If we increase one leg of a right triangle by 5 and decrease the other leg by 5, the area of the triangle increases by 5. How does the area of the square drawn on the hypotenuse change in the meantime?
Solution. The sides of the first right-angled triangle are: $a, b, c$, where $c$ is the hypotenuse.

The sides of the second right-angled triangle are: $a^{\prime}=a-5, b^{\prime}=b+5, c^{\prime}$, where $c^{\prime}$ is the hypotenuse.

Their areas:

$$
\begin{aligned}
& T_{1}=\frac{a b}{2} \\
& T_{2}=\frac{a^{\prime} b^{\prime}}{2}=\frac{(a-5)(b+5)}{2}=T_{1}+5
\end{aligned}
$$

From (1) and (2):

$$
\frac{(a-5)(b+5)}{2}=\frac{a b}{2}+5
$$

which gives

$$
a-b=7
$$

Using the Pythagorean theorem for both triangles, we can find the area of the squares drawn on the hypotenuses. For the first triangle:

$$
a^{2}+b^{2}=c^{2}
$$

For the other triangle: $(a-5)^{2}+(b+5)^{2}=c^{\prime 2}$. This can also be written as:

$$
a^{2}+b^{2}-10(a-b)+50=c^{\prime 2}
$$

Substituting (3) and (4) into (5):

$$
c^{2}-10 \cdot 7+50=c^{\prime 2}, \quad c^{2}-20=c^{\prime 2}
$$

Thus, the area of the square drawn on the hypotenuse decreases by 20 square units compared to the original.