Question

The given figure shows a triangle PQR in which XY is parallel to QR. If $PX:XQ=1:3$ and $QR=9\text{ cm}$ , find the length of XY. Further, if the area of $\Delta PXY=x\text{ c}{{\text{m}}^{2}}$ , find in terms of x the area of
(a) triangle PQR
(b) trapezium XQRY

Answer 1

Hint: To find the length of XY, we have to prove that triangles PQR and PXY are similar which will result in $\dfrac{PX}{PQ}=\dfrac{XY}{QR}$ using the property of similar triangles. From the figure, $PQ=PX+XQ$ and using the given ratio, we will get the value of $\dfrac{PX}{PQ}$ . Simplifying the previous result will give the value of XY. To find the area of triangle PQR, we have to apply the property of areas of similar triangles and simplify. To find the area of trapezium XQRY, we have to subtract the area of triangle PXY from the area of triangle PQR.

Complete step by step answer:
We are given triangles PQR and PXY. We are also given that XY is parallel to QR. We know that if the transversal intersects two parallel lines, then the corresponding angles will be equal. Therefore,
$\begin{align}
  & \angle PXY=\angle PQR \\
 & \angle PYX=\angle PRQ \\
\end{align}$
Hence, by Angle-Angle (AA) property, $\Delta PXY\sim \Delta PQR$ .
We know that if two triangles are similar, then their corresponding sides are in equal proportion.
$\Rightarrow \dfrac{PX}{PQ}=\dfrac{XY}{QR}...\left( i \right)$
We are given that $\dfrac{PX}{XQ}=\dfrac{1}{3}$ . We can write the value of XQ from this ratio as
$XQ=3PX...\left( ii \right)$
From the figure, we can see that PQ will be the sum of PX and XQ.
$\Rightarrow PQ=PX+XQ$
Let us substitute (ii) in the above equation.
$\begin{align}
  & \Rightarrow PQ=PX+3PX \\
 & \Rightarrow PQ=4PX \\
\end{align}$
Let us find $\dfrac{PX}{PQ}$ from the above equation.
$\Rightarrow \dfrac{PX}{PQ}=\dfrac{1}{4}...\left( iii \right)$
We are given that $QR=9\text{ cm}$ . Let us substitute this value and (iii) in the equation (i).
$\begin{align}
  & \Rightarrow \dfrac{1}{4}=\dfrac{XY}{9} \\
 & \Rightarrow XY=\dfrac{9}{4} \\
 & \Rightarrow XY=2.25\text{ cm} \\
\end{align}$
Therefore, the length of XY is 2.25 cm.
Now, let us find the areas of triangle PQR and trapezium XQRY.
(a) We know that if two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
$\Rightarrow \dfrac{ar\left( \Delta PXY \right)}{ar\left( \Delta PQR \right)}={{\left( \dfrac{PX}{PQ} \right)}^{2}}$
We are given the area of $\Delta PXY=x\text{ c}{{\text{m}}^{2}}$ . Therefore, the above equation becomes
$\begin{align}
  & \Rightarrow \dfrac{x}{ar\left( \Delta PQR \right)}={{\left( \dfrac{1}{4} \right)}^{2}} \\
 & \Rightarrow \dfrac{x}{ar\left( \Delta PQR \right)}=\dfrac{1}{16} \\
\end{align}$
Let us cross multiply.
$\Rightarrow ar\left( \Delta PQR \right)=16x\text{ c}{{\text{m}}^{2}}$
Therefore, the area of triangle PQR is $16x\text{ c}{{\text{m}}^{2}}$ .
(b) From the figure, we can find the area of trapezium XQRY by subtracting the area of triangle PXY from the area of triangle PQR.
$\begin{align}
  & \Rightarrow ar\left( XQRY \right)=ar\left( PQR \right)-ar\left( PXY \right) \\
 & \Rightarrow ar\left( XQRY \right)=16x-x \\
 & \Rightarrow ar\left( XQRY \right)=15x\text{ c}{{\text{m}}^{2}} \\
\end{align}$
Therefore, the area of trapezium XQRY is $15x\text{ c}{{\text{m}}^{2}}$ .

Note: Students must be thorough with the properties of lines and angles, triangles, similar and congruent triangles. They must prove the similarity between two triangles either using Angle-Angle Similarity (AA or AAA) or Side-Angle-Side Similarity (SAS) or Side-Side-Side Similarity (SSS).