Points $A,B,C,D$ and $E$ are located in 3-dimensional space with $AB=BC=CD=DE=EA=2$ and $\angle ABC=\angle CDE=\angle DEA=90^o$. The plane of $\triangle ABC$ is parallel to $\overline{DE}$. What is the area of $\triangle BDE$?
$\mathrm {(A)} \sqrt{2}\qquad \mathrm {(B)} \sqrt{3}\qquad \mathrm {(C)} 2\qquad \mathrm {(D)} \sqrt{5}\qquad \mathrm {(E)} \sqrt{6}$
[2007 AMC 12B Problem 25.png](https://artofproblemsolving.com/wiki/index.php/File:2007_AMC_12B_Problem_25.png)
Link to graph: [https://www.math3d.org/pHFSD6vRi](https://artofproblemsolving.comhttps://www.math3d.org/pHFSD6vRi)

Let $A=(0,0,0)$, and $B=(2,0,0)$. Since $EA=2$, we could let $C=(2,0,2)$, $D=(2,2,2)$, and $E=(2,2,0)$. Now to get back to $A$ we need another vertex $F=(0,2,0)$. Now if we look at this configuration as if it was two dimensions, we would see a square missing a side if we don't draw $FA$. Now we can bend these three sides into an equilateral triangle, and the coordinates change: $A=(0,0,0)$, $B=(2,0,0)$, $C=(2,0,2)$, $D=(1,\sqrt{3},2)$, and $E=(1,\sqrt{3},0)$. Checking for all the requirements, they are all satisfied. Now we find the area of triangle $BDE$. The side lengths of this triangle are $2, 2, 2\sqrt{2}$, which is an isosceles right triangle. Thus the area of it is $\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}$.