The base of the triangle is 20; the medians drawn to the lateral sides are 18 and 24. Find the area of the triangle.

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If $M$ is the point of intersection of the medians of triangle $ABC$, then its area is three times the area of triangle $BMC$.

## Solution

Let $BB_1$ and $CC_1$ be the medians of triangle $ABC$, $M$ be their point of intersection, $BB_1 = 18$, $CC_1 = 24$, and $BC = 20$. Then the sides of triangle $BMC$ are:

$$
BC = 20, BM = \frac{2}{3} BB_1 = 12, CM = \frac{2}{3} CC_1 = 16
$$

This triangle is a right triangle ( $BC^2 = BM^2 + CM^2$ ). Its area is 96, and the area of triangle $ABC$ is three times larger, i.e., 288.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-51.jpg?height=391&width=483&top_left_y=222&top_left_x=793)

## Answer

## Problem