4. The side lengths of the squares drawn on the sides of the triangle in the figure are $4 \mathrm{~cm}$ and $5 \mathrm{~cm}$. The area of the triangle is $8 \mathrm{~cm}^{2}$. What is the area of the shaded parallelogram?

![](https://cdn.mathpix.com/cropped/2024_06_03_2716aa3ceafccc2b473cg-1.jpg?height=425&width=580&top_left_y=2152&top_left_x=744)
4. 

![](https://cdn.mathpix.com/cropped/2024_05_30_cbd21e468efb4566cfb2g-4.jpg?height=445&width=531&top_left_y=2199&top_left_x=708)

Completed sketch:

$1 \text{ POINT}$

Let the angle in the triangle between the sides of length $4 \text{ cm}$ and $5 \text{ cm}$ be $\alpha$.

Then, the angle of the parallelogram at the vertex shared with the triangle is

$360^{\circ} - 90^{\circ} - 90^{\circ} - \alpha = 180^{\circ} - \alpha$, and the adjacent angle of the parallelogram is of size $\alpha$.

$2 \text{ POINTS}$

The diagonal of the parallelogram from the vertex shared with the triangle divides the parallelogram into two congruent triangles congruent to triangle $ABC$.

$1 \text{ POINT}$

The triangles in the parallelogram are congruent to triangle $ABC$ by the SAS theorem because they have sides of length $4 \text{ cm}$ and $5 \text{ cm}$ and an angle of size $\alpha$ between those sides.

Therefore, the area of the shaded parallelogram is $2 \cdot 8 = 16 \text{ cm}^2$. $\qquad$
TOTAL 6 POINTS