Difference between revisions of "2015 AMC 12B Problems/Problem 16"
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==Problem== | ==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? | ||
| + | <math>\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}</math> | ||
| + | ==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 <math>\sqrt{8^2 - 6^2} = \sqrt{64 - 36} = \sqrt{28} = 2\sqrt{7}</math>. | ||
| − | == | + | The area of the hexagon is |
| + | |||
| + | <cmath>\frac{3\sqrt{3}}{2} \cdot (\text{side})^2 = \frac{3\sqrt{3}}{2} \cdot 6^2 = 54\sqrt{3}</cmath> | ||
| + | |||
| + | 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>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|num-a=17|num-b=15}} | {{AMC12 box|year=2015|ab=B|num-a=17|num-b=15}} | ||
| + | |||
| + | [[Category:3D Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:25, 6 November 2020
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 | |
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