2005 AMC 10A Problems/Problem 20

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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?

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

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 $\frac72\Rightarrow\mathrm{(A)}$ .

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2005 AMC 10A Problems/Problem 20

From AoPSWiki

Problem

Solution

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