Difference between revisions of "2007 AMC 12B Problems/Problem 7"
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| − | Since <math>A</math> and <math>B</math> are [[right angle]]s, and <math>AE</math> equals <math>BC</math>, <math>AECB</math> is a [[square]], and <math>EC</math> is 5. Since <math>ED</math> and <math>CD</math> are also 5, triangle <math>CDE</math> is [[equilateral triangle|equilateral]]. Angle <math>E</math> is therefore <math>90+60= | + | Since <math>A</math> and <math>B</math> are [[right angle]]s, and <math>AE</math> equals <math>BC</math>, <math>AECB</math> is a [[square]], and <math>EC</math> is 5. Since <math>ED</math> and <math>CD</math> are also 5, triangle <math>CDE</math> is [[equilateral triangle|equilateral]]. Angle <math>E</math> is therefore <math>90+60=150 \Rightarrow \mathrm {(E)}</math> |
==See Also== | ==See Also== | ||
Revision as of 21:03, 5 December 2018
Problem
All sides of the convex pentagon 
are of equal length, and 
. What is the degree measure of 
?
Solution
Since 
and 
are right angles, and 
equals 
, 
is a square, and 
is 5. Since 
and 
are also 5, triangle 
is equilateral. Angle 
is therefore 
See Also
| 2007 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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. 
