2008 AMC 12A Problems/Problem 18

Problem

A triangle $\triangle ABC$ with sides $5$ , $6$ , $7$ is placed in the three-dimensional plane with one vertex on the positive $x$ axis, one on the positive $y$ axis, and one on the positive $z$ axis. Let $O$ be the origin. What is the volume if $OABC$ ?

$\textbf{(A)}\ \sqrt{85} \qquad \textbf{(B)}\ \sqrt{90} \qquad \textbf{(C)}\ \sqrt{95} \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ \sqrt{105}$

Solution

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Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lenghts of 5, 6, and 7. Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively. Then by the Pythagorean Theorem, $\begin{align*} a^2+b^2 &=5^2 , \\ b^2+c^2&=6^2, \\ c^2+a^2 &=7^2 , \end{align*}$ so $a^2 = (5^2+7^2-6^2)/2 = 19$ ; similarly, $b^2 = 6$ and $c^2 = 30$ . Since $OA$ , $OB$ , and $OC$ are mutually perpendicular, the tetrahedron's volume is $abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},$ which is answer choice C. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2008 AMC 12A Problems/Problem 18

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