Difference between revisions of "2007 AMC 12B Problems/Problem 7"
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==Problem== | ==Problem== | ||
| − | All sides of the convex pentagon <math>ABCDE</math> are of equal length, and <math>\angle A = \angle B = 90^{\circ}</math>. What is the degree measure of <math>\angle E</math>? | + | All sides of the [[convex polygon|convex]] [[pentagon]] <math>ABCDE</math> are of equal length, and <math>\angle A = \angle B = 90^{\circ}</math>. What is the degree measure of <math>\angle E</math>? |
| − | <math>\mathrm {(A)} 90 | + | <math>\mathrm {(A)}\ 90 \qquad \mathrm {(B)}\ 108 \qquad \mathrm {(C)}\ 120 \qquad \mathrm {(D)}\ 144 \qquad \mathrm {(E)}\ 150</math> |
==Solution== | ==Solution== | ||
| + | [[Image:2007_12B_AMC-7.png]] | ||
| − | + | 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> | |
| − | |||
| − | Since <math>A</math> and <math>B</math> are right | ||
| − | |||
==See Also== | ==See Also== | ||
| + | {{AMC12 box|year=2007|ab=B|num-b=6|num-a=8}} | ||
| − | + | [[Category:Introductory Geometry Problems]] | |
Revision as of 17:44, 17 October 2007
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 | |
