Difference between revisions of "2015 AMC 12B Problems/Problem 16"
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<cmath>\frac{3\sqrt{3}}{2} \cdot (\text{side})^2 = \frac{3\sqrt{3}}{2} \cdot 6^2 = 54\sqrt{3}</cmath> | <cmath>\frac{3\sqrt{3}}{2} \cdot (\text{side})^2 = \frac{3\sqrt{3}}{2} \cdot 6^2 = 54\sqrt{3}</cmath> | ||
| − | Thus, | + | Thus, the volume of the pyramid is |
<cmath>\frac{1}{3} \times \text{base} \times \text{height} = \frac{ 54\sqrt{3} \cdot 2\sqrt{7}}{3} = \boxed{\textbf{(C)}\; 36\sqrt{21}}</cmath>. | <cmath>\frac{1}{3} \times \text{base} \times \text{height} = \frac{ 54\sqrt{3} \cdot 2\sqrt{7}}{3} = \boxed{\textbf{(C)}\; 36\sqrt{21}}</cmath>. | ||
Revision as of 19:55, 16 January 2018
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?
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 
.
The area of the hexagon is
![\[\frac{3\sqrt{3}}{2} \cdot (\text{side})^2 = \frac{3\sqrt{3}}{2} \cdot 6^2 = 54\sqrt{3}\]](http://latex.artofproblemsolving.com/2/f/b/2fb97fa93c4994fd942b2dae2168bdbdb81ec6e7.png)
Thus, the volume of the pyramid is
![\[\frac{1}{3} \times \text{base} \times \text{height} = \frac{ 54\sqrt{3} \cdot 2\sqrt{7}}{3} = \boxed{\textbf{(C)}\; 36\sqrt{21}}\]](http://latex.artofproblemsolving.com/d/7/f/d7f855aa1137990641e2f567f2b1b3613ed760fa.png)
.
See Also
| 2015 AMC 12B (Problems • Answer Key • Resources) | |
| 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. 
