Difference between revisions of "2003 AMC 10B Problems/Problem 23"
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==See Also== | ==See Also== | ||
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Revision as of 11:10, 4 July 2013
Contents
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 1
Here is an easy way to look at this, where 
is the perimeter, and 
is the apothem:
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 
.
Solution 2
Here is a less complicated way than that of the user above. If you draw lines connecting opposite vertices and draw the rectangle ABEF, you can see that two of the triangles share the same base and height with half the rectangle. Therefore, the rectangle's area is the same as 4 of these triangles, and is the area of the octagon
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
