Two congruent right triangles have legs measuring $4 \mathrm{~cm}$ and $7 \mathrm{~cm}$. In the figure below, on the left, the triangles were 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 were 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_a444074f1f8ae50444d5g-02.jpg?height=248&width=1076&top_left_y=1156&top_left_x=492)

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_a444074f1f8ae50444d5g-15.jpg?height=282&width=504&top_left_y=1277&top_left_x=781)

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 $A B C D$. Therefore, the shaded area is equal to $\frac{3}{4}(7 \times 4)=21 \mathrm{~cm}^{2}$.

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

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-15.jpg?height=265&width=510&top_left_y=1911&top_left_x=775)

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

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

So, we just need to calculate $x$. We have that $x+y=7$ and, by the Pythagorean Theorem applied to triangle $A E D, y^{2}=x^{2}+4^{2}$. Substituting $y=7-x$ in this last equation, we get:

$$
(7-x)^{2}=x^{2}+16 \Rightarrow 49-14 x+x^{2}=x^{2}+16 \Rightarrow x=\frac{49-16}{14}=\frac{33}{14}
$$

Finally, the shaded area is:

$$
2 \times \frac{33}{14}+14=\frac{33}{7}+14=4 \frac{5}{7}+14=18 \frac{5}{7} .
$$