The area of a square is 12 more than its perimeter.
Let the side of the square be x.
Then area of the square =[tex] \\
\
x*x=x^2\\ [/tex]
and the perimeter=4x
As per the problem
Area-Perimeter=12
[tex] \\
\
x^2-4x=12\\
\\
\Rightarrow x^2-4x-12=0\\
\\
\Rightarrow x^2-6x+2x-12=0\\
\\
\Rightarrow x(x-6)+2(x-6)=0\\
\\
\Rightarrow (x-6)(x+2)=0\\
\\
\Rightarrow x=6,-2 [/tex]
x can not be negative. Hence x=6.
side length=x=6
Perimeter=4x=24
