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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