Difference between revisions of "2005 AMC 10A Problems/Problem 20"
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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. | + | <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 12:15, 8 November 2015
Problem
An equiangular octagon has four sides of length 1 and four sides of length , arranged so that no two consecutive sides have the same length. What is the area of the octagon?
Solution
The area of the octagon can be divided up into 5 squares with side 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 squares.
The area of each square is , so the area of 7 squares is .
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.