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Difference between revisions of "2002 AIME II Problems/Problem 2"
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== Problem ==
 
== Problem ==
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Three vertices of a cube are <math>P=(7,12,10)</math>, <math>Q=(8,8,1)</math>, and <math>R=(11,3,9)</math>. What is the surface area of the cube?
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Three [[vertex|vertices]] of a [[cube]] are <math>P=(7,12,10)</math>, <math>Q=(8,8,1)</math>, and <math>R=(11,3,9)</math>. What is the [[surface area]] of the cube?
  
 
== Solution ==
 
== Solution ==

Revision as of 07:47, 17 April 2008

Problem

Three vertices of a cube are $P=(7,12,10)$, $Q=(8,8,1)$, and $R=(11,3,9)$. What is the surface area of the cube?

Solution

$PQ=\sqrt{(8-7)^2+(8-12)^2+(1-10)^2}=\sqrt{98}$

$PR=\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\sqrt{98}$

$QR=\sqrt{(11-8)^2+(3-8)^2+(9-1)^2}=\sqrt{98}$

So, PQR is an equilateral triangle. Let the side of the cube is $a$. $a\sqrt{2}=\sqrt{98}$

So, $a=7$, and hence the surface area=$6a^2=294$.

See also

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