Difference between revisions of "2005 AMC 12A Problems/Problem 22"
Riflescoop (talk | contribs) m (→Solution) |
Riflescoop (talk | contribs) (→Solution) |
||
Line 12: | Line 12: | ||
<cmath>(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400</cmath> | <cmath>(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400</cmath> | ||
<cmath>\sqrt{l^2 + w^2 +h^2} = 20 = diameter</cmath> | <cmath>\sqrt{l^2 + w^2 +h^2} = 20 = diameter</cmath> | ||
− | <cmath>r=\frac{20}{2} = B 10</cmath> | + | <cmath>r=\frac{20}{2} = \boxed{\textbf{(B)} 10}</cmath> |
== See also == | == See also == |
Revision as of 18:41, 4 September 2018
Problem
A rectangular box is inscribed in a sphere of radius . The surface area of is 384, and the sum of the lengths of its 12 edges is 112. What is ?
Solution
Box P has dimensions , , and . Surface area = Sum of all edges =
The diameter of the sphere is the space diagonal of the prism, which is
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.