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KC Sinha Mathematics Solution Class 9 Chapter 8 Lines and Angles exercise 8.1

Page 8.12

Exercise 8.1

Question 1 

Fill in the blanks in each of the following to make the statement true:

(i) Two distinct points in a plane determine __ line .

Sol : unique

Page 8.13

(ii) A line separates a plane into __ parts namely the __ and the itself.

Sol : three , two half planes , line

(iii) Two distinct __ in  a plane cannot have more than one point in common.

Sol :  lines

(iv) If any ray stands on a line , sum of the two adjacent angles are __

Sol :  180°

(v) If two lines intersect each other, vertically opposite angles are ___

Sol :  equal

Question 2

Which of the following statements are true (T) and which are false (F) . Give reasons.

(i) Angles forming a linear pair can both be acute angles

Sol : F

(ii) Angles forming a linear pair are supplementary

Sol : T

(iii) Two distinct lines in a plane can have two points in common

Sol : F

(iv) If two lines intersect and one of the angles so formed is of measure 90° then each of the other three angles is of measure 90°

Sol : T

(v) If angles forming a linear pair are equal, then each of these angles is of measure 90° ?

Sol : T

Question 3

Give answer to the following questions :

(i) If a ray stand on a line , what will be the sum of the two adjacent angles ?

Sol : 180°

(ii) If sum of two adjacent angles is two right angles , what type of angles will these be ?

Sol : supplementary

(iii) If two lines intersect , what is a the relation between vertically opposite angles ?

Sol : They are equal

Question 4

Write the sum of all angles (in right angles) formed at any point in a plane.

Sol :

Four right angles or 360°

Question 5

Lines AB and CD intersect each other at a point O such that ∠AOC=∠COB. What is the relation between these lines ?

Sol :


Question 6

If lines AB and CD intersect each other at a point O and ∠AOC=135° , then

(a) ∠AOD = __

Sol : 45°

(b) ∠BOD = __

Sol : 135°

(c) ∠COB = __

Sol : 45°

Question 7

If a ray stands on a line, then angles formed between the bisectors of adjacent angles is __

Sol :

Right angle

Question 8

Lines AB and CD intersect at point O, Write in degree the measure of the angles between bisectors of ∠AOC and ∠BOC .

Sol :



Type 1

Question 9

In the given f‌igure, find the value of x

In the given f‌igure, find the value of x

Question 10

In the following f‌igure, f‌ind the value of y.


In the following f‌igure, f‌ind the value of y.


Sol :

Question 11

In the given f‌igure, f‌ind the value of y

<fig to be added>

Sol :


Question 12

In the given f‌igure, f‌ind ∠AOB in degree

<fig to be added>


Question 13

In the given f‌igure, a is greater than b by one third of a right angles. Find the values of a and b .

<fig to be added>

[Hint: Given, a-b=\dfrac{1}{3}\times 90^{\circ}


Also ⇒a+b=180°..(ii)

On solving (i) and (ii) , we get


⇒a=105° , b=180°-105° = 75° ]


Question 14

If a ray stands on a line such that difference of adjacent angles so formed is 30° , then find the measure of each adjacent angle in degree.

[Hint: Let a and b be the two adjacent angle. Then , a-b=30° and a+b=180°

Solving , we get a=105° , b=75°]

Sol :

Question 15

In the given figure , what value of x will make POQ a straight line ?

<fig to be added>

[Hint : For POQ to be a line , we must have


Hence x=\dfrac{170^{\circ}}{5}=34^{\circ} ]



Question 16

In the given figures (i) and (ii) , find the values of x in each case


<fig to be added>



<fig to be added>


Question 17

What is the measure of the angle (in degree) which is twice of its supplementary angles ?

Sol :


Type 2

Question 18

Ray OE bisects ∠AOB and ray OF is opposite to ray OE. Show that ∠FOB=∠FOA

Sol :



Question 19

If from any point O on a line PQ, two lines OR and OS are drawn in the opposite sides of PQ, such that ∠POR=∠QOS , then prove that OR and OS lie in a line.

