2008 AMC 12A Problems/Problem 18

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 C. $\blacksquare$

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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