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?

$\dfrac{7}{2}$

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

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

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

$7$

答案：A

知识点：等角多边形面积分割特殊直角三角形

难度评级：1760

解答：

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.

2005 AMC 10A 第 20 题

答案：A

难度评级：1760

解答：

其他年份的第 20 题