Q2 of 105 Page 598

In the given figure, RQ is a tangent to the circle with center O. If SQ = 6 cm and QR = 4 cm, then OR is equal to

As SQ is diameter and OQ is radius in the given circle,


∴ 2OQ = SQ [As 2 × radius) = diameter]


2OQ = 6 cm


OQ = 3 cm


Now, QR is tangent


∴ OQ ⏊ QR


In right - angled △OQR,


By Pythagoras Theorem,


[i.e. (Hypotenuse)2 = (Base)2 + (Perpendicular)2 ]


(QR)2 + (OQ)2 = (OR)2


(4)2 + (3)2 = (OR)2


16 + 9 = (OR)2


(OR)2 = 25


OR = 5 cm

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