document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #228781 by ankor@dixie-net.com(22740)\"\" \"About 
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Let A be the ratio of the volume of a sphere to the volume of a cube each of
\n" ); document.write( " whose face is tangent to the sphere, and let B be the ratio of the surface
\n" ); document.write( " area of this sphere to the surface area of the cube.
\n" ); document.write( " Then find the sum A and B.
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\n" ); document.write( "From the description, the sphere is enclosed in a cube
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\n" ); document.write( "Let r = radius of the sphere
\n" ); document.write( "then
\n" ); document.write( "2r = side of the cube
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\n" ); document.write( "Volume Ratio; sphere/cube
\n" ); document.write( "A = \"%28%284%2F3%29pi%2Ar%5E3%29%2F%282r%29%5E3\" = \"%28%284%2F3%29pi%2Ar%5E3%29%2F%288r%5E3%29\"
\n" ); document.write( "cancel r^3
\n" ); document.write( "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
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\n" ); document.write( "Surface area ratio; sphere/cube
\n" ); document.write( "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\"
\n" ); document.write( "Cancel 4, and r^2
\n" ); document.write( "B = \"pi%2F6\"
\n" ); document.write( ":
\n" ); document.write( "A + B: \"pi%2F6\"+\"pi%2F6\" = \"%282pi%29%2F6\" = \"pi%2F3\"
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