A [cube](https://artofproblemsolving.com/wiki/index.php/Cube) with side length $1$ is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the [midpoints](https://artofproblemsolving.com/wiki/index.php/Midpoint) $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of [quadrilateral](https://artofproblemsolving.com/wiki/index.php/Quadrilateral) $ABCD$?

$\mathrm{(A)}\ \frac{\sqrt{6}}{2}\qquad\mathrm{(B)}\ \frac{5}{4}\qquad\mathrm{(C)}\ \sqrt{2}\qquad\mathrm{(D)}\ \frac{5}{8}\qquad\mathrm{(E)}\ \frac{3}{4}$
Since $AB = AD = CB = CD = \sqrt{\left(\frac{1}{2}\right)^2+1^2}$, it follows that $ABCD$ is a [rhombus](https://artofproblemsolving.com/wiki/index.php/Rhombus). The area of the rhombus can be computed by the formula $A = \frac 12 d_1d_2$, where $d_1,\,d_2$ are the diagonals of the rhombus (or of a [kite](https://artofproblemsolving.com/wiki/index.php/Kite) in general). $BD$ has the same length as a face diagonal, or $\sqrt{1^2 + 1^2} = \sqrt{2}$. $AC$ is a space diagonal, with length $\sqrt{1^2+1^2+1^2} = \sqrt{3}$. Thus $A = \frac 12 \times \sqrt{2} \times \sqrt{3} = \frac{\sqrt{6}}{2}\ \mathrm{(A)}$.