Exercise 6.1

Q.1    In figure , lines AB and CD intersect at O. If ∠AOC+∠BOE=70o and ∠BOD=40o, find ∠BOE and reflex ∠COE.

Sol.

Since OA and OB are opposite rays. Therefore AB is a line.
Since ray OC stands on AB. Therefore,
∠AOC+∠COB=180o [Linear Pairs]
⇒ ∠AOC+∠COE+∠BOE=180o [Since ∠COB=∠COE+∠BOE]
⇒ (∠AOC+∠BOE)+∠COE=180o
⇒ 70o+∠COE=180o [Since ∠AOC+∠BOE=70o (Given)]
⇒ ∠COE=180o−70o=110o
Therefore Reflex ∠COE=360o−110o=250o
Since OC and OD are opposite rays. Therefore CD is a line.
Since ray OE stands on CD. Therefore –
∠COE+∠EOD=180o [Linear Pairs]
⇒ ∠COE+∠BOE+∠BOD=180o[Since ∠EOD=∠BOE+∠BOD]
⇒ 110o+∠BOE+40o=180o
[Since ∠COE=110o(provedabove),∠BOD=40o (Given)]
⇒ ∠BOE=180o−110o−40o=30o
Hence, ∠BOE=30o
and reflex∠COE=250o

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Q.2       In figure , lines XY and MN intersect at O. If ∠POY=90o and a : b = 2 : 3, find c.

Sol.

Since a : b = 2 : 3 and a + b = ∠POX=∠POY = 90º and sum of ratios = 2 + 3 = 5
Therefore a=25×90o=2×18o=36o
and b=35×90o=3×18o=54o
Since OM and ON are opposite rays. Therefore MN is a line.
Since ray OX stands on MN. Therefore,
∠MOX+∠XON=180o [Linear Pairs]
⇒b+c=180o
⇒54∘+c=180o
⇒c=180o−54o=126o
Hence , c=126o

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Q.3       In figure ∠PQR=∠PRQ, then prove that ∠PQS=∠PRT.

Sol.

Since QS and QR are opposite rays. Therefore, SR is a line.
Since QP stands on the line SR.
Therefore ∠PQS+∠PQR=180o [Linear Pair] …(1)
Again RQ and RT are opposite rays. Therefore, QT is a line.
Since PR stands on the line QT.
Therefore ∠PRQ+∠PRT=180o [Linear Pair] … (2)
From (1) and (2), we have
∠PQS+∠PQR=∠PRQ+∠PRT [Since Each side = 180º] … (3)
Also ∠PQR=∠PRQ       (given)      …………….(4)
Subtracting (4) from (3), we have
∠PQS=∠PRT

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Q.4       In figure, if x + y = w + z, then prove that AOB is a line.

Sol.

We know that the sum of all angles around a point is equal to  360º
Therefore (∠BOC+∠COA)+(∠BOD+∠AOD)=360o
⇒ (x+y)+(w+z)=360o
But x+y=w+z [Given]
Therefore x+y=w+z=360o2=180o
Thus , ∠BOCand∠COA,∠BODand∠AOD form linear pairs. Consequently OA and OB are two opposite rays.
Therefore AOB is a straight line.

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Q.5       In figure POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS=12(∠QOS−∠POS)

Sol.

Since OR is perpendicular to the line PQ.
Therefore ∠POR=∠ROQ [Since Each = 90º]
⇒ ∠POS+∠ROS=∠QOS−∠ROS
⇒ 2∠ROS=∠QOS−∠POS
⇒ ∠ROS=12(∠QOS−∠POS)

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Q.6      It is given that ∠XYZ=64o and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP,find∠XYQ and reflex ∠QYP.
Sol.

Since XY is produced to point P. Therefore XP is a straight line.
Since YZ stands on XP.
Therefore ∠XYZ+∠ZYP=180o [Linear Pair]
⇒ 64o+∠ZYP=180o [Since ∠XYZ=64o]
⇒ ∠ZYP=180o−64o=116o
Since ray YQ bisects ∠ZYP

Therefore , ∠QYP=∠ZYQ=116o2=58∘
Now , ∠XYQ=∠XYZ+∠ZYQ
⇒ ∠XYQ=64o+58o=122o
and reflex ∠QYP=360o−∠QYP=360o−58o=302o

Exercise 6.2

Q.1     In figure, find the values of x and y and then show that AB || CD.

Sol.

Since AB || CD and transversal PQ intersects them at R and S respectively.
Since ∠ARS=∠RSD [Alternate angles]

⇒ x=y
But ∠RSD=∠CSQ [Vertically opp. angles]
⇒ y=130o [Since ∠CSQ=130o]
Hence , x = y = 130º

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Q.2      In figure , if AB || CD, CD|| EF and y : z = 3 : 7 , find x.

