2015 AMC 12B Problems/Problem 16

Problem

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?

$\textbf{(A)}\; 18 \qquad\textbf{(B)}\; 162 \qquad\textbf{(C)}\; 36\sqrt{21} \qquad\textbf{(D)}\; 18\sqrt{138} \qquad\textbf{(E)}\; 54\sqrt{21}$

Solution

The distance from a corner to the center is 6, and from the corner to the top of the pyramid is 8, so the height is $\sqrt{8^2 - 6^2} = \sqrt{64 - 36} = \sqrt{28} = 2\sqrt{7}$ .

The area of the hexagon is

$\frac{3\sqrt{3}}{2} \cdot (\text{side})^2 = \frac{3\sqrt{3}}{2} \cdot 6^2 = 54\sqrt{3}$

Thus, The volume of the pyramid is

$\frac{1}{3} \times \text{base} \times \text{height} = \frac{ 54\sqrt{3} \cdot 2\sqrt{7}}{3} = \boxed{\textbf{(C)}\; 36\sqrt{21}}$ .

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2015 AMC 12B Problems/Problem 16

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