Difference between revisions of "2005 AMC 12A Problems/Problem 22"
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<math>\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16</math> | <math>\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16</math> | ||
− | == Solution == | + | == Solution== |
− | + | Box P has dimensions <math>l</math>, <math>w</math>, and <math>h</math>. | |
− | <cmath> | + | Surface area = <cmath>2lw+2lh+2wl=384</cmath> |
− | <cmath> | + | Sum of all edges = <cmath>4l+4w+4h=112 \Longrightarrow l + w + h = 28</cmath> |
− | + | The diameter of the sphere is the space diagonal of the prism, which is <cmath>\sqrt{l^2 + w^2 +h^2}</cmath> | |
− | <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> | |
− | + | <math>r=\frac{20}{2} = 10</math>$ | |
− | <cmath>( | ||
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− | <cmath> | ||
== See also == | == See also == |
Revision as of 20:05, 19 June 2017
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 |
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