1973 AHSME Problems/Problem 32

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Problem

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

$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 9/2 \qquad \textbf{(C)}\ 27/2 \qquad \textbf{(D)}\ \frac{9\sqrt3}{2} \qquad \textbf{(E)}\ \text{none of these}$

Solution

[asy] import three; unitsize(1cm); size(200); draw((0,0,0)--(6,0,0)--(3,5.196,0)--(0,0,0)); draw((3,1.732,1.732)--(0,0,0)); draw((3,1.732,1.732)--(6,0,0)); draw((3,1.732,1.732)--(3,5.196,0)); draw((3,1.732,1.732)--(3,1.732,0)--(0,0,0),dotted); label("6",(4.5,2.598,0),SW); label("$\sqrt{15}$",(4.5,0.866,0.866),N);  currentprojection=orthographic(1/6,1/2,1/3);  [/asy]

Draw an altitude towards the equilateral triangle base. By symmetry (this can also be proved by HL), the base of the altitude is equidistant from the three points of the equilateral triangle. This means that the distance from the base of the altitude to one of the points of the equilateral triangle is $2\sqrt{3}$.

[asy] draw((0,1.732)--(0,0)--(3.464,0),dotted); draw((0,1.732)--(3.464,0)); label("$2\sqrt{3}$",(1.732,0),S); label("$\sqrt{15}$",(1.732,0.866),NE); [/asy]

Using the Pythagorean Theorem, the length of the altitude is $\sqrt{3}$, so the volume of the triangular pyramid is $\tfrac13 \cdot \tfrac{6^2 \cdot \sqrt{3}}{4} \cdot \sqrt{3} = \boxed{\textbf{(A)}\ 9}$.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 31
Followed by
Problem 33
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