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2015 AMC 12B Problem 16

Problem 16 of 25IntermediateGeometry

A regular hexagon with sides of length 66 has an isosceles triangle attached to each side. Each of these triangles has two sides of length 8.8. 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?

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Solution

The distance from the hexagon’s center to a vertex is 6.6. A lateral edge has length 8,8, so the pyramid’s height is 82−62=28=27.\sqrt{8^2 - 6^2} = \sqrt{28} = 2\sqrt7. The hexagon’s area is 332⋅62=543.\dfrac{3\sqrt3}{2}\cdot 6^2 = 54\sqrt3. Thus the volume is 13⋅543⋅27=3621.\dfrac13 \cdot 54\sqrt3 \cdot 2\sqrt7 = 36\sqrt{21}. Thus, the correct answer is C.
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Tagged: pyramid · volume · Pythagorean Theorem

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