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 \(ABCD\).
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 \(AD\times DC = (2x)(3y) = 6xy=6(15)=90\text{ cm}^2\).