Dividing a Square

Let $x$ , $y$ , and $z$ represent the distances shown in the diagram. Then $z = \sqrt{x^2 + y^2}$ . Since the area of the parallelogram is $1 \cdot z = z$ and it is also $x \cdot (x - y)$ , we get that $z = x^2 - xy$ . Since the area of the parallelogram is also $\frac{1}{3}$ of the area of the square, we get $z = \frac{1}{3} x^2$ . Since $x (x-y) = \frac{1}{3} x^2$ , we get $x - y = \frac{1}{3} x$ , or $y = \frac{2}{3} x$ . Plugging this into the equation $z = x^2 - xy$ yields $z = x^2/3$ , whereas plugging it into the equation $z = \sqrt{x^2+y^2}$ yields $z = \sqrt{13} x /3$ . Setting $x^2/3$ equal to $\sqrt{13} x / 3$ gives us $x = \sqrt{13}$ . Hence, the area of the square is 13 square units.