Question

In the given figure $PS\perp segRQsegQT\perp segPR$, If $RQ=6,PS=6$ and $PR=12$, then find $QT$

Answer 1

Given $PQ=6cm,PR=12cm,PS=6cm$

${PR}^2={PS}^2+{RS}^2$

$LetQS=xcm$

$\therefore RS=(6+x)cm$

$⟹{12}^2=6^2+{(6+x)}^2$

$\sqrt{108}=6+x⟹x=4.39cm$

$\therefore QS=x=4.39cm$

Now,consider$△PSQ$

${PQ}^2={PS}^2+{QS}^2$

$⟹PQ=\sqrt{6^2+{4.39}^2}=7.43cm$

Consider$△PTQ$

$AsPR=12cm$ and $T$ divides $PR$ in some unknown ratio.

$LetPT=(12-x)cmandRT{=}^{\prime }{x}^{\prime }cm.$

$In△PTQ,{:PQ}^2={PT}^2+{QT}^2$

${QT}^2={PQ}^2-{PT}^2$

$⟹{QT}^2={7.43}^2-{(12-x)}^2-\left(1\right)$

From$△RQT$

${RQ}^2={RT}^2+{QT}^2$

$⟹{QT}^2={RQ}^2-{RT}^2$

$\therefore {QT}^2=6^2-x^2-\left(2\right)$

equating$\left(1\right)\&\left(2\right)$

${7.43}^2-{(12-x)}^2=6^2-x^2$

$19.205=144-24x+x^2-x^2$

$\therefore x=5.2cm$

$\therefore QT=\sqrt{6^2-x^2}=2.99cm$