Find the length of AC in right angled triangle ABC which has a circle inscribed in it, that touches the sides AB, BC and CA at D, E and F respectively, and has AD = 6 cm and BE = 5 cm.

41 cm

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55 cm

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44 cm

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61 cm

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Solution

The correct option is D 61 cm
AD = AF = 6 cm; BD = BE = 5 cm; EC = FC = x.
Applying Pythagoras theorem,
$A C^{2} = B C^{2} + A C^{2} i . e ., 121 + (5 + x)^{2} = (6 + x)^{2}$
Solving, we get x = 55.
AC = 55 + 6 = 61 cm

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