A circle touches all the four sides of a quadrilateral ABCD at P, Q, R and S respectively as shown in the figure. If AB= $10$ cm, BC= $8$ cm and DC= $5$ cm, Find AD.

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Solution

Let P,Q,R,S be the points where the circle touches the quadrilateral .From a point out side of the circle the two tangents to the circle are equal.
AP = AS,BP = BQ , CR = CQ , and DR = DS
Adding AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AD + BC
$10 + 5 = A D + 8$
$A D = 7 c m .$

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A circle touches all the four sides of a quadrilateral ABCD at P, Q, R and S respectively as shown in the figure. If AB=10 cm, BC=8 cm and DC= 5 cm, Find AD.

Let P,Q,R,S be the points where the circle touches the quadrilateral .From a point out side of the circle the two tangents to the circle are equal.AP = AS,BP = BQ , CR = CQ , and DR = DSAdding AP + BP + CR + DR = AS + BQ + CQ + DSAB + CD = AD + BC10+5=AD+8AD=7cm.

A circle touches all the four sides of a quadrilateral ABCD at P, Q, R and S respectively as shown in the figure. If AB= $10$ cm, BC= $8$ cm and DC= $5$ cm, Find AD.

Let P,Q,R,S be the points where the circle touches the quadrilateral .From a point out side of the circle the two tangents to the circle are equal.
AP = AS,BP = BQ , CR = CQ , and DR = DS
Adding AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AD + BC
$10 + 5 = A D + 8$
$A D = 7 c m .$