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Chapter 6 Lines and Angles Ex-6.3 Interview Questions Answers

Question 1 :

In the given figure, sidesQP and RQ of ΔPQR are produced to points S and T respectively. If SPR = 135º and PQT = 110º, find PRQ.

Answer 1 :

It is given that,

SPR = 135º and PQT = 110º

SPR + QPR = 180º (Linear pair angles)

135º + QPR = 180º

QPR = 45º

Also, PQT + PQR = 180º (Linear pair angles)

110º + PQR = 180º

PQR = 70º

As thesum of all interior angles of a triangle is 180º, therefore, for ΔPQR,

QPR + PQR + PRQ =180º

45º + 70º + PRQ = 180º

PRQ = 180º − 115º

PRQ = 65º

Question 2 :

In the given figure, X = 62º, XYZ = 54º. If YO and ZO are the bisectors of XYZ and XZY respectively of ΔXYZ, find OZY and YOZ.

Answer 2 :

As the sum of all interiorangles of a triangle is 180º, therefore, for ΔXYZ,

X + XYZ + XZY =180º

62º + 54º+ XZY =180º

XZY = 180º − 116º

XZY = 64º

OZY = = 32º (OZ is the anglebisector of XZY)

Similarly,OYZ= = 27º

Usingangle sum property for ΔOYZ, we obtain

OYZ + YOZ + OZY =180º

27º + YOZ + 32º = 180º

YOZ = 180º − 59º

YOZ = 121º

Question 3 :

In the given figure, if AB|| DE, BAC = 35ºand CDE =53º, find DCE.


Answer 3 :

AB || DE and AE is atransversal.

BAC = CED (Alternate interior angles)

CED = 35º

In ΔCDE,

CDE + CED + DCE =180º (Angle sum property of a triangle)

53º + 35º+ DCE =180º

DCE = 180º − 88º

DCE = 92º

Question 4 :

In the given figure, iflines PQ and RS intersect at point T, such that PRT = 40º, RPT = 95ºand TSQ =75º, find SQT.

Answer 4 :

Using angle sum propertyfor ΔPRT, we obtain

PRT + RPT + PTR =180º

40º + 95º+ PTR =180º

PTR = 180º − 135º

PTR = 45º

STQ = PTR = 45º (Vertically opposite angles)

STQ = 45º

By usingangle sum property for ΔSTQ, we obtain

STQ + SQT + QST =180º

45º + SQT + 75º = 180º

SQT = 180º − 120º

SQT = 60º

Question 5 :

In the given figure, if PQ PS, PQ|| SR, SQR = 28ºand QRT =65º, then find the values of x and y.

Answer 5 :

It is given that PQ || SRand QR is a transversal line.

PQR = QRT (Alternate interior angles)

x + 28º = 65º

= 65º − 28º

x = 37º

By usingthe angle sum property for ΔSPQ, we obtain

SPQ + x + y = 180º

90º + 37º+ y = 180º

y = 180º − 127º

= 53º

x =37º and y = 53º

Question 6 : In the given figure, the side QR of ΔPQR is producedto a point S. If the bisectors of PQRand PRS meet at point T, thenprove that QTR=QPR.

Answer 6 :

In ΔQTR, TRS is an exterior angle.

QTR + TQR = TRS

QTR = TRS − TQR (1)

For ΔPQR,PRS is anexternal angle.

QPR + PQR = PRS

QPR + 2TQR = 2TRS (AsQT and RT are angle bisectors)

QPR = 2(TRS − TQR)

QPR = 2QTR [By using equation (1)]

QTR = QPR


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Chapter 6 Lines and Angles Ex-6.3 Contributors

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