Square \(PQRS\) has \(W\) on \(PQ\), \(U\) on \(QR\), \(T\) on \(PS\), and \(V\) on \(TU\) such that \(QUVW\) is a square, and \(PWVT\) and \(RSTU\) are rectangles.
The side length of square \(PQRS\) is \(9\) cm, and \[\text{area of } QUVW - \text{ area of } RSTU = \text{ area of } RSTU - \text{ area of }PWVT\] If square \(QUVW\) has side length equal to \(x\) cm, determine the value of \(x\) and the areas of rectangles \(PWVT\) and \(RSTU\).
We know \(SR=PQ= 9\) cm and \(WQ=QU = x\) cm.
Therefore, \(PW = PQ -WQ = (9-x)\) cm.
Similarly, \(UR = (9-x)\) cm.
Thus, we have that the area of \(QUVW\) is equal to \(x^2\) cm\(^2\), the area of \(RSTU\) is equal to \(9(9-x)\) cm\(^2\), and the area of \(PWVT\) is equal to \(x(9-x)\) cm\(^2\).
Therefore, we know that \[\begin{aligned} \text{area of } QUVW - \text{ area of } RSTU &= \text{ area of } RSTU - \text{ area of }PWVT\\ x^2 - 9(9-x)&=9(9-x) - x(9-x)\\ x^2 - 81 + 9x &=81 - 9x -9x + x^2\\ 27x&=162\\ x&=6 \end{aligned}\]
Therefore, \(x = 6\) cm, the area of \(PWVT\) is equal to \(x(9-x) = 6(9 -6) = 18\) cm\(^2\), and the area of \(RSTU = 9(9-x)=9(9-6) =27\ \mbox{cm}^2\).