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2005 AMC 10A Problems/Problem 20

Revision as of 09:45, 31 May 2021 by Mobius247 (talk | contribs) (→‎Solution 2)

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 1

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 $\frac72\Rightarrow\mathrm{\boxed{A}}$.

[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(A--F, blue);draw(E--B,blue);draw(C--H, blue); draw(D--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); [/asy]

Solution 2

Using the diagram from above, we can extend the sides of length $1$ to form four right triangles and the octagon, all inside a square. The right triangles are 45-45-90 triangles with hypotenuse $\frac{\sqrt{2}}{2}$, so the side length is $\frac{1}{2}$. Thus, the area of the larger square is $2 \cdot 2 = 4$, and the area of the four right triangles combined is $\frac{1}{2}$, so the area of the octagon is $\frac{7}{2}$, or $\boxed{\mathrm{(A). \ }}$


$4$ - $\frac{1}{2}$ $4-\left(\frac12\right)=7$

Video Solution

CHECK OUT Video Solution: https://youtu.be/rwPFZnYk9V8

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AMC 10 Problems and Solutions

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