Difference between revisions of "2007 AMC 10B Problems/Problem 7"
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| − | <math>AB = EC</math> because they are opposite sides of a square. Also, <math>ED = DC = AB</math> because all sides of the convex pentagon are of equal length. Since <math>ABCE</math> is a square and <math>\triangle CED</math> is an equilateral triangle, <math>\angle AEC = 90</math> and <math>\angle CED = 60.</math> Use angle addition | + | <math>AB = EC</math> because they are opposite sides of a square. Also, <math>ED = DC = AB</math> because all sides of the convex pentagon are of equal length. Since <math>ABCE</math> is a square and <math>\triangle CED</math> is an equilateral triangle, <math>\angle AEC = 90</math> and <math>\angle CED = 60.</math> Use angle addition: |
<cmath>\angle E = \angle AEC + \angle CED = 90 + 60 = \boxed{\textbf{(E)} 150}</cmath> | <cmath>\angle E = \angle AEC + \angle CED = 90 + 60 = \boxed{\textbf{(E)} 150}</cmath> | ||
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{{AMC10 box|year=2007|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2007|ab=B|num-b=6|num-a=8}} | ||
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Latest revision as of 11:54, 4 June 2021
Problem
All sides of the convex pentagon 
are of equal length, and 
What is the degree measure of 
Solution
![[asy] unitsize(1.5cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A=(0,2), B=(0,0), C=(2,0), D=(2+sqrt(3),1), E=(2,2); draw(A--B--C--D--E--cycle); draw(E--C,gray); draw(rightanglemark(B,A,E)); draw(rightanglemark(A,B,C)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,E); label("$E$",E,NE); [/asy]](http://latex.artofproblemsolving.com/b/1/a/b1adac4a39fceaa691b5d5346eee30cd5e1094fc.png)
because they are opposite sides of a square. Also, 
because all sides of the convex pentagon are of equal length. Since 
is a square and 
is an equilateral triangle, 
and 
Use angle addition:
![\[\angle E = \angle AEC + \angle CED = 90 + 60 = \boxed{\textbf{(E)} 150}\]](http://latex.artofproblemsolving.com/7/6/7/7676dfaafe286eed98677436adb6d221a5797049.png)
See Also
| 2007 AMC 10B (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
