Q. 28

In Fig. 10.90, PR =


A. 20 cm

B. 26 cm

C. 24 cm

D. 28 cm

Answer :

Given:


QP = 4 cm


OQ = 3 cm


SR = 12 cm


SO’ = 5 cm


Property: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.


By above property, ∆OPQ is right-angled at OQP (i.e., OQP = 90°) and OSR is right-angled at OSR (i.e., OSR = 90°).


By Pythagoras theorem in ∆OPQ,


OP2 = QP2 + OQ2


OP2 = 42 + 32


OP2 = 16 + 9


OP2 = 25


OP = 25


OP = 5 cm


By Pythagoras theorem in ∆O’SR,


O’R2 = SR2 + O’S2


OR2 = 122 + 52


OR2 = 144 + 25


OR2 = 169


OR = 169


OR2 = 13 cm


Now,


PR = PO + ON + NO’ + O’R


PR = 5 cm + 3 cm + 5 cm + 13 cm


PR = 26 cm


Hence, PR = 26 cm

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