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2007 AMC 12B Problems/Problem 7: Difference between revisions

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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>  <math>\mathrm {(B)} 108</math>  <math>\mathrm {(C)} 120</math>  <math>\mathrm {(D)} 144</math>  <math>\mathrm {(E)} 150</math>
<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]]


{{image}}
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>
 
Since <math>A</math> and <math>B</math> are right angles, 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. Angle <math>E</math> is therefore <math>90+60=150 \Rightarrow \mathrm {(E)}</math>
 


==See Also==
==See Also==
{{AMC12 box|year=2007|ab=B|num-b=6|num-a=8}}


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

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

Solution

Since $A$ and $B$ are right angles, and $AE$ equals $BC$, and $AECB$ is a square. Since the length of $ED$ and $CD$ are also 5, triangle $CDE$ is equilateral. Angle $E$ is therefore $90+60=150 \Rightarrow \mathrm {(E)}$

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

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