Question

In adjoining figure, $PS\perp RQ$ and $QT\perp PR$. If $RQ=6,PS=6$ and $PR=12$, then Find $QT$.

Answer 1

Given:

$PS\perp RQ$
$QT\perp PR$
$RQ=6,PS=6$ and $PR=12$
With base $PR$ and height $QT$,

Area of $△PQR=\frac{1}{2}\times PR\times QT$ ---(1)
With base $QR$ and height $PS$,

Area of $△PQR=\frac{1}{2}\times QR\times PS$ ---(2)

From (1) and (2),

$\frac{1}{2}\times PR\times QT=\frac{1}{2}\times QR\times PS$

$\Rightarrow PR\times QT=QR\times PS$

Now, substitute the given values,

$\Rightarrow 12\times QT=6\times 6$

$\Rightarrow QT=\frac{36}{12}$

$\Rightarrow QT=3$
Hence, the measure of side $QT$ is $3$ units.