SOLUTION: Let A be the ratio of the volume of a sphere to the volume of a cube each of whose face is tangent to the sphere, and let B be the ratio of the surface area of this sphere to the s

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Question 319494: Let A be the ratio of the volume of a sphere to the volume of a cube each of whose face is tangent to the sphere, and let B be the ratio of the surface area of this sphere to the surface area of the cube. Then find the sum A and B.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Let A be the ratio of the volume of a sphere to the volume of a cube each of
whose face is tangent to the sphere, and let B be the ratio of the surface
area of this sphere to the surface area of the cube.
Then find the sum A and B.
:
From the description, the sphere is enclosed in a cube
:
Let r = radius of the sphere
then
2r = side of the cube
:
Volume Ratio; sphere/cube
A = %28%284%2F3%29pi%2Ar%5E3%29%2F%282r%29%5E3 = %28%284%2F3%29pi%2Ar%5E3%29%2F%288r%5E3%29
cancel r^3
A = %28%284%2F3%29pi%29%2F8 = %28%284%2F3%29pi%29%2A%281%2F8%29 = %28%281%2F3%29pi%29%2A%281%2F2%29 = pi%2F6; vol ratio
:
Surface area ratio; sphere/cube
B = %284%2Api%2Ar%5E2%29%2F%286%282r%29%5E2%29 = %284%2Api%2Ar%5E2%29%2F%286%2A4r%5E2%29 = %284%2Api%2Ar%5E2%29%2F%2824r%5E2%29
Cancel 4, and r^2
B = pi%2F6
:
A + B: pi%2F6+pi%2F6 = %282pi%29%2F6 = pi%2F3