AMC12 2015 12B, Problem 16

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?

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 \[\frac{1}{3} \times \text{base} \times \text{height} = \frac{ 54\sqrt{3} \cdot 2\sqrt{7}}{3} = \fbox{36\sqrt{21}}\].

Answer: 36\sqrt{21}