## Task B-2.2.

In triangle $A B C$, $|A B|=12 \mathrm{~cm},|B C|=25 \mathrm{~cm},|C A|=17 \mathrm{~cm}$. A rectangle $K L M N$ is inscribed in the triangle such that vertices $M$ and $N$ are on side $\overline{B C}$, vertex $K$ is on side $\overline{A B}$, and vertex $L$ is on side $\overline{C A}$.

Determine the lengths of the sides of the rectangle if its area is equal to $\frac{216}{5}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_1ebddb66d71be8305afcg-06.jpg?height=502&width=746&top_left_y=660&top_left_x=655)

Using Heron's formula, we can determine the area of triangle $ABC$, and then the height to side $\overline{BC}$

$$
\begin{gathered}
s=\frac{12+25+17}{2}=27 \Longrightarrow P=\sqrt{27 \cdot 15 \cdot 10 \cdot 2}=90 \\
v_{a}=\frac{2 P}{|B C|}=\frac{36}{5}
\end{gathered}
$$

Using the similarity of triangles $AKL$ and $ABC(K K)$, we get

$$
\begin{gathered}
v_{a}:\left(v_{a}-y\right)=|B C|: x \\
\frac{36}{5}:\left(\frac{36}{5}-y\right)=25: x \\
y=\frac{900-36 x}{125}
\end{gathered}
$$

Since the area of the rectangle is $P=x y$, we have

$$
x \cdot \frac{900-36 x}{125}=\frac{216}{5}
$$

which, after simplification, gives the quadratic equation

$$
x^{2}-25 x+150=0
$$

Its solutions are $x=10$ and $x=15$. Then, $y=\frac{108}{25}=4.32$ or $y=\frac{72}{25}=2.88$.

Thus, two rectangles satisfy the given conditions: one with side lengths 10 and 4.32, and the other with side lengths 15 and 2.88.