14. In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$, and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. Find the area of pentagon $A B C D E$ if it is known that quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas param1 and param2, respectively, and the area of triangle $B C P$ is param3.

| param1 | param2 | param3 | Answer |
| :---: | :---: | :---: | :---: |
| 5 | 8 | 10 | 25 |
| 6 | 10 | 3 | 29 |
| 7 | 9 | 7 | 27.5 |
| 8 | 11 | 20 | 41.2 |
| 9 | 12 | 4 | 38.5 |

## Condition

In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$, and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. Find the area of pentagon $A B C D E$ if it is known that quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas 5 and 8, respectively, and the area of triangle $B C P$ is 10.

## Answer.

25
# Solution

Let the areas of parallelograms ABP M and DCPN, triangle BCP, and quadrilateral

![](https://cdn.mathpix.com/cropped/2024_05_06_a839f63dd79b7140018dg-12.jpg?height=63&width=1559&top_left_y=1279&top_left_x=288)
$A E \sqcup B N N C D$. Similarly, $D E \sqcup C M \sqcup\{B$. Then M PNE is also a parallelogram, and thus $S_{4}: S_{1}=N P: P B$ (parallelograms ABPM and M PNE have equal heights). Similarly, triangle $B C P$ and parallelogram $D C P N$ have equal heights, so $S_{3}: S_{2}=$ $\frac{1}{2} B P: P N$. Multiplying the written equalities, we get: $\frac{S_{4}}{S_{1}} \cdot \frac{S_{3}}{S_{2}}=\frac{1}{2} \cdot$ Therefore, $S=S_{1}+S_{2}+$ $S_{3}+S_{4}=S_{1}+S_{2}+S_{3}+\frac{S_{1} S_{2}}{2 S_{3}}=25$