In the figure below, we have a square divided into two congruent triangles and a right triangle whose larger leg has the same measure as the side of the square and the smaller leg has half the measure of the side of the square.
![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-04.jpg?height=310&width=596&top_left_y=590&top_left_x=612)

If the area of the square is $4 k$, determine:

a) The light gray area in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-04.jpg?height=386&width=389&top_left_y=1264&top_left_x=736)

b) The dark gray area in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-04.jpg?height=388&width=391&top_left_y=1979&top_left_x=738)

#
Solution

a) The light gray area is half the area of the square $(2 k)$ minus the area of the triangle, which is a quarter of the area of the square, that is, $k$. Therefore, the light gray area is $2 k-k=k$.

b) Extending one side of the square and the hypotenuse of the triangle, as shown in the figure, we will mark the points $A, B, C, D$ and $E$.

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-05.jpg?height=748&width=441&top_left_y=800&top_left_x=911)

The triangles $\triangle A B E$ and $\triangle C D E$ are similar with a ratio of 4, and consequently, the ratio of the areas is 16. Moreover, $\frac{E B}{E D}=4$ (ratio), which implies that the ratio of the areas of the triangles $\triangle B C E$ and $\triangle C D E$ is 4. If the area of $\triangle C D E$ is equal to $x$, the area of $\triangle B C E$ is equal to $4 x$. Since the area of $\triangle A B C$ is equal to the area of the square ($4 k$), we have:

$$
\begin{aligned}
\frac{A_{A B E}}{A_{C D E}} & =16 \\
\frac{A_{A B C}-A_{B C E}}{A_{C D E}} & =16 \\
\frac{4 k-4 x}{x} & =16 \\
16 x & =4 k-4 x \\
20 x & =4 k \\
x & =\frac{k}{5}
\end{aligned}
$$