2015 AMC 12B Problem 16

Below is the professionally curated solution for Problem 16 of the 2015 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 12B solutions, or check the answer key.

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Concepts:pyramid volume Pythagorean Theorem

Difficulty rating: 1900

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?

$18$

$162$

$36\sqrt{21}$

$18\sqrt{138}$

$54\sqrt{21}$

Solution:

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.

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