1. ## Calculate the area

Can somone help me solve this equation. I've already tried with
(x-1)²=x²/2, but I failed.

2. Originally Posted by kevin3000
Can somone help me solve this equation. I've already tried with
(x-1)²=x²/2, but I failed.

I would let $x$ represent the length of the square.

Therefore the base and height of the triangle are $x + 1$.

Since the areas are the same...

$x^2 = \frac{1}{2}(x + 1)^2$

$x^2 = \frac{1}{2}x^2 + x + \frac{1}{2}$

$\frac{1}{2}x^2 - x - \frac{1}{2} = 0$

$x^2 - 2x - 1 = 0$

$x^2 - 2x + (-1)^2 - (-1)^2 - 1 = 0$

$(x - 1)^2 - 2 = 0$

$(x - 1)^2 = 2$

$x - 1 = \pm \sqrt{2}$

$x = 1 \pm \sqrt{2}$.

Since you obviously can not have a negative length, the length of the square is

$1 + \sqrt{2}$ cm.

Thus the area is

$A= (1 + \sqrt{2})^2$

$= 1 + 2\sqrt{2} + 2$

$= 3 + 2\sqrt{2}\,\textrm{cm}^2$.

3. Hello, Kevin!

Can somone help me solve this equation?
I've already tried with : . $(x-1)^2 \:=\:\frac{x^2}{2}$ . . . but I failed.

Your equation is correct . . . It must be your algebra.

We have: . $(x-1)^2 \:=\:\frac{x^2}{2}$

Multiply by 2: . $2(x-1)^2 \:=\:x^2 \quad\Rightarrow\quad 2x^2 - 4x + 2 \:=\:x^2 \quad\Rightarrow\quad x^2 - 4x + 2 \:=\:0$

Quadratic Formula: . $x \;=\;\frac{\text{-}(\text{-}4) \pm \sqrt{(\text{-}4)^2 - 4(1)(2)}}{2(1)} \;=\;\frac{4 \pm\sqrt{8}}{2} \:=\:\frac{4 \pm2\sqrt{2}}{2} \:=\:2 \pm \sqrt{2}$

The side of the triangle is: . $2 + \sqrt{2}$ cm.

The side of the square is: . $1 + \sqrt{2}$ cm.

The area of the square is: . $(1 + \sqrt{2})^2 \;=\;1 + 2\sqrt{2} + 2 \;=\;\boxed{3 + 2\sqrt{2}\,\text{ cm}^2}$

4. Ok, thanks guys.
I understand better now