Two congruent right triangles have legs measuring $4 \mathrm{~cm}$ and $7 \mathrm{~cm}$. In the figure on the left, the triangles are drawn so that the legs of $7 \mathrm{~cm}$ coincide. Thus, $A B=7 \mathrm{~cm}$ and $A D=B C=4 \mathrm{~cm}$. In the figure on the right, they are drawn so that the hypotenuses coincide, where $A D=B C=4 \mathrm{~cm}$ and $A C=B D=7 \mathrm{~cm}$.
![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-029.jpg?height=266&width=1146&top_left_y=2288&top_left_x=540)

Calculate the shaded areas in both figures.
The points $A, B, C$ and $D$ form the rectangle $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-102.jpg?height=317&width=528&top_left_y=1766&top_left_x=798)

Since the diagonals of a rectangle divide it into four triangles of equal area, the shaded area is equal to three quarters of the area of the rectangle $A B C D$. Thus, the shaded area is equal to $\frac{3}{4}(7 \times 4)=21 \mathrm{~cm}^{2}$.

Now, let's consider the other figure. Let $x=D E=C E, y=A E=B E$ and $E$ be the intersection point of the segments $A C$ and $B D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-102.jpg?height=306&width=602&top_left_y=2394&top_left_x=767)

The shaded area is the sum of the areas of triangles $A D E$ and $A B C$, that is,

$$
\frac{4 \times x}{2}+\frac{4 \times 7}{2}=2 x+14
$$

Thus, it is enough to calculate $x$. We have that $x+y=7$ and, by the Pythagorean Theorem applied to triangle $A E D$, also $y^{2}=x^{2}+4^{2}$. Substituting $y=7-x$ in this last equation, we get $(7-x)^{2}=x^{2}+16$, so $49-14 x+x^{2}=x^{2}+16$, that is,

$$
x=\frac{49-16}{14}=\frac{33}{14}
$$

Finally, the shaded area is given by $2 \times \frac{33}{14}+14=\frac{33}{7}+14=\frac{131}{7} \mathrm{~cm}^{2}$.