, 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$$\text{266 cm 2 = 30 cm + (25-x) cm 2 (14 cm ) 266 = 55-x 2 14 532 = (55-x) 14 38 = 55 - x x = 55 - 38}$$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.