# Problem of the Week Problem D and Solution Square Parts

## Problem

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 $$RSTU$$.

## Solution

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 $$RSTU = 9(9-x)=9(9-6) =27\ \mbox{cm}^2$$.