Problem of the Week

Problem D and Solution

Problem

Rectangle $$STUV$$ has square $$PQRS$$ removed, leaving an area of $$92\mbox{ m}^2$$.
Side $$PT$$ is $$4\mbox{ m}$$ in length and side $$RV$$ is $$8\mbox{ m}$$ in length.
What is the area of rectangle $$STUV$$?

Solution

Let $$x$$ represent the side length of square $$PQRS$$. In the diagram, extend $$RQ$$ to intersect $$TU$$ at $$W$$. This creates rectangle $$PTWQ$$ and rectangle $$RWUV$$. Then $$UV=PT+SP=(4+x)$$ m and $$TW=RS=x$$ m.

\begin{aligned} \mbox{Area }PTWQ+\mbox{Area }RWUV&=\mbox{Remaining Area}\\ PT\times TW+RV\times UV&=92\\ 4x+8(4+x)&=92\\ 4x+32+8x&=92\\ 12x+32&=92\\ 12x&=60\\ x&=5\mbox{ m}\end{aligned}

Since $$x=5$$ m, $$SV=8+x=13$$ m and $$UV=4+x=9$$ m.
Therefore, the original area of rectangle $$STUV$$ is $$SV\times UV=13\times 9=117\mbox{ m}^2$$.