Answer :


Let sides AB, BC, CD, and AD touches circle at P, Q, R and S respectively.


As we know that tangents drawn from an external point to a circle are equal,


So, we have,


AP = AS = w (say)


[ Tangents from point A]


BP = BQ = x (say)


[Tangents from point B]


CP = CR = y (say)


[Tangents from point C]


DR = DS = z (say)


[Tangents from point D]


Now,


Given,


AB = 6 cm


AP + BP = 6


w + x = 6 …[1]


BC = 7 cm


BP + CP = 7


x + y = 7 …[2]


CD = 4 cm


CR + DR = 4


y + z = 4 …[3]


Also,


AD = AS + DS = w + z …[4]


Add [1] and [3] and substracting [2] from the sum we get,


w + x + y + z - (x + y) = 6 + 4 - 7


w + z = 3 cm


From [4]


AD = 3 cm

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