XXVII Olimpiada Matemática Rioplatense

Extend $BA$ and $DE$ so that they meet at point $P$. $AB=28$, $BC=15$, $DE=10$ and $EA=13$. The quadrilateral $PBCD$ has three right angles, so the fourth is also a right angle, hence it is a rectangle. Then $PD=BC=15$, therefore $PE=PD-DE=15-10=5$. Now we apply Pythagoras' theorem in the right-angled triangle $APE$ to determine the length of $AP$: $$AP=\sqrt{A{E}^2-P{E}^2}=\sqrt{13^2-5^2}=\sqrt{144}=12.$$ We conclude that $PB=PA+AB=12+28=40$. To calculate the area of the pentagon $ABCDE$, we subtract the area of the triangle $APE$ from the area of the rectangle $PBCD$, that is: $$ \text{area}(ABCDE) = 40 \cdot 15 - \frac{12 \cdot 5}{2} = 600 - 30 = 570.