# 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 toA. 2.5 cmB. 3 cmC. 5 cmD. 8 cm

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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