In Fig. 10.79, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR =

Given:


SQ = 6 cm


QR = 4 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, ∆ROQ is right-angled at OQR (i.e., OQR = 90°).


Diameter QS = 6 cm




Radius (OQ) = 3 cm


Now by Pythagoras theorem,


OR2 = OQ2 + QR2


OR2 = 32 + 42


OR2 = 9+ 16


OR2 = 25


OR= √25


OR= 5 cm


Hence, OR= 5 cm.

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