Difference between revisions of "2003 AMC 10B Problems/Problem 23"
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An easy way to look at this: | An easy way to look at this: | ||
− | Area of Octagon: <math> \frac{ap}{2}=1 </math> | + | |
− | Area of Rectangle: <math> \frac{p}{8}\times 2a=\frac{ap}{4} </math> | + | Area of Octagon: <math> \frac{ap}{2}=1 </math>. |
− | You can see from this that the octagon's area is twice as large as the rectangle's area is <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math> | + | |
+ | Area of Rectangle: <math> \frac{p}{8}\times 2a=\frac{ap}{4} </math>. | ||
+ | |||
+ | You can see from this that the octagon's area is twice as large as the rectangle's area is <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2003|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2003|ab=B|num-b=22|num-a=24}} |
Revision as of 14:06, 29 December 2011
Problem
A regular octagon has an area of one square unit. What is the area of the rectangle ?
Solution
An easy way to look at this:
Area of Octagon: .
Area of Rectangle: .
You can see from this that the octagon's area is twice as large as the rectangle's area is .
See Also
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |