The figure below shows a square PQRS. PW and SQ are straight lines. The height of triangle PXQ is twice the height of triangle SXW. SW = WR. Given that the area of triangle SXW is 14 $\text{cm}^2$, find the area of triangle PXQ. Ans: ___ $\text{cm}^2$

1. Determine the ratio of the bases of triangle PXQ and triangle SXW.

Since $PQRS$ is a square, $PQ = SR$.
Given $SW = WR$, $W$ is the midpoint of $SR$, so $SW = \frac{1}{2} SR = \frac{1}{2} PQ$.
Therefore, the ratio of the base of $\triangle PXQ$ to the base of $\triangle SXW$ is $PQ : SW = 2 : 1$.

2. Determine the ratio of the areas of triangle PXQ and triangle SXW.

Let the base of $\triangle SXW$ be $b$ and its height be $h$.
Area of $\triangle SXW = \frac{1}{2} \times b \times h = 14$.
The base of $\triangle PXQ$ is $2b$ and its height is $2h$ (given).
Area of $\triangle PXQ = \frac{1}{2} \times (2b) \times (2h) = 4 \times (\frac{1}{2} \times b \times h)$.
So, Area of $\triangle PXQ = 4 \times \text{Area of } \triangle SXW$.

3. Calculate the area of triangle PXQ.

Area of $\triangle PXQ = 4 \times 14 = 56 \text{ cm}^2$.