What is the perimeter of a square with a diagonal of the square root of 72

Diagonal, d = √72
Side Length = s
Perimeter = P

s² + s² = d²
2s² = (√72)²
2s² = 72
s² = 72 / 2
s² = 36
s = √36
s = 6

P = 4s
P = 4(6)
P = 24

Perimeter = 24
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consider a square ABCD.its diagonal say AC=sqrt72. now consider the triangle ABC,its a right angled triangle,
so AB^2+BC^2=AC^2
as we know that a square has each side equal say "a"
so using above equation v have
2a^2=AC^2
a=AC/sqrt2
and perimeter of a square is 4*a
hence the answer is 4*AC/sqrt2=4*3*2*sqrt2/sqt2=24units