Continuation of the previous question

Let $P =$ perimeter of a square, $A =$ area of a square and $s =$ side

$P = 4s$

$A = s^2$

If $P = A$ , then $s^2 = 4s$

$s^2 - 4s = 0$

$s(s - 4) = 0$

$Hence$ $s = 0, s = 4$ .

As $s = 4$ is a solution of a square ( $s = 0$ can be ignored), and a square is a shape, regardless of all other shapes the answer must be $TRUE$

For every polygon, there is a possibility that area is same as perimeter.

Take for example, a square with length 4, area = perimeter = 16 .

Take a equilateral triangle with length $a = \dfrac{\sqrt[]{3}}{12}$ , area = perimeter = $0.577\ldots$

consider a square with side length 4. 4X4=16. 4+4+4+4=16. Therefore p-a=0

Actually if you consider the unit of measurement, and ignore all line, then the answer must be false.

A circle with radius = 2 also works