The problem is based on similar triangles $\triangle PXY$ and $\triangle PRQ$. We need to: a) Name a correct order for 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 $\triangle PQR$. Given: $PJ=2, JR=4, PX=3, XQ=3$.

1. Problem Description

The problem is based on similar triangles $\triangle PXY$ and $\triangle PRQ$. We need to:

a) Name a correct order for 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 $\triangle PQR$.

Given: $PJ=2, JR=4, PX=3, XQ=3$.

2. Solution Steps

a) We are given that $\triangle PXY$ is similar to $\triangle PRQ$. Thus, a correct order is $\triangle PRQ$.

b) To find the length of $QR$, we can use the similarity of the triangles $\triangle PXY$ and $\triangle PRQ$.

We have $PJ = 2$ and $JR = 4$, so $PR = PJ + JR = 2 + 4 = 6$.

We also have $PX = 3$ and $XQ = 3$, so $PQ = PX + XQ = 3 + 3 = 6$.

Now, since $\triangle PXY \sim \triangle PRQ$, we have the following ratios:

$\frac{PX}{PQ} = \frac{PY}{PR} = \frac{XY}{RQ}$

From the image, it appears that $XY$ is parallel to $RQ$. Then, from the given information, we are given:

$\frac{PY}{PR} = \frac{YX}{RQ} = \frac{PX}{PQ}$.

Given $PX = 3$ and $PQ = 6$, then $\frac{PX}{PQ} = \frac{3}{6} = \frac{1}{2}$.

Also, we are given $\frac{YX}{RQ} = \frac{4}{RQ}$.

Therefore, $\frac{1}{2} = \frac{4}{RQ}$.

$RQ = 4 \times 2 = 8$.

Hence $QR = 8$.

c) To find the ratio of the area of $\triangle PXY$ to the area of $\triangle PQR$, we can use the property that the ratio of the areas of similar triangles is equal to the square of the ratio of their corresponding sides.

$\frac{Area(\triangle PXY)}{Area(\triangle PQR)} = (\frac{PX}{PQ})^2 = (\frac{PY}{PR})^2 = (\frac{XY}{QR})^2$

Since $\frac{PX}{PQ} = \frac{3}{6} = \frac{1}{2}$, then

$\frac{Area(\triangle PXY)}{Area(\triangle PQR)} = (\frac{1}{2})^2 = \frac{1}{4}$.

3. Final Answer

a) $\triangle PRQ$

b) $QR = 8$

c) $\frac{Area(\triangle PXY)}{Area(\triangle PQR)} = \frac{1}{4}$