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Revision as of 20:37, 3 July 2013

Problem

Triangle $ABC$, with sides of length $5$, $6$, and $7$, has 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 of tetrahedron $OABC$?

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

Solution

[asy] defaultpen(fontsize(8)); draw((0,10)--(0,0)--(8,0));draw((-3,-4)--(0,0));draw((0,10)--(-3,-4)--(8,0)--cycle); label("A",(8,0),(1,0));label("B",(0,10),(0,1));label("C",(-3,-4),(-1,-1));label("O",(0,0),(1,1)); label("$a$",(4,0),(0,1));label("$b$",(0,5),(1,0));label("$c$",(-1.5,-2),(1,0)); label("$5$",(4,5),(1,1));label("$6$",(-1.5,3),(-1,0));label("$7$",(2.5,-2),(1,-1)); [/asy]

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 lengths 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 \[abc/6\] because we can consider the tetrahedron to be a right triangular pyramid. \[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\] which is answer choice $\boxed{\text{C}}$.

See also

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