Question

Square ABCD has a diagonal of 8 inches long. How many square inches is the area of the shape?

Answer 1

The diagonal of any square always occurs in a ratio of $\sqrt2:1$ relative to the square's sides. That means if $s$ is the side length, then $\sqrt2s=8$ gives the length of the diagonal. It follows that $s=\dfrac8{\sqrt2}$.

Then the area of the square is

$s^2=\left(\dfrac8{\sqrt2}\right)^2=\dfrac{64}2=32$ square inches

Answer 2

If you look carefully, that diagonal would be the Hypotenuse and the arms of the Square (which is equal in magnitude) would be the shorter sides.

Then, According to Pythagoras Theorem, 
a² + b² = c²
2a² = 8²
a² = 64/2
a = √32

Now, Area = Side²

So, A = (√32)²
A = 32 in²

In short, Your Answer would be; 32 in²

Hope this helps!