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.