IMPORTANT QUESTIONS FOR CLASS 9 MATHS

Name Of The Chapter: Lines And Angles

Chapter – 06

Q.1Find out the two pairs of adjacent angles

Ans.Two pairs of Adjacent angles

1. <AOC and <COB

2. <AOD and <DOB

Q.2 In Fig, lines PQ and RS intersect each other at point O. If

ÐPOR : ÐROQ = 2 : 3 , find <POR and <ROQ(1 Marks)
Ans. ÐPOR + ÐROQ = 180! (Linear pair of angles)

But ÐPOR : ÐROQ = 2 : 3 (Given)

2
Therefore, ÐPOR = ´ 180! = 72!
5

3
Similarly, ÐROQ = ´ 180! = 108!
5

Q3. Two supplementary angles are in the ration 4:5. Find the angles.

Ans.

let A and B are two supplimentary angles

<A+<B = 180°
< A :< B = 4 : 5
ratio sum = 4+5 = 9
4
<A = ´180° = 80°
9
5
< B = ´180° = 100°
9

Q4. The measure of an angle is twice the measure of its supplementary angel.
Find its measure.

2
Ans.

let one angle = x

its supplimentary = 180° - x
AT. .Q
x = 2(180° - x)
3 x = 360°
x = 120°
supplimentary of x = 60°

Q.5Two adjacent angles are said to form a linear pair of angles, if there non-
common arms are ____________

Ans. Two opposite Rays

Q.6If a ray stands on a line, and then the sum of the adjacent angles so formed
is ______

Ans. 180°

Q7. Sum of all angles round a point is equal to ______

Ans. 360°

Q8. The angles between the bisector of a linear pair of angles is a ______

Ans. 90°

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Q9. If two parallel lines are intersected by a transversal, then bisectors of any
two corresponding angels ___________

Ans. Corresponding angles are parallel

Q10. If two lines are parallel to same line then these line will be _______ to
each other.

Ans. Parallel to each other

Q.1Match the following:(2 Marks)

a) Adjacent angles

b) Vertically opposite angles

c) Linear pair of angles

Fig-1Fig-2

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 Fig-3

Ans.a) Adjacent anglesFig-2

b) Vertically opposite anglesFig-1

c) Linear pair of angles Fig-3

Q.2 In Fig, lines PQ and RS intersect each other at point O. If

ÐPOR : ÐROQ = 5 : 7 , find all the angles.(2 Marks)

Ans. ÐPOR + ÐROQ = 180! (Linear pair of angles)

But ÐPOR : ÐROQ = 5 : 7 (Given)

5
Therefore, ÐPOR = ´ 180! = 75!
12

7
Similarly, ÐROQ = ´ 180! = 105!
12

Now, ÐPOS = ÐROQ = 105! (Vertically opposite angles)

and ÐSOQ = ÐPOR = 75! (Vertically opposite angles)

5
Q.3 In the figure, find the value of y .(2 Marks)

Ans. 2 y
!
+ 3 y ! + 5 y ! = 180!

10 y ! = 180!
180!
y= = 18
10!

Q.4 In the figure, side QR of DPQR has been produced S , if ÐP : ÐQ : ÐR = 3 : 2 :1

and RT ^ PR, Find ÐTRS (2 Marks)

Ans.Given: DPQR, side QR is us extended upto S and TR ^ PR,

ÐP : ÐQ : ÐR = 3 : 2 :1

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 P Q R
or = = =k
3 2 1
P = 3k , Q = 2k , R = k
ÐP + ÐQ + ÐR = 1800
6k = 1800
k = 300
ÐP = 3k = 900
ÐQ = 2k = 600
ÐR = k = 300
Þ ÐP = 90! , ÐQ = 60! and ÐR = 30!

Hence, we have to find the value of ÐTRS .

From the figure, we have

ÐPRT = 90!
\ÐTRS = 180! - ( ÐPRQ + ÐPRT )
= 180! - ( 30! + 90! )
= 180! - 120! = 60!

Q.5In figure if x + y = w + z , then prove that AOB is a line. ( 2 Marks)

Ans. x + y + z + w = 360!

