Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "A regular octagon <math> ABCDEFGH </math> has an area of one square unit. What is the area of the rectangle <math> ABEF </math>? <asy> unitsize(8mm); defaultpen(linewidth(.8pt)+...")
 
Line 1: Line 1:
 +
==Problem==
 +
 
A regular octagon <math> ABCDEFGH </math> has an area of one square unit. What is the area of the rectangle <math> ABEF </math>?
 
A regular octagon <math> ABCDEFGH </math> has an area of one square unit. What is the area of the rectangle <math> ABEF </math>?
  
Line 4: Line 6:
  
 
<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==
 +
 +
An easy way to look at this:
 +
Area of Octagon: <math> \frac{ap}{2}=1 </math>
 +
Area of Rectangle: <math> \frac{p}{8}\times 2a=\frac{ap}{4} </math>
 +
You can see from this that the octagon's area is twice as large as the rectangle's area is <math>\frac{1}{2}</math>

Revision as of 09:45, 14 August 2011

Problem

A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$?

[asy] unitsize(8mm); defaultpen(linewidth(.8pt)+fontsize(6pt)); 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

An easy way to look at this: 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 $\frac{1}{2}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_23&oldid=41409"