OX to ray OA, which make a straight lineAX.

From figure:

AOB andBOX arelinear pair angles, therefore,

AOB +BOX = 1800

Or, AOB + BOC + COX = 1800 —–—–(1)

Also,

AOE andEOX arelinear pair angles, therefore,

AOE+EOX =180°

Or, AOE + DOE + DOX = 1800 —–(2)

By adding equations, (1) and (2), we get;

AOB + BOC + COF + AOE + DOE + DOX = 1800 + 1800

AOB + BOC + COD + DOE + EOA = 3600

Hence Proved.

Question 5 : In figure, AOC and BOC form a linear pair. If a – 2b= 30°, find a and b?

Given : AOC andBOCform a linear pair.

=> a + b = 180…..(1)

a – 2b = 300 …(2) (given)

On subtracting equation (2) from (1), we get

a + b – a + 2b = 180 – 30

3b = 150

b = 150/3

b = 500

Since, a – 2b = 300

a – 2(50) = 30

a = 30 + 100

a = 1300

Therefore, the values of a and b are 130° and 50° respectively.

Question 6 :

How many pairs ofadjacent angles are formed when two lines intersect at a point?

Four pairs of adjacent angles are formed when two linesintersect each other at a single point.

For example, Let two lines AB and CD intersect at point O.

The 4 pair of adjacent angles are :

(AOD,DOB),(DOB,BOC),(COA, AOD) and (BOC,COA).

Question 7 :

How many pairs ofadjacent angles, in all, can you name in figure given?

Number of Pairs of adjacent angles, from the figure, are :

EOC andDOC

EOD andDOB

DOC andCOB

EOD andDOA

DOC andCOA

BOC andBOA

BOA andBOD

BOA andBOE

EOC andCOA

EOC andCOB

Hence, there are 10 pairs of adjacent angles.

Question 8 :

In figure,determine the value of x.

The sum of all the angles around a point O is equal to 360°.

Therefore,

3x + 3x + 150 + x = 3600

7x = 3600 – 1500

7x = 2100

x = 210/7

x = 300

Hence, the value of x is 30°.

Question 9 : In figure, AOC is a line, find x.

From the figure, AOB and BOC arelinear pairs,

AOB +BOC =180°

70 + 2x = 180

2x = 180 – 70

2x = 110

x = 110/2

x = 55

Therefore, the value of x is 550.

Question 10 :

In figure, POS is aline, find x.

From figure, POQ andQOS arelinear pairs.

Therefore,

POQ + QOS=1800

POQ + QOR+SOR=1800

600 + 4x +400 = 1800

4x = 1800 -1000

4x = 800

x = 200

Hence, the value of x is 200.

krishan