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Difference between revisions of "2008 AMC 12A Problems/Problem 18"

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Triangle <math>ABC</math>, with sides of length <math>5</math>, <math>6</math>, and <math>7</math>, has one [[vertex]] on the positive <math>x</math>-axis, one on the positive <math>y</math>-axis, and one on the positive <math>z</math>-axis. Let <math>O</math> be the [[origin]]. What is the volume of [[tetrahedron]] <math>OABC</math>?
 
Triangle <math>ABC</math>, with sides of length <math>5</math>, <math>6</math>, and <math>7</math>, has one [[vertex]] on the positive <math>x</math>-axis, one on the positive <math>y</math>-axis, and one on the positive <math>z</math>-axis. Let <math>O</math> be the [[origin]]. What is the volume of [[tetrahedron]] <math>OABC</math>?
  
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<math>\textbf{(A)}\ \sqrt{85} \qquad \textbf{(B)}\ \sqrt{90} \qquad \textbf{(C)}\ \sqrt{95} \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ \sqrt{105}</math>
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<math>\mathrm{(A)}\ \sqrt{85}\qquad\mathrm{(B)}\ \sqrt{90}\qquad\mathrm{(C)}\ \sqrt{95}\qquad\mathrm{(D)}\ 10\qquad\mathrm{(E)}\ \sqrt{105}</math>
  
 
==Solution==
 
==Solution==

Revision as of 01:45, 26 April 2008

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

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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. $\blacksquare$

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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