10.087. A square of unit area is inscribed in an isosceles triangle, one side of which lies on the base of the triangle. Find the area of the triangle, given that the centers of mass of the triangle and the square coincide (the center of mass of the triangle lies at the intersection of its medians).
Solution.

$B D$ is the median, $B O: O D=2: 1 ; O D=0.5$ (cm), $B O=1($ cm $), B D=$ $=O D+B O=1.5$ (cm) (Fig.10.84). $\triangle B E F$ and $\triangle B D C$ are similar, $\frac{B D}{B E}=\frac{D C}{E F}$. Therefore, $D C=\frac{B D \cdot E F}{B E}, D C=\frac{1.5 \cdot 0.5}{0.5}=1.5(\mathrm{~cm}), A C=2 \times$ $\times D C=3$ (cm), $S_{\triangle A B C}=\frac{1}{2} A C \cdot B D=\frac{1}{2} \cdot 3 \cdot 1.5=\frac{3}{2} \cdot \frac{3}{2}=\frac{9}{4}$ (sq. units).

Answer: $\frac{9}{4}$ sq. units.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-598.jpg?height=404&width=428&top_left_y=90&top_left_x=150)

Fig. 10.85

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-598.jpg?height=412&width=386&top_left_y=78&top_left_x=783)

Fig. 10.86