Difference between revisions of "2007 AMC 12B Problems/Problem 7"

Revision as of 22:15, 1 October 2018

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 $BC$ , $AECB$ is a square, and $EC$ is 5. Since $ED$ and $CD$ are also 5, triangle $CDE$ is equilateral. Angle $E$ is therefore $90+60=151 \Rightarrow \mathrm {(E)}$

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.

@@ Line 7: / Line 7: @@
 [[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>, <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=151 \Rightarrow \mathrm {(E)}</math>
 ==See Also==

Page

Toolbox

Search

Difference between revisions of "2007 AMC 12B Problems/Problem 7"

Revision as of 22:15, 1 October 2018

Problem

Solution

See Also