CEMC Cayley, 2023

Solution 1

Since $AB$ and $ED$ are parallel, quadrilateral $ABDE$ is a trapezoid.

We know that $AB = 30$ cm.

Since $ABCF$ is a rectangle, then
$FC = AB = 30$ cm.

Suppose that $DC = x$ cm.

Then $ED = FC - FE - DC = (30\text{ cm}) - (5\text{ cm}) - (x\text{ cm}) = (25-x)\text{ cm}$.

The height of the trapezoid is the length of $AF$, which is 14 cm.

Since the area of the trapezoid is $266\text{ cm}^2$, then $$\begin{align*} 266\text{ cm}^2 & = \dfrac{30\text{ cm} + (25-x)\text{ cm}}{2} \times (14\text{ cm}) \\ 266 & = \dfrac{55-x}{2} \times 14 \\ 532 & = (55-x) \times 14 \\ 38 & = 55 - x \\ x & = 55 - 38\end{align*}$$ and so $DE = x\text{ cm} = 17\text{ cm}$.

Solution 2

Let $DC = x$ cm.

Rectangle $ABCF$ has $AB = 30\text{ cm}$ and $AF = 14\text{ cm}$, and so the area of
$ABCF$ is $(30\text{ cm}) \times (14\text{ cm}) = 420\text{ cm}^2$.

The area of $\triangle AFE$, which is
right-angled at $F$, is $$\tfrac{1}{2} \times AF \times FE = \tfrac{1}{2} \times (14\text{ cm}) \times (5\text{ cm}) = 35\text{ cm}^2$$ The
area of quadrilateral $ABDE$ is $266\text{ cm}^2$.

The area of $\triangle BCD$, which is
right-angled at $C$, is $$\tfrac{1}{2} \times BC \times DC = \tfrac{1}{2} \times (14\text{ cm}) \times (x\text{ cm}) = 7x\text{ cm}^2$$
Comparing the area of rectangle $ABCF$ to the combined areas of the pieces,
we obtain $$\begin{align*} (35\text{ cm}^2) + (266\text{ cm}^2) + (7x\text{ cm}^2) & = 420\text{ cm}^2 \\ 301 + 7x & = 420 \\ 7x & = 119 \\ x & = 17\end{align*}$$ Thus, the length of $DC$ is 17 cm.