Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_c797caf821299ab18642g-4.jpg?height=253&width=460&top_left_y=765&top_left_x=1300)

Let $r$ be the radius of the circle and $a$ the side length of the larger square. The side length of the smaller square is 4.

![](https://cdn.mathpix.com/cropped/2024_05_30_c797caf821299ab18642g-4.jpg?height=419&width=729&top_left_y=1202&top_left_x=690)

From the right triangle $C D O$ we get: $r^{2}=\left(\frac{a}{2}\right)^{2}+a^{2}$,

(4 points)

and from the right triangle $E F O$: $r^{2}=\left(\frac{a}{2}+4\right)^{2}+4^{2}$.

(4 points)

By equating

$$
\left(\frac{a}{2}\right)^{2}+a^{2}=\left(\frac{a}{2}+4\right)^{2}+4^{2}
$$

we obtain the quadratic equation $a^{2}-4 a-32=0$, with solutions $a=8$ and $a=-4$. (7 points) We discard the negative solution, so the side length of the larger square is 8, and its area is 64.

(3 points)
4. Two squares are inscribed in a semicircle as shown in the figure.

If the area of the smaller square is 16, what is the area of the larger one? Justify your conclusion!