Problem:
Points A,B,C,D, and E are located in 3-dimensional space with AB=BC=CD=DE=EA=2 and ∠ABC=∠CDE=∠DEA=90∘. The plane of △ABC is parallel to DE. What is the area of △BDE ?
Answer Choices:
A. 2​
B. 3​
C. 2
D. 5​
E. 6​ Solution:
Introduce a coordinate system in which D=(−1,0,0),E=(1,0,0), and △ABC lies in a plane z=k>0. Because ∠CDE and ∠DEA are right angles, A and C are located on circles of radius 2 centered at E and D in the planes x=1 and x=−1, respectively. Thus A=(1,y1​,k) and C=(−1,y2​,k), where yj​=±4−k2​ for j=1 and 2 . Because AC=22​, it follows that (1−(−1))2+(y1​−y2​)2=(22​)2. If y1​=y2​, there is no solution, so y1​=−y2​. It may be assumed without loss of generality that y1​>0, in which case y1​=1 and y2​=−1. It follows that k=3​, so A=(1,1,3​), C=(−1,−1,3​), and B is one of the points (1,−1,3​) or (−1,1,3​). In the first case, BE=2 and BE⊥DE. In the second case, BD=2 and BD⊥DE. In either case, the area of △BDE is (1/2)(2)(2)=2​.