Let
b------> the length side of a square
we know that
the area of a square is equal to
\(A=b^{2}\)
Find the length side of the diagonal applying the Pythagoras Theorem
\(d^{2}=b^{2}+b^{2}\)
\(d^{2}=2b^{2}\)
\(d=b\sqrt{2}\ units\)
Remember that
\(d=x\ units\) -----> given problem
substitute
\(d=b\sqrt{2}\ units\)
\(x=b\sqrt{2}\ units\)
Solve for b
\(b=\frac{x\sqrt{2}}{2}\ units\)
Substitute in the formula of area
\(A=(\frac{x\sqrt{2}}{2})^{2}\)
\(A=\frac{x^{2}}{2}\ units^{2}\)
therefore
the answer is
\(\frac{x^{2}}{2}\ units^{2}\)
