2015 AMC 12B 第 16 题

Question

一个边长为

6

的正六边形，每条边的外侧都连接着一个等腰三角形。每个等腰三角形有两条边长为

8

。将这些等腰三角形折起，形成一个以该六边形为底面的棱锥。这个棱锥的体积是多少？

A regular hexagon with sides of length

6

has an isosceles triangle attached to each side. Each of these triangles has two sides of length

8.

The isosceles triangles are folded to make a pyramid with the hexagon as the base of the pyramid. What is the volume of the pyramid?

18

162

36\sqrt{21}

18\sqrt{138}

54\sqrt{21}

边长为

6

的正六边形中，中心到顶点的距离也为

6

。

In a regular hexagon of side

6,

the distance from the center to a vertex is also

6

顶点位于中心正上方，高为

\sqrt{8^2 - 6^2}

；再使用体积公式

V = \dfrac13 \cdot (\text{底面积}) \cdot h

。

The apex sits above the center at height

\sqrt{8^2 - 6^2};

use

V = \dfrac13 \cdot (\text{base area}) \cdot h

六边形中心到顶点的距离为

6

。侧棱长为

8

，所以棱锥高为

\sqrt{8^2 - 6^2} = \sqrt{28} = 2\sqrt7

。

六边形面积为

\dfrac{3\sqrt3}{2}\cdot 6^2 = 54\sqrt3

。因此体积为

\dfrac13 \cdot 54\sqrt3 \cdot 2\sqrt7 = 36\sqrt{21}

。

因此，正确选项是 C。

The distance from the hexagon’s center to a vertex is

6.

A lateral edge has length

8,

so the pyramid’s height is

\sqrt{8^2 - 6^2} = \sqrt{28} = 2\sqrt7.

The hexagon’s area is

\dfrac{3\sqrt3}{2}\cdot 6^2 = 54\sqrt3.

Thus the volume is

\dfrac13 \cdot 54\sqrt3 \cdot 2\sqrt7 = 36\sqrt{21}.

Thus, the correct answer is C.

其他年份的第 16 题