Difference between revisions of "2003 AMC 10B Problems/Problem 23"

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{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #15]] and [[2003 AMC 10B Problems|2003 AMC 10B #23]]}}
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==Problem==
 
==Problem==
  
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<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==
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==Solution==
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===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]]:
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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==
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===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 01:01, 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 $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$?

[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]

$\textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4}$

Solution

Solution 1

Here is an easy way to look at this, where $p$ is the perimeter, and $a$ is the apothem:

Area of Octagon: $\frac{ap}{2}=1$.

Area of Rectangle: $\frac{p}{8}\times 2a=\frac{ap}{4}$.

You can see from this that the octagon's area is twice as large as the rectangle's area is $\boxed{\textbf{(D)}\ \frac{1}{2}}$.

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 $\boxed{\textbf{(D)}\ \frac{1}{2}}$ the area of the octagon

See Also

2003 AMC 12B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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