3+ A square is inscribed in a right triangle such that one of its sides lies on the hypotenuse. The lateral segments of the hypotenuse are equal to m and $n$. Find the area of the square.
Through the vertex of the square, belonging to the leg of the given triangle, draw a line parallel to the other leg.

## Solution

Let the vertices $M$ and $N$ of the square MNKL be located respectively on

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-16.jpg?height=54&width=641&top_left_y=2870&top_left_x=14)
[ Rectangles and squares. Criteria and properties ] Classes: 8,9

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-16.jpg?height=49&width=45&top_left_y=2215&top_left_x=635)

Table

The

I

First method.

Through point $N$, draw a line parallel to $AC$, up to the intersection with the hypotenuse $AB$ at point $P$. From the equality of the right triangles $AML$ and $PNK$, it follows that $PK = AL = m$.

The side $NK$ of the specified square is the height of the right triangle $PNB$, drawn from the vertex of the right angle. Therefore,

$$
NK^2 = PK \cdot BK = mn
$$

Thus, the area of the square $MNKL$ is $mn$.

Second method.

Let $\angle BAC = \alpha$. Then $\angle$

$BKN = \alpha$. From the right triangles $ALM$ and $NKB$, we find that

$$
\tan \alpha = \frac{ML}{AL}, \tan \alpha = \frac{BK}{KN}
$$

therefore

$$
\frac{ML}{AL} = \frac{KB}{KN}
$$

Thus,

$$
\begin{gathered}
S_{\mathrm{MNKL}} = ML^2 = ML \cdot KL = AL \cdot BK = \\
mn.
\end{gathered}
$$

Let the vertices $M$ and $N$ of the square $MNKL$ be located respectively on the legs $AC$ and $BC$ of the right triangle $ABC$, and the vertices $K$ and $L$ - on the hypotenuse $AB$. Here, $AL = m$, $BK = n$.

First method.

Through point $N$, draw a line parallel to $AC$, up to the intersection with the hypotenuse $AB$ at point $P$. From the equality of the right triangles $AML$ and $PNK$, it follows that $PK = AL = m$.

The side $NK$ of the specified square is the height of the right triangle $PNB$, drawn from the vertex of the right angle. Therefore,

$$
NK^2 = PK \cdot BK = mn.
$$

Thus, the area of the square $MNKL$ is $mn$.

Second method.

## Let $\angle BAC = \alpha$. Then

$BKN = \alpha$. From the right triangles $ALM$ and $NKB$, we find that

$$
\tan \alpha = \frac{ML}{AL}, \tan \alpha = \frac{BK}{KN}
$$

therefore

$$
\frac{ML}{AL} = \frac{KB}{KN}
$$

Thus,

$S_{\mathrm{MNKL}} = ML^2 = ML \cdot KL = AL \cdot BK = mn$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-19.jpg?height=303&width=688&top_left_y=0&top_left_x=-1)

## Answer

## $mn$.