1973 AMC 12 Problem 32

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.