2007 AMC 12B Problems/Problem 25

Problem

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 $\triangle BDE$ ?

$\mathrm {(A)} \sqrt{2}\qquad \mathrm {(B)} \sqrt{3}\qquad \mathrm {(C)} 2\qquad \mathrm {(D)} \sqrt{5}\qquad \mathrm {(E)} \sqrt{6}$

Solution

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 $BDE$ . It is a $2-2-2\sqrt{2}$ triangle, which is an isosceles right triangle. Thus the area of it is $\frac{2*2}{2}=2\Rightarrow \mathrn{(C)}$ (Error compiling LaTeX. Unknown error_msg).

Solution 2

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2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Final Question
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2007 AMC 12B Problems/Problem 25

Contents

Problem

Solution

Solution 2

See also