[ Pythagorean Theorem (direct and inverse). ] Complex [ The ratio of the areas of triangles with a common base or a common height] Class

In trapezoid $A B C D$, the lateral side $A D$ is perpendicular to the bases and equals $9, C D=12$, and the segment $A O$, where $O-$ is the point of intersection of the diagonals of the trapezoid, equals 6. Find the area of triangle $B O C$.

#
Triangles $B O C$ and $A O D$ are equal in area.

## Solution

By the Pythagorean theorem from the right triangle $A D C$, we find that

$$
A C=\sqrt{A D^{2}+C D^{2}}=\sqrt{81+144}=\sqrt{225}=15
$$

Then

$$
\begin{gathered}
O C=A C-A O=15-6=9 \\
S_{\triangle \mathrm{AOD}}=\frac{A O}{A C} \cdot S_{\triangle \mathrm{ADC}}=\frac{6}{15} \cdot \frac{1}{2} A D \cdot C D=\frac{2}{5} \cdot \frac{1}{2} \cdot 9 \cdot 12=\frac{108}{5}
\end{gathered}
$$

Triangle $B O C$ is equal in area to triangle $A O D$, since

$$
S_{\triangle \mathrm{BOC}}=S_{\triangle \mathrm{ACB}}-S_{\triangle \mathrm{AOB}}=S_{\triangle \mathrm{ADB}}-S_{\triangle \mathrm{AOB}}=S_{\triangle \mathrm{AOD}}
$$

Therefore, $S_{\Delta \text { BOC }}=\frac{108}{5}$.

## Answer

$\frac{108}{5}$.