Q. 245.0( 1 Vote )

In Fig. 10.86, if quadrilateral PQRS circumscribes a circle, then PD + QB =


A. PQ

B. QR

C. PR

D. PS

Answer :

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.


By the above property,


PD = PA (tangent from P)


QB = QA (tangent from Q)


RC = RB (tangent from R)


SC = SD (tangent from S)


Now,


PD + QB = PA + QA


PD + QB = PQ [PQ = PA + QA]


Hence, PD + QB = PQ

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