Difference between revisions of "2007 AMC 12B Problems/Problem 7"
m |
|||
| Line 1: | Line 1: | ||
| − | All sides of the convex pentagon ABCDE are of equal length, and <A = | + | ==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>? | ||
| − | A) 90 B) 108 C) 120 D) 144 E) 150 | + | <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> |
| + | ==Solution== | ||
| − | {{ | + | {{image}} |
| − | {{ | + | |
| + | 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== | ||
| + | |||
| + | {{AMC12 box|year=2007|ab=B|num-b=6|num-a=8}} | ||
Revision as of 09:50, 17 October 2007
Problem
All sides of the convex pentagon 
are of equal length, and 
. What is the degree measure of 
?
Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Since 
and 
are right angles, and 
equals 
, 
is a square, and 
is 5. Since 
and 
are also 5, triangle 
is equilateral. Angle 
is therefore 
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 | |
