2005 AMC 10A Problems/Problem 20

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

$[asy] pair A=(0.7, 0), B=(0, 0.7), C=(0, 1.4), D=(0.7, 2.1), E=(1.4, 2.1), F=(2.1, 1.4), G=(2.1, 0.7), H=(1.4, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H); draw(H--A);draw(D--A, blue);draw(E--H,blue);draw(C--F, blue); draw(B--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); [/asy]$

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2005 AMC 10A Problems/Problem 20

Problem

Solution

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