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

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 $90+60=150 \Rightarrow \mathrm {(E)}$

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 12 Problems and Solutions

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

Latest revision as of 12:41, 22 August 2024

Problem

Solution

See Also