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