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

(→‎Solution 3)
(→‎Solution 2)
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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===
 
===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 above. If you draw a line segment from each vertex to the center of the octagon 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 the 8 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 a line segment from each vertex to the center of the octagon and draw the rectangle ABEF( In red), you can see that two of the triangles (In blue) share the same base and height with half the rectangle. Therefore, the rectangle's area is the same as 4 of the 8 triangles, and is  <math>\boxed{\textbf{(D)}\ \frac{1}{2}}</math> the area of the octagon.
  
 
===Solution 3===
 
===Solution 3===

Revision as of 10:35, 1 September 2018

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

[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 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 two of the triangles (In blue) share the same base and height with half the rectangle. Therefore, the rectangle's area is the same as 4 of the 8 triangles, and is $\boxed{\textbf{(D)}\ \frac{1}{2}}$ the area of the octagon.

Solution 3

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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