## Task 1 - 060811

In the isosceles right triangle $\triangle A B C$ with $\overline{A B}=$ $c=7 \mathrm{~cm}$ and $\angle A C B=\gamma=90^{\circ}$, arcs with a radius of $\frac{7}{2} \mathrm{~cm}$ are drawn around the points $A$ and $B$ (see figure).

Determine the area $I_{F}$ of the shaded region shown in the figure!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0606.jpg?height=297&width=452&top_left_y=454&top_left_x=1316)
If the area of triangle $\triangle A B C$ is denoted by $I_{F_{1}}$, and the areas of the circular segments inside the triangle are denoted by $I_{F_{2}}$ and $I_{F_{3}}$, then the desired area $I_{F}$ is given by:

$$
I_{F}=I_{F_{1}}-\left(I_{F_{2}}+I_{F_{3}}\right)
$$

If $M$ is the midpoint of $A B$, then the height of triangle $\triangle A B C$ is $M C$, and its length is $\frac{7}{2}$ $\mathrm{cm}$. Therefore, for $I_{F_{1}}$ we have:

$$
I_{F_{1}}=\frac{1}{2} \cdot 7 \cdot \frac{7}{2} \mathrm{~cm}^{2}=\frac{49}{4} \mathrm{~cm}^{2}
$$

The two circular segments with areas $I_{F_{2}}$ and $I_{F_{3}}$ can be combined to form a quarter circle. Thus, $I_{F_{2}}+I_{F_{3}}=\frac{1}{4} \cdot \frac{49}{4} \pi \mathrm{cm}^{2}$. Therefore, for $I_{F}$ we have:

$$
I_{F}=I_{F_{1}}-\left(I_{F_{2}}+I_{F_{3}}\right)=\frac{49}{4}\left(1-\frac{\pi}{4}\right) \mathrm{cm}^{2} \approx 2.63 \mathrm{~cm}^{2}
$$

The area of the shaded region in the figure is thus $I_{F} \approx 2.63 \mathrm{~cm}^{2}$.