Sol.

Since CD|| EF and transversal PQ intersects them at S and T respectively.
Since ∠CST=∠STF [Alternate Angles ]
⇒ 180o−y=z [Since ∠y+∠CST=180o being linear pair]
⇒ y+z=180o
Given y : z = 3 : 7
So, the sum of ratios = 3 + 7 = 10

Therefore y=310×180o
=3×18o=54o
and z=710×180o=7×18o=126o
Since AB || CD and transversal PQ intersects them at R and S respectively.
Since ∠ARS+∠RSC=180o [Consecutive interior angles are supplementary]
⇒ x+y=180o
⇒ x=180o−y
=180o−54o=126o [Since y = 54º]
Hence, x = 126º

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Q.3     In figure , if AB || CD, EF ⊥ CD and ∠GED = 126º, find ∠AGE,∠GEFand∠FGE.

Sol.

Since AB || CD and transversal GE cuts them at G and E respectively.
Since ∠AGE=∠GED [Alternate angles]
⇒ ∠AGE=126o [Since ∠GED=126o (given)]
∠GEF=∠GED−∠FED=126o−90o=36o
and , ∠FGE=∠GEC [Alternate angles]
⇒ ∠FGE=90o−∠GEF
=90o−36o=54o
Hence, ∠AGE=126o,∠GEF=36oand∠FGE=54o

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Q.4     In figure , if PQ || ST, ∠PQR=110o and ∠RST=130o , find ∠QRS.

Sol.         Produce PQ to intersect SR at point M.

Now, PM || ST and transversal SM intersects them at M and R respectively.
Since ∠SMQ=∠TSM  [Alternate angles]
⇒ ∠SMQ= 130º  [Since ∠TSM= 130º(given)]
⇒ ∠QMR= 180º – 130º = 50º [Since ∠SMQ+∠QMR= 180º( linear pairs)]
Since ray RQ stands at Q on PM.
Since ∠PQR+∠RQM= 180º  (Linear pair)
⇒ 110º + ∠RQM = 180º
⇒ ∠RQM= 70º
Since ∠QRS = 180º – (70º + 50º) = 60º  [Since sum of the angles of a triangle is 180º]

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Q.5      In figure , if AB || CD, ∠APQ= 50º and ∠PRD= 127º , find x and y.

Sol.

Therefore AB || CD and transversal PQ intersects them at P and Q respectively.
Therefore ∠PQR=∠APQ [Alternate angles]
⇒ x = 50º [Since ∠APQ= 50º (given)]
Since AB || CD and transversal PR intersects them at P and R respectively.
Therefore ∠APR=∠PRD [Alternate angles]
⇒ ∠APQ+∠QPR = 127º [Since ∠PRD= 127º]
⇒ 50º + y = 127º [Since ∠APQ= 50º]
⇒ y = 127º – 50º = 77º
Hence , x = 50º and y = 77º

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Q.6      In figure , PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB|| CD.

Sol.

Two plane mirrors PQ and RS are placed parallel to each other i.e., PQ || RS. An incident ray AB after reflections takes the path BC and CD.

BN and CM are the normals to the plane mirrors PQ and RS respectively.
Since BN ⊥ PQ, CM ⊥ RS and PQ || RS
Therefore BN ⊥ RS
⇒ BN || CM
Thus BN and CM are two parallel lines and a transversal BC cuts them at B and C respectively.
Therefore ∠2=∠3 [Alternative interior angles]
But , ∠1=∠2and∠3=∠4 [By laws of reflection]
Therefore, ∠1+∠2=∠2+∠2and∠3+∠4=∠3+∠3
⇒ ∠1+∠2=2(∠2)and∠3+∠4=2(∠3)
⇒ ∠1+∠2=∠3+∠4 [Since ∠2=∠3⇒2(∠2)=2(∠3)]
⇒ ∠ABC=∠BCD
Thus, lines AB and CD are intersected by transversal BC such that –
∠ABC=∠BCD
i.e., alternate interior angles are equal.
Therefore, AB || CD.

Exercise 6.3

Q.1     In figure sides QP and RQ of Δ PQR are produced to points S and T respectively. If ∠SPR= 135º and ∠PQT= 110º, find ∠PRQ.

Sol.

We have ,
∠QPR+∠SPR= 180º [Linear pair]
⇒ ∠QPR + 135º = 180º
⇒ ∠QPR= 180º – 135º = 45º
Now , ∠TQP=∠QPR+∠PRQ [By exterior angle theorem]
⇒ 110º = 45º + ∠PRQ
⇒ ∠PRQ = 110º – 45º = 65º
Hence ∠PRQ = 65º

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Q.2      In figure, ∠X= 62º, ∠XYZ= 54º. If YO and ZO are the bisectors of ∠XYZand∠XZY respectively of Δ XYZ, find ∠OZYand∠YOZ

Sol.

