4. The cities A, B, and C are connected by straight roads. Along the road A-B, there is a square field with a side length of $0.5 \overline{A B}$, and along the road B-C, there is a square field with a side length of $\overline{B C}$; along the road A-C, there is a rectangular forest with a length of $\overline{A C}$ and a width of 4 kilometers. Find the area of the forest, if it is 20 square kilometers larger than the sum of the areas of the square fields.
Solution. Let $\overline{A B}=a, \overline{B C}=b, \overline{A C}=x$. Then $a+b \geq x$. According to the condition

$$
\begin{aligned}
& 4 x=\frac{a^{2}}{4}+b^{2}+20 \\
& x=\frac{a^{2}}{16}+\frac{b^{2}}{4}+5 \\
& a+b \geq \frac{a^{2}}{16}+\frac{b^{2}}{4}+5 \\
& 16 a+16 b \geq a^{2}+4 b^{2}+80 \\
& a^{2}-16 a+4 b^{2}-16 b+80 \leq 0 \\
& (a-8)^{2}+4(b-2)^{2} \leq 0 \\
& a=8, b=2 \\
& x=10 \\
& P=40 m^{2}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_134071f890a711206f81g-2.jpg?height=567&width=622&top_left_y=1096&top_left_x=876)

## Second Year