In the diagram, $A B C D E F G$ is a room having square corners, with $E F=20 \mathrm{~m}, A B=10 \mathrm{~m}$, and $A G=G F$. The total area of the room is $280 \mathrm{~m}^{2}$. A wall is built from $A$ to $D$ creating two rooms of equal area. What is the distance, in metres, from $C$ to $D$ ?
(A) 15
(B) $\frac{50}{3}$
(C) 12
(D) 13
(E) $\frac{40}{3}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-119.jpg?height=510&width=495&top_left_y=636&top_left_x=1358)
First, we determine the lengths of the sides of the rooms.

Suppose that $A G=x$. Then $F G=x$.

So the room can be thought of as a rectangular room of width $F E=20$ and length $A B+F G=10+x$, with a rectangular corner of dimensions $A G=x$ by $A B=10$ removed.

Equating the area of the entire room with this way of visualizing it,

$$
\begin{aligned}
20(10+x)-10 x & =280 \\
10 x+200 & =280 \\
x & =8
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-217.jpg?height=512&width=499&top_left_y=888&top_left_x=1366)

Therefore, the lengths of the sides of the room are (starting from $B$ and proceeding clockwise) $10,8,8,20,18$ and 12 .

Let $y$ now be the distance from $C$ to $D$, the point where the new wall touches $C E$. Now $C B A D$ can be viewed as a trapezoid with base $C D$ and parallel side $A B$ (since the room has square corners).

Also, the height of the trapezoid is $B C=12$. The area of this trapezoid is supposed to be 140 , or half of the total area of the large room.

Therefore, since the area of the trapezoid is

$\frac{1}{2}(B C)(A B+C D)$, we have

$\frac{1}{2}(12)(10+y)=140$

$$
\begin{aligned}
6(10+y) & =140 \\
10+y & =\frac{70}{3} \\
y & =\frac{40}{3}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-217.jpg?height=515&width=575&top_left_y=1775&top_left_x=1317)

Also, the height of the trapezoid is $B C=12$. The area of this trapezoid is supposed to be 140 , or half of the total area of the large room.

Therefore, since the area of the trapezoid is $\frac{1}{2}(B C)(A B+C D)$, we have

Thus, the distance from $C$ to $D$ is $\frac{40}{3}$.

Answer: (E)