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

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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>
  
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==Solution==
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==Video Solution==
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==Video Solution
 
 
https://www.youtube.com/watch?v=LREcUjK-56U&feature=youtu.be
 
https://www.youtube.com/watch?v=LREcUjK-56U&feature=youtu.be
  
  
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===Solution 1===
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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>.
  
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===Solution 2===
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==Solution 2==
 
<asy>
 
<asy>
 
unitsize(1cm);
 
unitsize(1cm);
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</asy>
 
</asy>
  
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Here is a less complicated way than that of the user above. If you draw a line segment from each vertex to the center of the octagon and draw the rectangle ABEF (in red), you can see that <math>2</math> of the triangles (in blue) share the same base and height with <math>\dfrac{1}{2}</math> the rectangle. Therefore, the rectangle's area is the same as <math>2\cdot2</math> of the <math>8</math> triangles, and is  <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math> the area of the octagon.
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Here is a less complicated way than that of the user below. If you draw a line segment from each vertex to the center of the octagon and draw the rectangle ABEF (in red), you can see that <math>2</math> of the triangles (in blue) share the same base and height with <math>\dfrac{1}{2}</math> the rectangle. Therefore, the rectangle's area is the same as <math>2\cdot2</math> of the <math>8</math> triangles, and is  <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math> the area of the octagon.
  
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===Solution 3===
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==Solution 3==
 
<asy>
 
<asy>
 
unitsize(1cm);
 
unitsize(1cm);

Revision as of 16:08, 13 August 2022

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]

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

Video Solution

https://www.youtube.com/watch?v=LREcUjK-56U&feature=youtu.be


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: $\frac{p}{8}\times 2a=\dfrac{2ap}{8}=\frac{ap}{4}$.

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

[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); draw(A--E--F--B--C--G--H--D); draw(A--E--F--B--A,blue); draw(A--F--E--B--A,red); [/asy]

Here is a less complicated way than that of the user below. If you draw a line segment from each vertex to the center of the octagon and draw the rectangle ABEF (in red), you can see that $2$ of the triangles (in blue) share the same base and height with $\dfrac{1}{2}$ the rectangle. Therefore, the rectangle's area is the same as $2\cdot2$ of the $8$ triangles, and is $\boxed{\textbf{(D)}\ \frac{1}{2}}$ the area of the octagon.

Solution 3

[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); draw(A--D--E--H--G--B--C--F); [/asy] Drawing lines $AD$, $BG$, $CF$, and $EH$, we can see that the octagon is comprised of $1$ square, $4$ rectangles, and $4$ triangles. The triangles each are $45-45-90$ triangles, and since their diagonal is length $x$, each of their sides is $\frac{\sqrt{2}}{2}x$. The area of the entire figure is, likewise, $x^2$ (the square)$+4x^2\frac{\sqrt{2}}{2}$ (the 4 rectangles)$+2\cdot(\frac{\sqrt{2}}{2}x)^2$ (the triangles), which simplifies to $2x^2 + 2\sqrt{2}x^2$. The area of $ABEF$ is just $x(x+\frac{2\sqrt{2}}{2}x)$, or $x^2$ + $x^2\sqrt{2}$, which we can see is the area of $\frac{ABCDEFGH}{2} = \boxed{\textbf{(D)}\ \frac{1}{2}}$ 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

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