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Problem of the Week

Problem D and Solution

Shady Square

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 \(STUV\)?

Square P Q R S has vertex P on side S T of rectangle S T U V, and vertex R on side S V. Vertex Q in the interior of rectangle S T U V.

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 \(SV\times UV=13\times 9=117\mbox{ m}^2\).