In the given figure, a circle touches all the four sides of a quadrilateral ABCD with AB = 12cm, BC = 10 cm and CD = 14cm. Find AD.
A
10
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B
12
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C
15
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D
16
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Solution
The correct option is D 16
Let AD =x, then BD = (12-x)
AD' = AG = x [ Tangents drawn from the same external point are equal to each other.]
Similarly,
BD = BE = (12-x),
Then CE = 10- (12-x) =x-2
CE = CF =(x-2) [ Tangents drawn from the same external point are equal to each other.]
FD = 14 - (x-2) = (16 - x)
Similarly,
FD = DG = (16 - x)
AD = AG + GD = x + 16-x =16 cm