Q. 225.0( 1 Vote )

In the given figure, PQ and PR are tangents to a circle with center A. If QPA = 27° then QAR equals


A. 63°

B. 117°

C. 126°

D. 153°

Answer :

In Given Figure,


PQ = PR…[1]


[Tangents drawn from an external point are equal]


In AOP and BOP


PQ = PR [By 1]


AP = AP [Common]


AQ = AR [radii of same circle]


AQP ≅△ARP [By Side - Side - Side Criterion]


QPA = RPA


[Corresponding parts of congruent triangles are congruent]


Now,


QPA + RPA = QPR


QPA + QPA = QPR


2 QPA = QPR


QPR = 2(27) = 54°


As PQ and PQ are tangents to given circle,


We have,


AQ PQ and AR PR


[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]


So, AQP = ARP = 90°


In quadrilateral AQRP, By angle sum property of quadrilateral, we have


AQP + ARP + QAR + QPR = 360°


90° + 90° + QAR + 54° = 360°


QAR = 126°

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