Difference between revisions of "2007 AMC 12B Problems/Problem 25"
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==Problem== | ==Problem== | ||
| − | Points <math>A,B,C,D</math> and <math>E</math> are located in 3-dimensional space with <math>AB= | + | Points <math>A,B,C,D</math> and <math>E</math> are located in 3-dimensional space with <math>AB=BC=CD=DE=EA=2</math> and <math>\angle ABC=\angle CDE=\angle DEA=90^o</math>. The plane of <math>\triangle ABC</math> is parallel to <math>\overline{DE}</math>. What is the area of <math>\triangle BDE</math>? |
<math>\mathrm {(A)} \sqrt{2}\qquad \mathrm {(B)} \sqrt{3}\qquad \mathrm {(C)} 2\qquad \mathrm {(D)} \sqrt{5}\qquad \mathrm {(E)} \sqrt{6}</math> | <math>\mathrm {(A)} \sqrt{2}\qquad \mathrm {(B)} \sqrt{3}\qquad \mathrm {(C)} 2\qquad \mathrm {(D)} \sqrt{5}\qquad \mathrm {(E)} \sqrt{6}</math> | ||
Revision as of 19:31, 17 December 2013
Problem
Points 
and 
are located in 3-dimensional space with 
and 
. The plane of 
is parallel to 
. What is the area of 
?
Solution
Let 
, and 
. Since 
, we could let 
, 
, and 
. Now to get back to 
we need another vertex 
. 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 
. Now we can bend these three sides into an equilateral triangle, and the coordinates change: , , , 
, and 
. Checking for all the requirements, they are all satisfied. Now we find the area of triangle 
. It is a 
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).
See also
| 2007 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 24 |
Followed by Final Question |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
