 # Problem of the Week Problem D and Solution The Whole Rectangle

## Problem

In the diagram, $$ABCD$$ is a rectangle. Points $$F$$ and $$G$$ are on $$DC$$ (with $$F$$ closer to $$D$$) such that $$DF=FG=GC$$. Point $$E$$ is the midpoint of $$AD$$. If the area of $$\triangle BEF$$ is $$30\text{ cm}^2$$, determine the area of rectangle $$ABCD$$.

## Solution

Let $$DF=FG=GC=y$$. Then $$AB=DC=3y$$ and $$FC=2y$$.

Since $$E$$ is the midpoint of $$AD$$, let $$AE=ED=x$$. Then $$AD=BC=2x$$. We will formulate an equation connecting the areas of the four triangles inside the rectangle to the area of the entire rectangle. \begin{aligned} \text{Area }ABCD&=\text{Area }\triangle ABE+\text{Area }\triangle BCF+\text{Area }\triangle FDE+\text{Area }\triangle BEF\\ AD\times DC&=\frac{AE\times AB}{2} +\frac{BC\times FC}{2}+\ \frac{DF\times ED}{2}+ 30\\ (2x)(3y)&=\frac{x\times 3y}{2} + \frac{2x\times 2y}{2}+\frac{y\times x}{2} + 30\\ 6xy &=\frac{3xy}{2}+2xy+\frac{xy}{2}+30\\ 12xy&=3xy+4xy+xy+60\\ 4xy&=60\\ xy&=15\end{aligned}

Therefore, the area of rectangle $$ABCD$$ is $$AD\times DC = (2x)(3y) = 6xy=6(15)=90\text{ cm}^2$$.