Let the circle touches all the sides of quadrilateral ABCD at the point P, Q, R and S.
We know that tangents to a circle from an external point are equal.
AP = AS …..(i)
BP= BQ …(ii)
CR = CQ …..(iii)
DR = DS ……(iv)
Adding equation (i), (ii), (iii) and (iv)
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ AP + BP + CR + DR = AS + DS + BQ + CQ
⇒ AB + CD = AD + BC
⇒ 8 + CD = 5 + 6
⇒ 8 + CD = 11
⇒ CD = 11 – 8
∴ CD = 3 cm
