10.326. The medians of a triangle are 5, 6, and 5 m. Find the area of the triangle.
Solution.

The area of $\triangle ABC$ is 3 times the area of $\triangle AOC$ (where $O$ is the point of intersection of the medians; see Fig. 10.114). We have $S_{\triangle AOC} = \frac{1}{2} AC \cdot OK = KC \cdot OK$. But $OK = \frac{1}{2} BK = 2M$, $KC = \sqrt{OC^2 - OK^2}$, where $OC = \frac{2}{3} MC = \frac{10}{3} M \Rightarrow$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0794.jpg?height=563&width=461&top_left_y=1067&top_left_x=801)

Fig. 10.114

$$
\Rightarrow KC = \sqrt{\left(\frac{10}{3}\right)^2 - 2^2} = \frac{8}{3} M, \text{ and } S_{\triangle AOC} = \frac{16}{3} M^2, \text{ i.e., } S_{\triangle ABC} = 16 M^2
$$

Answer: $16 M^2$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0795.jpg?height=275&width=523&top_left_y=256&top_left_x=122)

Fig. 10.115

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0795.jpg?height=468&width=575&top_left_y=63&top_left_x=673)

Fig. 10.116