Difference between revisions of "2003 AMC 10B Problems/Problem 23"
m |
Mariekitty (talk | contribs) m (→Solution) |
||
| Line 10: | Line 10: | ||
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 13:06, 29 December 2011
Problem
A regular octagon 
has an area of one square unit. What is the area of the rectangle 
?
![[asy] unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW);[/asy]](http://latex.artofproblemsolving.com/5/0/e/50ef8af0ec716cfeea89017a030c7d57d3b40e4c.png)
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 | ||
