3. Two equal rectangles $P Q R S$ and $P_{1} Q_{1} R_{1} S_{1}$ are inscribed in triangle $A B C$ (with points $P$ and $P_{1}$ lying on side $A B$, points $Q$ and $Q_{1}$ lying on side $B C$, and points $R, S, R_{1}$ and $S_{1}$ lying on side $A C$). It is known that $P S=12, P_{1} S_{1}=3$. Find the area of triangle $A B C$.
Answer: $\frac{225}{2}$.

Solution: Draw the height $B F$ of triangle $A B C$. Let it intersect segments $P Q$ and $P_{1} Q_{1}$ at points $H$ and $M$ respectively. Note that $H M=9$. From the similarity of triangles $B P Q$ and $B P_{1} Q_{1}$, it follows that $\frac{B H}{P Q}=\frac{B M}{P_{1} Q_{1}}$ (the ratio of height to base is written), from which $\frac{B H}{3}=\frac{9+B H}{12}, B H=3$. Therefore, the area of triangle $B P Q$ is $\frac{1}{2} \cdot 3 \cdot 3=\frac{9}{2}$. But triangles $A B C$ and $B P Q$ are similar (the similarity ratio is $B F: B H=15: 3=5$), hence the area of triangle $A B C$ is $25 \cdot \frac{9}{2}=\frac{225}{2}$.