Question

In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Answer 1

Let the circle touches the sides AB, BC, CD and DA at P, Q, R, S respectively
We know that the length of tangents drawn from an exterior point to a circle are equal
Given,

$AB=6cm,BC=7cm,CD=4cm,CD=4cm$

AP = AS —-(1) {tangents from A}
BP = BQ —(2) {tangents from B}
CR = CQ —(3) {tangents from C}
DR = DS—-(4) {tangents from D}
Adding (1), (2) and (3) we get
∴ AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
⇒ AD = (AB + CD) – BC = {(6 + 4) – 7} cm = 3 cm
Hence, $AD=3cm$