2015 AMC 12B Problems/Problem 16

Revision as of 20:55, 16 January 2018 by Mathislife16 (talk | contribs) (Solution)

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

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

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS