In the diagram, $A B C D$ is a rectangle with $A D=13, D E=5$ and $E A=12$. The area of $A B C D$ is
(A) 39
(D) 30
(B) 60
(E) 25
(C) 52

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-139.jpg?height=266&width=401&top_left_y=2198&top_left_x=1317)
In the diagram, $A B C D$ is a rectangle with $A D=13, D E=5$ and $E A=12$. The area of $A B C D$ is
(A) 39
(B) 60
(C) 52
(D) 30
(E) 25

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-258.jpg?height=312&width=458&top_left_y=1476&top_left_x=1386)

## Solution

Since $13^{2}=12^{2}+5^{2}$ we use the converse of Pythagorus' Theorem to conclude that $\angle A E D=90^{\circ}$.

The area of $\triangle A E D$ is then $\frac{1}{2}(5)(12)=30$. Through $E$, we draw a line parallel to $C D$ and $B A$. Since the area of $\triangle F D E$ equals the area of $\triangle C D E$ we label each of these areas $A$. Similarly, the area of $\triangle A F E$ equals the area of $\triangle B A E$ and so each of these areas can be labelled $B$. Since $A+B=30$, the area of the rectangle is $2(A+B)$ or $2(30)=60$.

ANSWER: (B)