8,9}

The medians of a triangle are 3, 4, and 5. Find the area of the triangle.

#
Prove that the area of the triangle formed by the medians of a given triangle is $\frac{3}{4}$ of the area of the given triangle.

## Solution

Let $B_{1}$ be the midpoint of side $A C$ of triangle $A B C$, and $M$ be the point of intersection of its medians. On the extension of median $B B_{1}$ beyond point $B_{1}$, lay off segment $B_{1} K$ equal to $M B_{1}$. Then $A M C K$ is a parallelogram, $C K=A M$. The sides of triangle $K M C$ are $\frac{2}{3}$ of the corresponding medians of triangle $A B C$. Therefore, triangle $K M C$ is similar to the triangle whose sides are equal to the medians of triangle $A B C$. Thus, the area of triangle $K M C$ is $\frac{4}{9}$ of the area of the triangle with sides $3, 4, 5$, i.e., $\frac{4}{9} \cdot 6=\frac{8}{3}$. Therefore,

$$
S_{\Delta \mathrm{ABC}}=6 S_{\Delta \mathrm{B}_{1} \mathrm{MC}}=6 \cdot \frac{1}{2} S_{\Delta \mathrm{KMC}}=8
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f3ff818e52eab4ca192bg-05.jpg?height=506&width=535&top_left_y=107&top_left_x=767)

## Answer

8.