Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME Prove it Argentina

The pentagon $ABCDE$ , with sides $AB$ , $BC$ , $CD$ , $DE$ and $EA$ , satisfies the following conditions:
$\bullet \angle ABC = \angle BCD = \angle CDE = 90^\circ$
$\bullet$ $CD$ is longer than $AB$ .
$\bullet$ $AB = 28$ , $BC = 15$ , $DE = 10$ and $EA = 13$ .
Calculate the area of the pentagon.

Solution

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 $AP$ :
$AP = \sqrt{AE^2 - PE^2} = \sqrt{13^2 - 5^2} = \sqrt{144} = 12.$
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.

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