In the diagram, $A B C F$ is a rectangle with $A B=30 \mathrm{~cm}$ and $A F=14 \mathrm{~cm}$. Points $E$ and $D$ are on $F C$ so that $F E=5 \mathrm{~cm}$ and the area of quadrilateral $A B D E$ is $266 \mathrm{~cm}^{2}$. The length of $D C$ is
(A) $17 \mathrm{~cm}$
(B) $16 \mathrm{~cm}$
(C) $19 \mathrm{~cm}$
(D) $20 \mathrm{~cm}$
(E) $18 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-004.jpg?height=298&width=452&top_left_y=602&top_left_x=1300)
Solution 1

Since $A B$ and $E D$ are parallel, quadrilateral $A B D E$ is a trapezoid.

We know that $A B=30 \mathrm{~cm}$.

Since $A B C F$ is a rectangle, then $F C=A B=30 \mathrm{~cm}$.

Suppose that $D C=x \mathrm{~cm}$.

Then $E D=F C-F E-D C=(30 \mathrm{~cm})-(5 \mathrm{~cm})-(x \mathrm{~cm})=(25-x) \mathrm{cm}$.

The height of the trapezoid is the length of $A F$, which is $14 \mathrm{~cm}$.

Since the area of the trapezoid is $266 \mathrm{~cm}^{2}$, then

$$
\begin{aligned}
266 \mathrm{~cm}^{2} & =\frac{30 \mathrm{~cm}+(25-x) \mathrm{cm}}{2} \times(14 \mathrm{~cm}) \\
266 & =\frac{55-x}{2} \times 14 \\
532 & =(55-x) \times 14 \\
38 & =55-x \\
x & =55-38
\end{aligned}
$$

and so $D E=x \mathrm{~cm}=17 \mathrm{~cm}$.

## Solution 2

Let $D C=x \mathrm{~cm}$.

Rectangle $A B C F$ has $A B=30 \mathrm{~cm}$ and $A F=14 \mathrm{~cm}$, and so the area of $A B C F$ is $(30 \mathrm{~cm}) \times(14 \mathrm{~cm})=420 \mathrm{~cm}^{2}$.

The area of $\triangle A F E$, which is right-angled at $F$, is

$$
\frac{1}{2} \times A F \times F E=\frac{1}{2} \times(14 \mathrm{~cm}) \times(5 \mathrm{~cm})=35 \mathrm{~cm}^{2}
$$

The area of quadrilateral $A B D E$ is $266 \mathrm{~cm}^{2}$.

The area of $\triangle B C D$, which is right-angled at $C$, is

$$
\frac{1}{2} \times B C \times D C=\frac{1}{2} \times(14 \mathrm{~cm}) \times(x \mathrm{~cm})=7 x \mathrm{~cm}^{2}
$$

Comparing the area of rectangle $A B C F$ to the combined areas of the pieces, we obtain

$$
\begin{aligned}
\left(35 \mathrm{~cm}^{2}\right)+\left(266 \mathrm{~cm}^{2}\right)+\left(7 x \mathrm{~cm}^{2}\right) & =420 \mathrm{~cm}^{2} \\
301+7 x & =420 \\
7 x & =119 \\
x & =17
\end{aligned}
$$

Thus, the length of $D C$ is $17 \mathrm{~cm}$.

ANSWER: (A)