Given a square side of length $s$. On a diagonal as base a triangle with three unequal sides is constructed so that its area
equals that of the square. The length of the altitude drawn to the base is:
$\textbf{(A)}\ s\sqrt{2} \qquad \textbf{(B)}\ s/\sqrt{2} \qquad \textbf{(C)}\ 2s \qquad \textbf{(D)}\ 2\sqrt{s} \qquad \textbf{(E)}\ 2/\sqrt{s}$
The area of the square is $s^2$.  The diagonal of a square with side $s$ bisects the square into two $45-45-90$ right triangles, so the diagonal has length $s\sqrt{2}$.
The area of the triangle is $\frac{1}{2}bh$.  The base $b$ of the triangle is the diagonal of the square, which is $b = s\sqrt{2}$.  If the area of the triangle is equal to the area of the square, we have:
$s^2 = \frac{1}{2}bh$
$s^2 = \frac{1}{2}s\sqrt{2}\cdot h$
$s = \frac{\sqrt{2}}{2}h$
$h = \frac{2}{\sqrt{2}}s$
$h = s\sqrt{2}$
This is option $\boxed{\textbf{(A)}}$