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 "2005 AMC 10A Problems/Problem 20"

(→‎See Also)
(→‎Solution)
Line 11: Line 11:
 
The area of each square is <math>\left(\frac{\sqrt2}2\right)^2=\frac12</math>, so the area of 7 squares is <math>\frac72\Rightarrow\mathrm{(A)}</math>.
 
The area of each square is <math>\left(\frac{\sqrt2}2\right)^2=\frac12</math>, so the area of 7 squares is <math>\frac72\Rightarrow\mathrm{(A)}</math>.
  
−
<asy> pair A=(0.7, 0), B=(0, 0.7), C=(0, 1.4), D=(0.7, 2.1), E=(1.4, 2.1), F=(2.1, 1.4), G=(2.1, 0.7), H=(1.4, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H);  draw(H--A);draw(D--A, blue);draw(E--H,blue);draw(C--F, blue); draw(B--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); </asy>
+
<asy> pair A=(0.5, 0), B=(0, 0.5), C=(0, 1.5), D=(0.5, 2), E=(1.5, 2), F=(2, 1.5), G=(2, 0.5), H=(1.5, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H);  draw(H--A);draw(D--A, blue);draw(E--H,blue);draw(C--F, blue); draw(B--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); </asy>
  
 
==See Also==
 
==See Also==

Revision as of 11:15, 8 November 2015

Problem

An equiangular octagon has four sides of length 1 and four sides of length $\frac{\sqrt{2}}{2}$, arranged so that no two consecutive sides have the same length. What is the area of the octagon?

$\mathrm{(A) \ } \frac72\qquad \mathrm{(B) \ } \frac{7\sqrt2}{2}\qquad \mathrm{(C) \ } \frac{5+4\sqrt2}{2}\qquad \mathrm{(D) \ } \frac{4+5\sqrt2}{2}\qquad \mathrm{(E) \ } 7$

Solution

The area of the octagon can be divided up into 5 squares with side $\frac{\sqrt2}2$ and 4 right triangles, which are half the area of each of the squares.

Therefore, the area of the octagon is equal to the area of $5+4\left(\frac12\right)=7$ squares.

The area of each square is $\left(\frac{\sqrt2}2\right)^2=\frac12$, so the area of 7 squares is $\frac72\Rightarrow\mathrm{(A)}$.

[asy] pair A=(0.5, 0), B=(0, 0.5), C=(0, 1.5), D=(0.5, 2), E=(1.5, 2), F=(2, 1.5), G=(2, 0.5), H=(1.5, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H); draw(H--A);draw(D--A, blue);draw(E--H,blue);draw(C--F, blue); draw(B--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); [/asy]

See Also

2005 AMC 10A (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_20&oldid=72823"