In the given figure, $A B = 27, A D = 38 c m, E D = 24 c m$ and $∠ E = 90^{o}$ , then radius of the circle is equal to:

11 c m

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15 c m

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13 c m

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17 c m

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Solution

The correct option is C $13 c m$
AB = AC = $27$ cm (length of tangent is equal from a point)
AD - AC = $38 - 27$ = $11$ cm = CD =DM
ED - DM = $24 - 11$ = $13$ cm
ED and EF touch the circle at M and N respectively EMNO is a square
therefore OM is $13$ cm

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