14. In the pentagon $A B C D E$, the angles at vertices $A, C, E$ are right angles. It is also known that $\overline{A B}=15 \mathrm{~m}, \overline{B C}=12 \mathrm{~m}, \overline{C D}=5 \mathrm{~m}, \overline{D E}=20 \mathrm{~m}$. By how many $\mathrm{m}^{2}$ is the area of the pentagon?
(A) 180
(B) 210
(C) 240
(D) 200
(E) 270
Problem 14. The answer is (C).

Drawing the perpendicular $B H$ from $B$ to the side $D E$, it can be easily observed that the right triangles $B C D$ and $B H D$ are congruent, so $B H$ measures $12 \, \text{m}$.

Alternatively, the same result can be obtained by applying the Pythagorean theorem first to determine the diagonal $B D$ and then for $B H$.

The area of the pentagon is thus given by the sum of the area of the rectangle $A B H E$ and the area of the quadrilateral $B C D H$ (which is twice the area of the right triangle $B C D$).

Therefore, the required area is given (in $\text{m}^2$) by $15 \cdot 12 + 5 \cdot 12 = 240$.

(Problem proposed by P. Francini)

![](https://cdn.mathpix.com/cropped/2024_04_17_2506d961b89bc3754c53g-3.jpg?height=661&width=418&top_left_y=1632&top_left_x=1409)