In trapezoid $A B C D, A D$ is parallel to $B C$ and $B C: A D=5: 7$. Point $F$ lies on $A D$ and point $E$ lies on $D C$ so that $A F: F D=4: 3$ and $C E: E D=2: 3$. If the area of quadrilateral $A B E F$ is 123 , determine the area of trapezoid $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_04_30_790b6557ee0304df033eg-4.jpg?height=336&width=547&top_left_y=2144&top_left_x=846)
Since $B C: A D=5: 7$, then we let $B C=5 k$ and $A D=7 k$ for some real number $k>0$.

Since $A D=7 k$ and $F$ divides $A D$ in the ratio $4: 3$, then $A F=4 k$ and $F D=3 k$.

Drop a perpendicular from $D$ to $P$ on $B C$ extended, and let the height of trapezoid $A B C D$ be $5 h$, for some real number $h>0$.

Draw a line through $E$ parallel to $A D$ and $B C$ meeting $D P$ at $Q$. Note that $E Q$ is perpendicular to $D P$.

Now $\triangle D Q E$ is similar to $\triangle D P C$ since each is right-angled, and the two triangles share a common angle at $D$.

Since $C E: E D=2: 3$, then $D E: D C=3: 5$.

By similarity, $D Q: D P=3: 5$; since $D P=5 h$, then $D Q=3 h$.

This means that $Q P=D P-D Q=2 h$.

![](https://cdn.mathpix.com/cropped/2024_04_30_264fdb419065898dfdf2g-08.jpg?height=350&width=572&top_left_y=2270&top_left_x=836)

Trapezoid $A D B C$ has parallel sides $A D=7 k$ and $B C=5 k$, and height $5 h$; thus, its area is $\frac{1}{2}(7 k+5 k)(5 h)=30 h k$.

$\triangle F D E$ has base $F D=3 k$ and height equal to the distance from $F D$ to $E Q$, which is $3 h$; thus, its area is $\frac{1}{2}(3 k)(3 h)=\frac{9}{2} h k$.

$\triangle B C E$ has base $B C=5 k$ and height equal to the distance from $B C$ to $E Q$, which is $2 h$; thus, its area is $\frac{1}{2}(5 k)(2 h)=5 h k$.

We are given that the area of quadrilateral $A B E F$ is 123 .

Trapezoid $A B C D$ is made up of quadrilateral $A B E F, \triangle F D E$, and $\triangle B C E$.

Thus, the area of trapezoid $A B C D$ equals the sum of the areas of quadrilateral $A B E F, \triangle F D E$, and $\triangle B C E$.

This gives

$$
\begin{aligned}
30 h k & =123+\frac{9}{2} h k+5 h k \\
\frac{41}{2} h k & =123 \\
h k & =6
\end{aligned}
$$

Since the area of trapezoid $A B C D$ equals $30 h k$, then this area is $30(6)$ or 180 .

ANSWER: 180

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