Question

The volume of a pyramid whose base is an equilateral triangle of side length $6$ and whose other edges are each of length $\sqrt{15}$ is

$9$

$\dfrac92$

$\dfrac{27}{2}$

$\dfrac{9\sqrt3}{2}$

none of these

Answer: A

Concepts:pyramid equilateral triangle volume Pythagorean Theorem

Difficulty rating: 2040

Small Hint:

The altitude from the apex meets the base at the equilateral triangle’s circumcenter

Big Hint:

Use the base circumradius $2\sqrt3$ and a lateral edge $\sqrt{15}$ to find the height

Solution:

The base area is $\frac{\sqrt3}{4}(6^2)=9\sqrt3.$ Because the apex is equally distant from all three base vertices, its perpendicular projection is the base circumcenter. The circumradius of the equilateral base is $\frac{6}{\sqrt3}=2\sqrt3.$ If $h$ is the pyramid’s height, then $h^2+(2\sqrt3)^2=(\sqrt{15})^2,$ so $h=\sqrt3.$ The volume is $\frac13(9\sqrt3)(\sqrt3)=9.$

Therefore, the correct answer is A.

1973 AMC 12 Problem 32

Answer: A

Difficulty rating: 2040

Small Hint:

Big Hint:

Solution:

Problem 32 in Other Years