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The area of a square is equal to twice the square's perimeter. Find the length of one side of the square.

Don't know how to do this or even where to start to solve this problem.

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Hint: If $s$ is the side-length of the square, the area is $s^2$ and the perimeter is $4s$.

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A: area

P: perimeter

x: length of side

$$A = 2\times P$$

$$\underbrace{x^2}_{A} = 2(\underbrace{4x}_{P}) = 8x$$

Now you've got a quadratic equation you can easily solve for side length (requiring $x >0$).

$$x^2 = 8x \iff x^2 - 8x = 0 \iff x(x - 8) = 0 \implies x = 0\;\;\text{or}\;\;\bf x = 8$$

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Let a be the side of a square Now Area $A = a^2$

and Perimeter $P = 4a$

we know $A=2P$

Putting the values of A and P respectively we get,

$$\begin{align*} a^2 & {} = 2*4a \\ a^2-8a & {} = 0 \\ a(a-8) & {} = 0 \\ \end{align*}$$

$a=8$ as a cannot be equal to zero (which is the other root)

Hence side is 8 units.

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Senseless question: Since the dimension of the area is some_unit$^2$, the perimeter's dimension is some_unit.


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