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>, and <math>AECB</math> is a [[square]]. 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> | + | Since <math>A</math> and <math>B</math> are [[right angle]]s, and <math>AE</math> equals <math>BC</math>, and <math>AECB</math> is a [[square]]. Since the length of <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== |
Latest revision as of 13:41, 22 August 2024
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
, and
is a square. Since the length of
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.