5. On the sides $\mathrm{AB}, \mathrm{BC}$, and $\mathrm{CA}$ of triangle $\mathrm{ABC}$, points $\mathrm{M}, \mathrm{N}$, and P are chosen, respectively, such that MP $\| \mathrm{BC}$ and MN $\| \mathrm{AC}$. If the area of $\triangle \mathrm{AMP}$ is $9 \mathrm{~cm}^{2}$, and the area of $\triangle \mathrm{MBN}$ is 16 $\mathrm{cm}^{2}$, calculate the area of triangle NPC.
Solution: Quadrilateral MNCP is a parallelogram, so $\mathrm{P}_{\mathrm{CPN}} = \mathrm{P}_{\text{MNP}}$. We have $\mathrm{P}_{\mathrm{CPN}} = \frac{1}{2} \left( \mathrm{P}_{\mathrm{ABC}} - \mathrm{P}_{\mathrm{AMP}} - \mathrm{P}_{\mathrm{MBN}} \right)$ ... (1). From the similarity of triangles AMP and ABC, and triangles MBN and ABC, it follows that: $x: c = \sqrt{\mathrm{P}_{\mathrm{AMP}}} : \sqrt{\mathrm{P}_{\mathrm{ABC}}}$

... (2) and $y: c = \sqrt{\mathrm{P}_{\mathrm{MBN}}} : \sqrt{\mathrm{P}_{\mathrm{ABC}}}$ ... (3) where $\mathrm{P}_{\text{AMP}} = 9 \mathrm{~cm}^{2}$, and $\mathrm{P}_{\text{MBN}} = 16 \mathrm{~cm}^{2}$. If we substitute and add the corresponding

![](https://cdn.mathpix.com/cropped/2024_06_05_e7266f2e82216ac5d245g-5.jpg?height=431&width=451&top_left_y=1455&top_left_x=1034)
sides of (2) and (3), we get: $\frac{\overline{\mathrm{AM}}}{\overline{\mathrm{AB}}} + \frac{\overline{\mathrm{MB}}}{\overline{\mathrm{AB}}} = \frac{3}{\sqrt{\mathrm{P}_{\mathrm{ABC}}}} + \frac{4}{\sqrt{\mathrm{P}_{\mathrm{ABC}}}}$, from which $\frac{3+4}{\sqrt{\mathrm{P}_{\mathrm{ABC}}}} = 1$, i.e., $\mathrm{P}_{\mathrm{ABC}} = 49 \mathrm{~cm}^{2}$.

If we substitute in (1), we get $\mathrm{P}_{\mathrm{PNC}} = \frac{1}{2} (49 - 16 - 9) = 12$, i.e., $\mathrm{P}_{\mathrm{PNC}} = 12 \mathrm{~cm}^{2}$.