Rectangle \(STUV\) has \(P\) on \(ST\), \(R\) on \(SV\), and \(Q\) inside the rectangle such that \(PQRS\) is a square. When square \(PQRS\) is removed from rectangle \(STUV\), the remaining shape has an area of \(92\text{ m}^2\).
If \(PT = 4\text{ m}\) and \(RV=8\text{ m}\), what is the area of rectangle \(STUV\)?
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 \(RWUV\). Also, \(UV=PT+SP=(4+x)\text{ m}\) and \(TW=RS=x\text{ m}\).
We are given that \(\text{area } PTWQ + \text{area } RWUV = 92\text{ m}^2\). That is, \[\begin{aligned} PT\times TW+RV\times UV&=92\\ 4x+8(4+x)&=92\\ 4x+32+8x&=92\\ 12x+32&=92\\ 12x&=60\\ x&=5 \end{aligned}\] Thus, \(x=5\text{ m}\), and so \(SV=8+x=13\text{ m}\), and \(UV=4+x=9\text{ m}\).
Therefore, the area of rectangle \(STUV\) is \(SV\times UV=13\times 9=117\text{ m}^2\).