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Difference between revisions of "2007 AMC 10B Problems/Problem 7"

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==Problem 7==
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==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 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>
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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
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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:
 
<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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== See Also ==
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{{AMC10 box|year=2007|ab=B|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 11:54, 4 June 2021

Problem

All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A= \angle B = 90^\circ.$ What is the degree measure of $\angle E?$

$\textbf{(A) } 90 \qquad\textbf{(B) } 108 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 150$

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]

$AB = EC$ because they are opposite sides of a square. Also, $ED = DC = AB$ because all sides of the convex pentagon are of equal length. Since $ABCE$ is a square and $\triangle CED$ is an equilateral triangle, $\angle AEC = 90$ and $\angle CED = 60.$ Use angle addition: \[\angle E = \angle AEC + \angle CED = 90 + 60 = \boxed{\textbf{(E)} 150}\]

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

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