Difference between revisions of "2007 AMC 12B Problems/Problem 7"
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| − | All sides of the convex pentagon ABCDE are of equal length, and <A = | + | ==Problem== |
| + | 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>? | ||
| − | A) 90 B) 108 C) 120 D) 144 E) 150 | + | <math>\mathrm {(A)}\ 90 \qquad \mathrm {(B)}\ 108 \qquad \mathrm {(C)}\ 120 \qquad \mathrm {(D)}\ 144 \qquad \mathrm {(E)}\ 150</math> |
| + | ==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>, 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== | ||
| + | {{AMC12 box|year=2007|ab=B|num-b=6|num-a=8}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:41, 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. 
