$\left[\begin{array}{l}\text { Properties of medians. The centroid of a triangle. } \\ {[\text { Pythagorean Theorem (direct and inverse). }]} \end{array}\right]$

Find the area of triangle $ABC$, if $AC=3$, $BC=4$, and the medians $AK$ and $BL$ are perpendicular to each other.
The medians of a triangle are divided by the point of intersection in the ratio 2:1, counting from the vertex.

## Solution

Let $M$ be the point of intersection of the medians of triangle $ABC$. Denote $M L=x, M K=y$. Then $M B=2 x, M A=2 y$. By the Pythagorean theorem from right triangles $B M K$ and $A M L$, we find:

$$
\left\{\begin{array}{l}
4 x^{2}+y^{2}=4 \\
x^{2}+4 y^{2}=\frac{9}{4}
\end{array}\right.
$$

From the obtained system, we find that

$$
x=\frac{\sqrt{11}}{2 \sqrt{3}}, y=\frac{1}{\sqrt{3}}
$$

$$
S_{\triangle \mathrm{AMB}}=\frac{1}{2} A M \cdot M B=\frac{1}{2} \cdot 2 x \cdot 2 y=2 x y=\frac{\sqrt{11}}{3}
$$

Therefore,

$$
S_{\triangle \mathrm{ABC}}=3 S_{\triangle \mathrm{AMB}}=\sqrt{11}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-25.jpg?height=443&width=828&top_left_y=356&top_left_x=615)

## Answer

$\sqrt{11}$.