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 ar

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(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 $r = 12$. Wait,re-evaluating: $BP = BQ$. If $BP = 25$,then $BQ = 25$. $BQ = BC - CQ = 38 - CQ = 25$,so $CQ = 13$. Since $CQ = CR$,$CR = 13$. $DR = CD - CR = 25 - 13 = 12$. Since $DR = r$,$r = 12$. Let's re-check the calculation: $BP = BQ = 25$. $CQ = BC - BQ = 38 - 25 = 13$. $CR = CQ = 13$. $DR = CD - CR = 25 - 13 = 12$. Since $DR = r$,$r = 12$. The provided options do not contain $12$. Let's re-read: $BP = 25$. If $BP = 25$,then $BQ = 25$. $CQ = 38 - 25 = 13$. $CR = 13$. $DR = 25 - 13 = 12$. The radius is $12$. Given the options,there might be a typo in the question values. If $r=14$,then $DR=14, CR=11, CQ=11, BQ=27, BP=27$. If $r=14$,$CD=25, CR=11, DR=14$. $BC=38, CQ=11, BQ=27$. $BP=27$. The calculation $r=12$ is correct based on the provided numbers.

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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