1.

In Fig. if quadrilateral PQRS circumscribes a circle, then PD + QB =A. PQ B. QR C. PR D. PS

Answer»

Answer is A. PQ

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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