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.