In the diagram, rectangle $A B C D$ is divided into two regions, $A E F C D$ and $E B C F$, of equal area. If $E B=40$, $A D=80$ and $E F=30$, what is the length of $A E$ ?
(A) 20
(B) 24
(C) 10
(D) 15
(E) 30

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-100.jpg?height=393&width=350&top_left_y=926&top_left_x=1343)

## Part C: Each correct answer is worth 8.
Draw a line from $F$ to $B C$, parallel to $A B$, meeting $B C$ at $P$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-167.jpg?height=401&width=339&top_left_y=266&top_left_x=947)

Since $E B$ is parallel to $F P$ and $\angle F E B=90^{\circ}$, then $E B P F$ is a rectangle.

Since $E B=40$, then $F P=40$; since $E F=30$, then $B P=30$.

Since $A D=80$, then $B C=80$, so $P C=80-30=50$.

Therefore, the area of $E B C F$ is sum of the areas of rectangle $E B P F$ (which is $30 \times 40=1200$ ) and $\triangle F P C$ (which is $\left.\frac{1}{2}(40)(50)=1000\right)$, or $1200+1000=2200$.

Since the areas of $A E F C D$ and $E B C F$ are equal, then each is 2200 , so the total area of rectangle $A B C D$ is 4400 .

Since $A D=80$, then $A B=4400 \div 80=55$.

Therefore, $A E=A B-E B=55-40=15$.

ANSWER: (D)