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==
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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.
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Therefore, the area of the octagon is equal to the area of <math>5+4\left(\frac12\right)=7</math> squares.
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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{(C)}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 16:43, 4 November 2006

Problem

An equiangular octagon has four sides of length 1 and four sides of length $\sqrt2/2$, 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 $\frac72\Rightarrow\mathrm{(C)}$.

See Also

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