Difference between revisions of "2002 AIME II Problems/Problem 2"
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<math>a\sqrt{2}=\sqrt{98}</math> | <math>a\sqrt{2}=\sqrt{98}</math> | ||
− | So, <math>a=7</math>, and hence the surface area is <math>6a^2=\ | + | So, <math>a=7</math>, and hence the surface area is <math>6a^2=\Box{294}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2002|n=II|num-b=1|num-a=3}} | {{AIME box|year=2002|n=II|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:35, 25 June 2015
Problem
Three vertices of a cube are , , and . What is the surface area of the cube?
Solution
So, is an equilateral triangle. Let the side of the cube be .
So, , and hence the surface area is .
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
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.