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 Þ ( x + y ) + ( z + w ) = 360!
Þ ( x + y ) + ( x + y ) = 360! [" x + y = z + w]
Þ 2 ( x + y ) = 360!
Þ x + y = 180!
Þ AOB is a straight line

Q.6 It is given that ÐXYZ= 64! and is produced to a point P. Draw a figure from
the given information. If ray YQ bisects ÐZYP, find ÐXYQ and reflex ÐQYP.
(3 Marks)

Ans. It is given that ray YQ bisects ÐZYP

So let ÐPYQ = ÐQYZ = x

Þ ÐPYZ = 2 x

Since QY stands on ray PX

\ ÐPYZ + ÐZYX = 180!

Þ 2 x + 64 = 180
Þ 2 x = 180 - 64
Þ 2 x = 116
Þ x = 58
\ ÐXYQ = ÐXYZ + ÐZYQ = 64! + 58! = 122!
and, reflex ÐQYP = 360! - ÐQYP = 360! - 58! = 302!

Q.7 If the angles of a triangle are in the ratio 2:3:4, find the three angles.(3
Marks)

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Ans. Let A,B and C are the angles of the triangle.

Then, A: B :C = 2:3: 4

A B C
or = = = k ( say )
2 3 4

\ A = 2k , B = 3k and C = 4k

" A + B + C = 180! ( Sum of the three angles of a D is 180! )

or 2k + 3k + 4k = 180!

Þ 9k = 180!
\ k = 20!
\ A = 2 ´ 20 = 40!
B = 3 ´ 20 = 60!
C = 4 ´ 20 = 80!

\Hence the angles are 40! , 60! and 80!

Q.8 In figure, lines l1 and l2 intersect at O forming angles as shown in the

figure. If a = 35 Find the value of b, c and d.(3 Marks)
!

Ans. Since lines l1 and l2 intersect at O

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 \ Ða = Ðc
Þ Ðc = 35!
Clearly, Ða + Ðb = 180!
[Since Ða and Ðb are angles of a linear pair ]
Þ 35! + Ðb = 180!
Þ Ðb = 180! - 35!
Þ Ðb = 145!
Since Ðb and Ðd are vertically opposite angle.
\ Ðd = Ðb Þ Ðd = 145!
Hence, b = 145! , c = 35! and d = 1450

Q.9 In Figure determine the value of y(3 Marks)

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Ans. Since ÐCOD and ÐEOF are vertically opposite angle

\ ÐCOD = ÐEOF
Þ ÐCOD = 5 y [" ÐEOF = 5Y ]
Now, OA and OB are opposite rays.
\ ÐAOD + ÐDOC + ÐCOB = 180!
Þ 2 y + 5 y + 5 y = 180
180
Þ 12 y = 180 Þ y = = 15
12
Hence y = 15

Q.10 In Figure two straight lines PQ and RS intersect each other at O . If

ÐPOT = 75 , find the values of a, b and c (3 Marks)
!

Ans. Since OR and OS in the same line.

\ ÐROP + ÐPOT + ÐTOS = 180!

Þ 4b! + 75! + b! = 180!
Þ 5b! + 75! = 180!
Þ 5b! = 105! Þ b = 21
Since PQ and RS intersect at O.

Therefore,ÐQOS = ÐPOR [ Vertically opp angles ]

Þ a = 4b
Þ a = 4 ´ 21 = 840 ëé! b = 21 ûù
0

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 Now, OR and OS are in the same line. Therefore
ÐROQ + ÐQOS = 180!
Þ 2c + a = 180
Þ 2c + 84 = 180
Þ 2c = 96
Þ c = 480
Hence, a = 840 , b = 210 and c = 480

Q.11 ABCDE is a regular pentagon and bisector of ÐBAE meets CD in M. IF

bisector of ÐBCD meets AM at P find ÐCPM (5 Marks)

