Question

In the given figure, In $\Delta PQRXY\parallel QR$, $PX:QX=1:3$ $YR=4.5cm$ and $QR=9cm$. Find $XY$.

Further if $Ar(\Delta PXY$)=$A{cm}^2$, find in terms of $A$ the area of $\Delta PQR$ and the area of trapezium $XYRQ$.

Answer 1

In $△PXY$ and $△PQR,XY$ is parallel to $QR$, so corresponding angles are equal.

$\angle PXY=\angle PQR$

$\angle PYX=\angle PRQ$

Hence, $△PXY\sim △PQR$ [By $AA$ similarity criterion]

$\frac{PX}{PQ}=\frac{XY}{QR}$

$\Rightarrow \frac{1}{1+3}=\frac{XY}{QR}$

$\Rightarrow \frac{1}{4}=\frac{XY}{9}$

$\Rightarrow XY=2.25$ $cm$

We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

$\frac{ar\left(△PXY\right)}{ar\left(△PQR\right)}=(\frac{PX}{PQ}{)}^2$

$\Rightarrow \frac{A}{ar\left(△PQR\right)}=\frac{1}{16}$

$\Rightarrow ar\left(△PQR\right)=16A$ $c{m}^2$

Now, $ar\left(trapeziumXYRQ\right)=(16A-A)c{m}^2=15A$ $c{m}^2$