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