2003 AMC 10B Problems/Problem 23: Difference between revisions
No edit summary |
No edit summary |
||
| Line 1: | Line 1: | ||
{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #15]] and [[2003 AMC 10B Problems|2003 AMC 10B #23]]}} | |||
==Problem== | ==Problem== | ||
| Line 7: | Line 9: | ||
<math> \textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4} </math> | <math> \textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4} </math> | ||
==Solution 1== | ==Solution== | ||
===Solution 1=== | |||
Here is an easy way to look at this, where <math>p</math> is the perimeter, and <math>a</math> is the [[apothem]]: | Here is an easy way to look at this, where <math>p</math> is the perimeter, and <math>a</math> is the [[apothem]]: | ||
| Line 16: | Line 19: | ||
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>. | 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>. | ||
==Solution 2== | ===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 <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math> the area of the octagon | 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 <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math> the area of the octagon | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2003|ab=B|num-b=14|num-a=16}} | |||
{{AMC10 box|year=2003|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2003|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:01, 5 January 2014
- The following problem is from both the 2003 AMC 12B #15 and 2003 AMC 10B #23, so both problems redirect to this page.
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
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 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