Sol :


Question 20

From any point O , four lines AO, OB , OC and OD are drawn respectively such that ∠AOB=∠COD and ∠BOC=∠DOA , prove that AOC and BOD are straight lines.

Sol :


Question 21

O is a point on line AB , OC and OD are perpendiculars drawn on AB in opposite directions. Prove that OC and OD lie in a straight line .

Sol :



Question 22

Two lines AB and CD intersect each other at point O. If line OP bisects ∠BOD , prove that if OP is produced backwards , then it bisects ∠AOC. If OP and OQ are respectively bisectors of ∠BOD and ∠AOC . Show that the rays OP and OQ are in the same line

[Hint : Since OP is the bisector of ∠BOD

∴ ∠1=∠6;

If OP is produced,

then ∠1=∠4

and ∠6=∠3 [vertically opposite angles]

∴ ∠3=∠4

Thus, OQ is the bisector of ∠AOC

Also ∠2=∠5 [vertically opposite angles]


Second part :

Since sum of the angles formed at a point is 360°

∴ ∠1+∠2++∠3+∠4+∠5+∠6=360°

⇒ (∠1+∠6)+(∠3+∠4)+(∠2+∠5)=360°

⇒ 2∠1+2∠3+2∠2=360° [Using above equations]

⇒ ∠1+∠3+∠2=180° ∴∠POQ=180°

Hence, OP and OQ are in the same line ]

Sol :



Question 23

(i) In the given figure, lines PQ and RS intersect at Point O. If ∠POR:∠ROQ= 5:7 , find all the angles.

[Hint: ∵∠POR+∠ROQ=180° [By linear pair axiom]

Given , \dfrac{\angle POR}{\angle ROQ}=\dfrac{5}{7} or \dfrac{\angle POR}{5}=\dfrac{\angle ROQ}{7} \dfrac{(\angle POR + \angle ROQ)}{12}=\dfrac{180^{\circ}}{12}

\angle POR=5\times \dfrac{180^{\circ}}{12}=75^{\circ} and \angle ROQ=7\times \dfrac{180^{\circ}}{12}=105^{\circ}

Now , ∠POS=∠ROQ=105° and ∠SOQ=∠POR=75°]


(ii) Three coplanar lines AB ,CD and EF intersect at a point O, forming angles as shown in the figure. Find the values of x, y , z and v.

[Hint: Clearly , ∠y=50° [Vertically opposite angles]

∠z=90° [Vertically opposite angles]

∠v=∠x [Vertically opposite angles]

Now, ∠x=40° ,∠y=50° , ∠z=90° and ∠v=40°]


(iii) In the given figure , find the value of x and then find ∠BOC, ∠ FOC , ∠COA

[Hint: ∵ ∠DOE=∠FOC=2x

Now ray OF stands in line AOB]









(iv) In the given f‌igure. two straight lines PQ and RS intersect eaach other at O.

If ∠POT=70°, find the value of a,b and c

<fig to be added>

[Hint : Since ray OT stands on line RS

∴ ∠ROP+∠POT+∠TOS=180°

or 4b+70°+b=180°


⇒b=22°; since PQ and RS intersected at O, so

∠QOS=∠POR or a=4b


Now ∠ROQ=∠POS  [Vertically opposite angles]

∴ 2c=70°+b=70°+22°=92°

or c=\dfrac{92}{2}=46^{\circ}]



2c+a=180° [∵Ray OQ stands on line RS]

⇒ 2c+88°=180°

⇒ 2c=180°-88°

⇒ c=46°


Question 24

If a ray OC stands on AB such that ∠AOC=∠COB , then show that ∠AOC=90°

Sol :


Question 25

Point O is the common end point of the rays OA , OB , OC , OD and OE. Show that ∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360°


[Hint: Draw a ray OP opposite to ray OA]


Question 26

In the given f‌igure, if each of ∠AOC and ∠AOB is 90° , show BOC is a line .

Sol :




Question 27

In the given f‌igure, OE and OF bisect ∠AOC and ∠COB respectively and OE⊥OF . Show that points A,O,B are collinear.

Sol :


Question 28

In the given f‌igure, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS , and ∠SOQ , if ∠POS=x , then find ∠ROT.

Sol :


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