Difference between revisions of "2005 AMC 10A Problems/Problem 20"
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<math> \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 </math> | <math> \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 </math> | ||
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| + | ==Solution== | ||
| + | The area of the octagon can be divided up into 5 squares with side <math>\frac{\sqrt2}2</math> 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 <math>5+4\left(\frac12\right)=7</math> squares. | ||
| + | |||
| + | 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{(C)}</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 15:43, 4 November 2006
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 
.
