2002 AIME II Problems/Problem 2


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?





So, $PQR$ is an equilateral triangle. Let the side of the cube be $a$.


So, $a=7$, and hence the surface area is $6a^2=\framebox{294}$.

See also

2002 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png