2003 AMC 10B Problems/Problem 23: Difference between revisions

Revision as of 13:12, 29 December 2011

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

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 $\boxed{\textbf{(D)}\ \frac{1}{2}}$ .

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AMC 10 Problems and Solutions

@@ Line 9: / Line 9: @@
 ==Solution==
-An easy way to look at this:
+Here is an easy way to look at this, where <math>p</math> is the perimeter, and <math>a</math> is the [[apothem]]:
 Area of Octagon: <math> \frac{ap}{2}=1 </math>.

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2003 AMC 10B Problems/Problem 23: Difference between revisions

Revision as of 13:12, 29 December 2011

Problem

Solution

See Also