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(A) For a quadrilateral $ABCD$ circumscribing a circle,the sum of the lengths of opposite sides is equal.This is because the lengths of tangents drawn

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$5$

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(A) For a quadrilateral $ABCD$ circumscribing a circle,the sum of the lengths of opposite sides is equal.This is because the lengths of tangents drawn from an external point to a circle are equal.Let the circle touch the sides $AB, BC, CD,$ and $DA$ at points $P, Q, R,$ and $S$ respectively.Then $AP = AS, BP = BQ, CQ = CR,$ and $DR = DS$.Sum of opposite sides: $AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS) = (AS + DS) + (BQ + CQ) = AD + BC$.Given $AB = 8, BC = 10, CD = 7$.Substituting the values: $8 + 7 = AD + 10$.$15 = AD + 10$.$AD = 15 - 10 = 5$.

$A$ circle touches all the sides of a quadrilateral $ABCD$. If $AB = 8$,$BC = 10$,and $CD = 7$,then find the length of $AD$.

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