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Problem of the Week
Problem D and Solution
The Whole Rectangle


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 \(AD=BC=2x\).

The sides of triangle BEF divide rectangle ABCD into four triangular regions, one of which is triangle BEF. Triangle BEF has an area of 30 square centimetres.

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