Question

In the following figure, quadrilateral $PQRV$ is a trapezium in which

$PQ\parallel VR,SR=6\text{cm},PQ=9\text{cm}$, then find $VR$.

Answer 1

Given that, $SR=6$ ...... (1)
$PQ=9$
$\square PQST$ is a rectangle as $PT$ and $QS$ are $\perp$ to $VR$
$\therefore TS=PQ=9$ .......... (2)
In right angled $\Delta QSR$
$\tan {45}^o=\frac{QS}{SR}$
$1=\frac{QS}{6}$
$QS=6$
$\therefore PT=QS=6$ ..... Prop. of a rectangle
Now In right angled $\Delta PTV$
$\tan {60}^o=\frac{VT}{PT}$
$\sqrt{3}=\frac{VT}{6}$
$VT=6\sqrt{3}$ ......... (iii)
$\therefore VR=VT+TS+SR$
$=6\sqrt{3}+9+6$
$=6\sqrt{3}+15$
$=3(2\sqrt{3}+5)$