The problem states that $\triangle PYX$ is similar to $\triangle PRQ$. Given the lengths $PY = 2$, $YR = 3$, and $YX = 4$, we are asked to find: a) A correct order of vertices for a triangle that is similar to $\triangle PYX$. b) The length of $QR$. c) The ratio of the area of $\triangle PXY$ to the area of $\triangle PQR$.

1. Problem Description

The problem states that $\triangle PYX$ is similar to $\triangle PRQ$. Given the lengths $PY = 2$, $YR = 3$, and $YX = 4$, we are asked to find:

a) A correct order of vertices for a triangle that is similar to $\triangle PYX$.

b) The length of $QR$.

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

2. Solution Steps

a) Since $\triangle PYX \sim \triangle PRQ$, a correct order of the vertices for a triangle similar to $\triangle PYX$ is $\triangle PRQ$.

b) We are given $PY=2$ and $YR=3$, so $PR = PY + YR = 2 + 3 = 5$. Also, $YX=4$. Since the triangles are similar, we have the ratios:

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

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

$2 \cdot RQ = 5 \cdot 4$

$2 \cdot RQ = 20$

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

$RQ = 10$

c) The ratio of the areas of similar triangles is the square of the ratio of corresponding sides. So,

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

3. Final Answer

a) $\triangle PRQ$

b) $QR = 10$

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