The problem consists of three parts related to similar triangles $\triangle PXY$ and $\triangle PQR$. (a) Name the correct order of the triangle that is similar to $\triangle PXY$. (b) Find the length of $QR$. (c) Find the ratio of the area of $\triangle PXY$ to the area of $\triangle PQR$.

1. Problem Description

The problem consists of three parts related to similar triangles $\triangle PXY$ and $\triangle PQR$.

(a) Name the correct order of the triangle that is similar to $\triangle PXY$.

(b) Find the length of $QR$.

(c) Find the ratio of the area of $\triangle PXY$ to the area of $\triangle PQR$.

2. Solution Steps

(a) Since $\triangle PXY$ is similar to $\triangle PQR$, we can directly write it as $\triangle PQR$.

(b) We are given $PY = 2$, $YR = 3$, so $PR = PY + YR = 2 + 3 = 5$. We are also given $YX = 4$.

Since $\triangle PXY \sim \triangle PQR$, we have the proportion:

$\frac{PY}{PR} = \frac{YX}{RQ}$

$\frac{2}{5} = \frac{4}{RQ}$

$2 \cdot RQ = 5 \cdot 4$

$2 \cdot RQ = 20$

$RQ = \frac{20}{2}$

$RQ = 10$

(c) The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides. So, the ratio of the area of $\triangle PXY$ to the area of $\triangle PQR$ is:

$\frac{Area(\triangle PXY)}{Area(\triangle PQR)} = \left( \frac{PY}{PR} \right)^2 = \left( \frac{2}{5} \right)^2 = \frac{4}{25}$

3. Final Answer

(a) $\triangle PQR$

(b) $RQ = 10$

(c) $\frac{Area(\triangle PXY)}{Area(\triangle PQR)} = \frac{4}{25}$