let the circle touches the side of quadrilateral AB, BC, CD and DA at P, Q, R and S. AB=AP+PB=9 BC=BQ+QC=8 CD=CR+RD=12 DA=DS+SA=? we know that the tangents from a point to the circle are equal. So, AP=AS BP=BQ CQ=CR DR=DS Now, DA=DS+SA =DR+AP =(12−CR)+(9−PB) =21−(CQ+BQ) =21−8 =13