A quadrilateral ABCD is drawn to circumscribe a circle. If AB = 12 cm, BC = 15 cm and CD = 14 cm, then find AD.


The sides AB, BC, CD and DA of the quadrilateral ABCD touches the circle at P, Q, R and S respectively.

We know that, the length of tangents drawn from an external point to a circle are equal.

∴ AP = AS   ...(1)

BP = BQ      ...(2)

CR = CQ     ...(3)

DR = DS      ...(4) 

Adding (1), (2), (3) and (4), we get

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

∴ AB + CD = AD + BC

⇒ 12 cm + 14 cm = AD + 15 cm

⇒ AD + 15 cm = 26 cm

⇒ AD = 26 cm – 15 cm = 11 cm

Thus, the length of AD is 11 cm.

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