Consider Δ XYZ,
∠YXZ+∠XYZ+∠XZY = 180º [Angle – sum property]
⇒ 62º + 54º + ∠XZY = 180º [Since ∠YXZ= 62º , ∠XYZ= 54º]
⇒∠XZY=180∘−62∘−54∘=64∘
Since YO and ZO are bisectors of ∠XYZand∠XZY
Therefore
∠OYZ=12×∠XYZ=12×54=27∘
and , ∠OZY=12×∠XZY=12×64∘=32∘
In Δ OYZ , we have
∠YOZ+∠OYZ+∠OZY= 180º [Angle sum property]
⇒ ∠YOZ+ 27º + 32º = 180º
⇒ ∠YOZ = 180º – 27º – 32º
= 180º – 59º = 121º
Hence, ∠OZY= 32º
and ∠YOZ= = 121º

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Q.3      In figure if AB || DE, ∠BAC = 35º and ∠CDE= 53º, find ∠DCE.

Sol.

Since AB || DE and transversal AE intersects them at A and E respectively.
Therefore ∠DEA=∠BAE [Alternate angles]
⇒ ∠DEC= 35º [Since ∠DEA=∠DECand∠BAE= 35º]
In Δ DEC, we have
∠DCE+∠DEC+∠CDE= 180º [Angle sum property]
⇒ ∠DCE+ 35º + 53º = 180º
⇒ ∠DCE= 180º – 35º – 53º
= 180º – 88º = 92º
Hence , ∠DCE= 92º

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Q.4      In figure, if lines PQ and RS intersect at point T, such that ∠PRT= 40º, ∠RPT= 95º and ∠TSQ= 75º, find ∠SQT.

Sol.

In Δ PRT, we have
∠PRT+∠RTP+∠TPR= 180º [Angle – sum property]
⇒ 40º + ∠RTP+ 95º = 180º
⇒ ∠RTP= 180º – 40º – 95º
= 180º – 135º = 45º
∠STQ=∠RTP [Vertically opp. angles]
⇒ ∠STQ= 45º [Since RTP = 45º (proved)]
In Δ TQS we have
∠SQT+∠STQ+∠TSQ= 180º [Angle – sum property]
⇒ ∠SQT+ 45º + 75º = 180º [Since ∠STQ=45∘(proved) and ∠TSQ=75∘]
⇒ ∠SQT= 180º – 45º – 75º
= 180º – 120º = 60º
Hence , ∠SQT= 60º

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Q.5      In figure if PQ ⊥ PS, PQ|| SR, ∠SQR = 28º and ∠QRT = 65º, then find the values of x and y.

Sol.

Using exterior angle property in Δ SRQ, we have
∠QRT=∠RQS+∠QSR
⇒ 65º = 28º + ∠QSR [Since ∠QRT= 65º , ∠RQS= 28º]
⇒ QSR = 65º – 28º = 37º
Since PQ|| SR and the transversal PS intersects them at P and S respectively.
Therefore ∠PSR+∠SPQ= 180º [Sum of consecutive interior angles is 180º]
⇒ (∠PSQ+∠QSR)+ 90º = 180º
⇒ y + 37º + 90º = 180º
⇒ y = 180º – 90º – 37º
= 180º – 127º = 53º
In the right ΔSPQ, we have
∠PQS+∠PSQ= 90º
⇒ x + 53º = 90º
⇒ x = 90º – 53º = 37º
Hence, x = 37º
and y = 53º

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Q.6    In figure , the side QR of Δ PQR is produced to a point S. If the bisectors of ∠PQRand∠PRS meet at point T, then prove that ∠QTR=12∠QPR.

Sol.

In ΔPQR we have ext. ∠PRS=∠P+∠Q
⇒ 12ext.∠PRS=12∠P+12∠Q   ( By exterior angle theorem )
⇒ ∠TRS=12∠P+∠TQR … (1)
[Since QT and RT are bisectors of ∠Qand∠PRS respectively therefore ∠Q=2∠TQRandext.∠PRS=2∠TRS]
In Δ QRT. we have ext. ∠TRS=∠TQR+∠T … (2)
From (1) and (2) , we get
12∠P+∠TQR=∠TQR+∠T
⇒ 12∠P=∠T
⇒ ∠QTR=12∠QPR [ Since ∠P=∠QPRand∠T=∠QTR]
Hence , ∠QTR=12∠QPR