C-08-Q228mediumsingle_mcq

In a circle with centre $O$ and chord $BC = 8$ cm, if the perpendicular $OP$ from $O$ has $OP = 3$ cm, then the radius equals—

The perpendicular from the centre bisects the chord, so $BP = \tfrac{8}{2} = 4$ cm. Then the radius is $\sqrt{O P^{2} + B P^{2}} = \sqrt{^{}}$ cm.

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73

\sqrt{73}

73

3

2

+

4^{2}

=

\sqrt{25}

=

5

\sqrt{OP^{2}+BP^{2}} = \sqrt{3^{2}+4^{2}} = \sqrt{25} = 5

O P^{2} + B P^{2} = 3^{2} + 4^{2} = 25 = 5