In trapezoid $A B C D, A D$ is parallel to $B C$. Also, $B D$ is perpendicular to $D C$. The point $F$ is chosen on line $B D$ so that $A F$ is perpendicular to $B D$. $A F$ is extended to meet $B C$ at point $E$. If $A B=41, A D=50$ and $B F=9$, what is the area of quadrilateral $F E C D$ ?
(A) 900
(D) 1560
(B) 1523.5
(C) 960

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-130.jpg?height=287&width=477&top_left_y=2120&top_left_x=1301)

continued ...
By Pythagoras in $\triangle A F B$,

$$
\begin{aligned}
A F^{2}+9^{2} & =41^{2} \\
A F & =40
\end{aligned}
$$

By Pythagoras in $\triangle A F D$,

$$
\begin{aligned}
F D^{2}+40^{2} & =50^{2} \\
F D & =30
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-235.jpg?height=345&width=572&top_left_y=1367&top_left_x=1237)

Since $A D$ is parallel to $B C, \angle F A D=\angle F E B$. Therefore, $\triangle A F D$ is similar to $\triangle E F B$, so $\frac{A F}{F D}=\frac{E F}{F B}$ or $\frac{40}{30}=\frac{E F}{9}$ or $E F=12$.

Since $A E$ is perpendicular to $B D$ and $D C$ is perpendicular to $B D$, then $A E$ is parallel to $D C$, so $A E C D$ is a parallelogram. Thus, $D C=A E=52$.

Now consider quadrilateral $F E C D$.

This quadrilateral is a trapezoid, so

Area of $F E C D=\frac{1}{2}(E F+C D)(F D)$

$$
\begin{aligned}
& =\frac{1}{2}(12+52)(30) \\
& =960
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-235.jpg?height=336&width=425&top_left_y=1962&top_left_x=1170)

ANSWER: (C)