In quadrilateral ABCD,m angle D = 90^circ. A | vedclass

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(C) $1$. Since the circle touches the sides $AB, BC, CD,$ and $DA$ at $P, Q, R,$ and $S$, the lengths of tangents from an external point to a circle a

Vedclass · Accepted Answer

$12$

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(C) $1$. Since the circle touches the sides $AB, BC, CD,$ and $DA$ at $P, Q, R,$ and $S$, the lengths of tangents from an external point to a circle are equal. Thus, $DR = DS = r$. Since $\angle D = 90^\circ$ and the radii $OR$ and $OS$ are perpendicular to the sides $CD$ and $DA$, $ORDS$ is a square of side $r$. Therefore, $DR = DS = r$.$2$. Given $CD = 25$, we have $CR = CD - DR = 25 - r$. Since tangents from a point are equal, $CQ = CR = 25 - r$.$3$. Given $BC = 38$, we have $BQ = BC - CQ = 38 - (25 - r) = 13 + r$.$4$. We are given $BP = 25$. Since tangents from point $B$ are equal, $BP = BQ$. Therefore, $25 = 13 + r$, which gives $r = 12$.

In quadrilateral $ABCD$,$m \angle D = 90^\circ$. $A$ circle with center $O$ and radius $r$ touches its sides $AB, BC, CD,$ and $DA$ at points $P, Q, R,$ and $S$ respectively. If $BC = 38, CD = 25,$ and $BP = 25,$ find the radius $r$ of the circle.

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