Mathematics, 22.04.2021 02:20 IsoSaysHi
In rectangle $ABCD$, $P$ is a point on $BC$ so that $\angle APD=90^{\circ}$. $TS$ is perpendicular to $BC$ with $BP=PT$, as shown. $PD$ intersects $TS$ at $Q$. Point $R$ is on $CD$ such that $RA$ passes through $Q$. In $\triangle PQA$, $PA=20$, $AQ=25$ and $QP=15$. Find $SD$. (Express your answer as a common fraction.)
Answers: 2
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In rectangle $ABCD$, $P$ is a point on $BC$ so that $\angle APD=90^{\circ}$. $TS$ is perpendicular t...
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