I would let $\displaystyle x$ represent the length of the square.
Therefore the base and height of the triangle are $\displaystyle x + 1$.
Since the areas are the same...
$\displaystyle x^2 = \frac{1}{2}(x + 1)^2$
$\displaystyle x^2 = \frac{1}{2}x^2 + x + \frac{1}{2}$
$\displaystyle \frac{1}{2}x^2 - x - \frac{1}{2} = 0$
$\displaystyle x^2 - 2x - 1 = 0$
$\displaystyle x^2 - 2x + (-1)^2 - (-1)^2 - 1 = 0$
$\displaystyle (x - 1)^2 - 2 = 0$
$\displaystyle (x - 1)^2 = 2$
$\displaystyle x - 1 = \pm \sqrt{2}$
$\displaystyle x = 1 \pm \sqrt{2}$.
Since you obviously can not have a negative length, the length of the square is
$\displaystyle 1 + \sqrt{2}$ cm.
Thus the area is
$\displaystyle A= (1 + \sqrt{2})^2$
$\displaystyle = 1 + 2\sqrt{2} + 2$
$\displaystyle = 3 + 2\sqrt{2}\,\textrm{cm}^2$.
Hello, Kevin!
Can somone help me solve this equation?
I've already tried with : .$\displaystyle (x-1)^2 \:=\:\frac{x^2}{2}$ . . . but I failed.
Your equation is correct . . . It must be your algebra.
We have: .$\displaystyle (x-1)^2 \:=\:\frac{x^2}{2}$
Multiply by 2: .$\displaystyle 2(x-1)^2 \:=\:x^2 \quad\Rightarrow\quad 2x^2 - 4x + 2 \:=\:x^2 \quad\Rightarrow\quad x^2 - 4x + 2 \:=\:0$
Quadratic Formula: .$\displaystyle x \;=\;\frac{\text{-}(\text{-}4) \pm \sqrt{(\text{-}4)^2 - 4(1)(2)}}{2(1)} \;=\;\frac{4 \pm\sqrt{8}}{2} \:=\:\frac{4 \pm2\sqrt{2}}{2} \:=\:2 \pm \sqrt{2}$
The side of the triangle is: .$\displaystyle 2 + \sqrt{2}$ cm.
The side of the square is: .$\displaystyle 1 + \sqrt{2}$ cm.
The area of the square is: .$\displaystyle (1 + \sqrt{2})^2 \;=\;1 + 2\sqrt{2} + 2 \;=\;\boxed{3 + 2\sqrt{2}\,\text{ cm}^2}$