AMC12 2015 B

AMC12 2015 B · Q16

AMC12 2015 B · Q16. It mainly tests Area & perimeter, 3D geometry (volume).

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?

一个边长为6的正六边形，每条边上都附着一个等腰三角形，每个等腰三角形有两条边长为8。这些等腰三角形被折叠起来，形成一个以六边形为底面的金字塔。这个金字塔的体积是多少？

(A) 18 18

(B) 162 162

(D) $18\sqrt{138}$ $18\sqrt{138}$

(E) $54\sqrt{21}$ $54\sqrt{21}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The distance from a vertex of the hexagon to its center is 6. The height of the pyramid can be calculated by the Pythagorean Theorem using the right triangle with other leg 6 and hypotenuse 8; it is $\sqrt{8^2-6^2}=2\sqrt{7}$. The volume is then $$ \frac{1}{3}Bh=\frac{1}{3}\cdot 6\left(6^2\cdot \frac{\sqrt{3}}{4}\right)\cdot 2\sqrt{7}=36\sqrt{21}. $$

答案（C）：六边形的一个顶点到其中心的距离是 6。金字塔的高可以用勾股定理计算：在一条直角三角形中，另一条直角边为 6，斜边为 8，因此高为 $\sqrt{8^2-6^2}=2\sqrt{7}$。体积为 $$ \frac{1}{3}Bh=\frac{1}{3}\cdot 6\left(6^2\cdot \frac{\sqrt{3}}{4}\right)\cdot 2\sqrt{7}=36\sqrt{21}. $$

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