2005 AMC 10A 第 20 题

Question

一个等角八边形有四条边长为

1

，四条边长为

\dfrac{\sqrt{2}}{2}

，并且任意两条相邻边长度不同。这个八边形的面积是多少？

An equiangular octagon has four sides of length

1

and four sides of length

\dfrac{\sqrt{2}}{2},

arranged so that no two consecutive sides have the same length. What is the area of the octagon?

\dfrac{7}{2}

\dfrac{7\sqrt{2}}{2}

\dfrac{5 + 4\sqrt{2}}{2}

\dfrac{4 + 5\sqrt{2}}{2}

7

等角八边形的每个内角为

135^\circ

；把八边形放入一个较大的正方形中。

Every interior angle of an equiangular octagon is

135^\circ;

enclose the octagon in a larger square

四条短边是等腰直角三角形的斜边，这些三角形直角边为

\frac{1}{2}

，从边长

2

的正方形四角切去。

The four short sides are hypotenuses of isosceles right triangles with legs

\frac{1}{2},

cut from a square of side

2

延长四条长为

1

的边可形成一个正方形。每条短边

\dfrac{\sqrt{2}}{2}

是等腰直角三角形的斜边，对应直角边为

\dfrac{1}{2}

，因此八边形可看作从边长

1 + 2 \cdot \frac{1}{2} = 2

的正方形四角切去四个这样的三角形，面积为

2^2 - 4 \cdot \frac{1}{2}\left(\frac{1}{2}\right)^2 = 4 - \frac{1}{2} = \dfrac{7}{2}

。

所以正确答案是 A。

Extend the four sides of length

1

to form a square. Each short side

\dfrac{\sqrt{2}}{2}

is the hypotenuse of an isosceles right triangle with legs

\dfrac{1}{2},

and cutting these four corners from a square of side

1 + 2 \cdot \frac{1}{2} = 2

gives the octagon. Its area is

2^2 - 4 \cdot \frac{1}{2}\left(\frac{1}{2}\right)^2 = 4 - \frac{1}{2} = \dfrac{7}{2}.

Thus, the correct answer is A.

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