2. If the sum of the perimeters of two squares is 8 and the difference of their areas is 3, determine the lengths of the sides of both squares.
2. If the sum of the perimeters of two squares is 8 and the difference of their areas is 3, determine the lengths of the sides of both squares.

## Solution.

Let the length of the side of the larger square be $x$, and the length of the side of the smaller square be $y$.

Then, according to the conditions of the problem,

$4 x + 4 y = 8$

1 POINT

and $x^{2} - y^{2} = 3$.

1 POINT

From the first equation, we conclude that $x + y = 2$.

Since $x^{2} - y^{2} = (x + y)(x - y)$, we get $3 = 2(x - y)$, i.e., $x - y = \frac{3}{2}$

1 POINT

From (1) and (2), we then have

$x = \frac{1}{2}(x + y + x - y) = \frac{1}{2}\left(2 + \frac{3}{2}\right) = \frac{7}{4}$

2 POINTS

(evaluate with 1 POINT in case of a calculation error)

and $y = 2 - x = 2 - \frac{7}{4} = \frac{1}{4}$.

1 POINT

TOTAL 6 POINTS