## Task B-1.3.

Over the sides of triangle $\triangle A B C$, semicircles have been constructed whose areas are equal to $9 \pi, 16 \pi$ and $25 \pi$. What is the area of triangle $\triangle A B C$?
## Solution.

The following holds:

$$
\begin{aligned}
& P_{1}=9 \pi, \quad \frac{r_{1}^{2} \pi}{2}=9 \pi, \quad r_{1}=\sqrt{18}=3 \sqrt{2}, \quad a=6 \sqrt{2} \\
& P_{2}=16 \pi, \quad \frac{r_{2}^{2} \pi}{2}=16 \pi, \quad r_{2}=\sqrt{32}=4 \sqrt{2}, \quad b=8 \sqrt{2} \quad 1 \text { point }
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_8e0ddf1d7bf4bfb115dcg-02.jpg?height=105&width=1424&top_left_y=553&top_left_x=453)

From Heron's formula, it follows that

$$
s=12 \sqrt{2}, \quad P=\sqrt{12 \sqrt{2} \cdot 6 \sqrt{2} \cdot 4 \sqrt{2} \cdot 2 \sqrt{2}}=\sqrt{2304}=48
$$

Note: If we observe that the triangle is right-angled $\left(a^{2}+b^{2}=72+128=200=(10 \sqrt{2})^{2}=c^{2}\right)$, the area can be calculated using the formula $P=\frac{a b}{2}=48$.