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

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==Solution==
 
==Solution==
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[[WLOG]], we let <math>AB</math> go between the <math>x</math> and <math>y</math> axes, <math>BC</math> between <math>y</math> and <math>z</math> axes<math>, </math>CA<math> between </math>z<math> and </math>x<math> axes. Let </math>x,y,z<math> be </math>OA,OB,OC,<math> respectively. By the [[Pythagorean Theorem]], </math>x^2+y^2=25<math>, </math>y^2+z^2=36<math>, </math>z^2+x^2=49<math>. Thus, </math>x^2 = 30<math>, </math>y^2 = 19<math>, and </math>z^2 = 6<math>. Thus the volume of the tetraehdron is </math>\frac{\sqrt{30\cdot 19\cdot 6}}{6}=\sqrt{95}\Rightarrow \boxed{C}$.
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[[WLOG]], we let <math>AB</math> go between the <math>x</math> and <math>y</math> axes, <math>BC</math> between <math>y</math> and <math>z</math> axes, <math>CA</math> between <math>z</math> and <math>x</math> axes. Let <math>x,y,z</math> be <math>OA,OB,OC,</math> respectively. By the [[Pythagorean Theorem]], <math>x^2+y^2=25</math>, <math>y^2+z^2=36</math>, <math>z^2+x^2=49</math>. Thus, <math>x^2 = 30</math>, <math>y^2 = 19</math>, and <math>z^2 = 6</math>. Thus the volume of the tetraehdron is <math>\frac{\sqrt{30\cdot 19\cdot 6}}{6}=\sqrt{95}\Rightarrow \boxed{C}</math>.
  
  

Revision as of 20:47, 17 February 2008

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

WLOG, we let $AB$ go between the $x$ and $y$ axes, $BC$ between $y$ and $z$ axes, $CA$ between $z$ and $x$ axes. Let $x,y,z$ be $OA,OB,OC,$ respectively. By the Pythagorean Theorem, $x^2+y^2=25$, $y^2+z^2=36$, $z^2+x^2=49$. Thus, $x^2 = 30$, $y^2 = 19$, and $z^2 = 6$. Thus the volume of the tetraehdron is $\frac{\sqrt{30\cdot 19\cdot 6}}{6}=\sqrt{95}\Rightarrow \boxed{C}$.


2008 AMC 12A (Problems • Answer Key • Resources)
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Problem 16
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All AMC 12 Problems and Solutions
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