Ans. Given: A regular pentagon bisector ÐA of meets CD in M and bisector ÐBCD of

meets AM in P

540!
To find : Each angle of a regular pentagon = = 108!
5

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 \ AM is the bisector of ÐA
1 1
\ ÐBAM = ´ ÐA = ´ 108! = 54!
2 2
Now is quadrilateral ABCM
ÐBAM + ÐB + ÐC + ÐAMC = 360!
54! + 108! + 108! + ÐAMC = 360!
ÐAMC = 360! - 270! = 90!
or ÐPMC = 90!
CP is the bisector of ÐBCD
1 1
\ ÐPCM = ´ ÐC = ´ 108! = 54!
2 2
Now in DPCM , we have
ÐPCM + ÐPMC + ÐCPM = 180!
54! + 90! + ÐCPM = 180!
ÐCPM = 180! - 144! = 36!

Q.12 In the given figure ÐQ > ÐR and M is a point QR such that PM is the
bisector of <P . If the perpendicular from P on QR meets QR at N, then
1
prove that ÐMPN = ( ÐQ - ÐR ) (5 Marks)
2

Ans. A DPQR in which ÐQ > ÐR , PM is the bisector of ÐQPR and PN ^ QR

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 1
To prove: ÐMPN = ( ÐQ - ÐR )
2
proof ÐQPM = ÐMPR ...(i)
[" PM bisects ÐP Sum of acute angles of a right D ]
Also in right DPNQ, ÐQ + ÐQPN = 90!
Þ ÐQ = 90! - ÐQPN ...(ii)
and in right DPNR, ÐR + ÐNPR = 90 !

Þ ÐR = 90! - ÐNPR ...(iii)

Now from (i), (ii) and (iii)
ÐQ - ÐR = 90! - ÐQPN - 90! + ÐNPR
Þ ÐQ - ÐR = ÐNPR - ÐQPN
= ÐMPN + ÐMPR - ( ÐQPM - ÐMPN )
Þ ÐQ - ÐR = ÐMPN + ÐMPR - ÐQPM + ÐMPN
Þ ÐQ - ÐR = 2ÐMPN [" ÐMPR = ÐQPM ]
1
( ÐQ - ÐR ) = ÐMPN éëSum of there anlges of a triangle is 180! ùû
2

Q.13 In Figure OP ! RS . Determine ÐPQR (5 Marks)

Ans. Produce OP to intersect RQ in a point T

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 Now, OT ! RS . and transversal RT intersects them at T and R respectively
\ ÐRTP = ÐSRT
Þ ÐRTP = 130!
Þ ÐPTQ = 180! - 130! = 50! éë" ÐRTP + ÐPTQ = 180! Linear Pairs ùû

Since, ray OP Stands at P on OT.

\ ÐOPQ + ÐQPT = 180!
Þ 110! + ÐQPT = 180!
Þ ÐQPT = 70!
\ ÐPQR = 180! - ( 70! + 50! ) = 60! ëé\ Sum of the Ðs of a triangle is 180! ûù

Q.14 In figure the sides AB and AC of are produced to points E and D

respectively. If bisectors BO and CO of ÐCBE and ÐBCD respectively
1
meet at point O, then prove that ÐBOC = 90 - ÐBAC
!

Ans. Ray BO is the bisector of ÐCBE

1
Therefore, ÐCBO = ÐCBE
2
1
= (180! - y )
2
y
= 90! - ...(1)
2

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 Similarly, ray CO is the bisector of ÐBCD
1
Therefore ÐBCO = ÐBCD
2
1
= (180! - z )
2
z
= 90! - ...(2)
2

in DBOC , ÐBOC + ÐBCO + ÐCBO = 180! ...(3)

Substituting (1)and (2) in (3) you get
z y
ÐBOC + 90! - + 90! - = 180!
2 2
z y
So, ÐBOC = +
2 2
1
or, ÐBOC = ( y + z )
2

But x + y + z = 180! ( Angle sum property of a triangle ) Deleted: ¶

! ¶
Therefore, y + z = 180 - x
!

Therefore,(4) becomes
1
ÐBOC =
2
(180! - x )
x
= 90! -
2
1
= 90! - ÐBAC
2

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