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What is the ratio of the volume of cube and the smallest sphere that could contain it?

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harleyrm | Student, Grade 9 | eNoter

Posted June 10, 2012 at 6:11 PM via web

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What is the ratio of the volume of cube and the smallest sphere that could contain it?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 10, 2012 at 6:41 PM (Answer #1)

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You need to fit the sphere inside the cube, hence, the diameter of sphere needs to be equal to the side of cube such that:

`D = a`

D represents the diameter of the sphere

a represents the side of the cube

You need to remember the formula of volume of sphere such that:

`V = (4pir^3)/3`

You need to remember that `r = D/2 =gt r^3 = (D^3)/8`

Substituting `(D^3)/8`  for`r^3`  in formula of volume of sphere yields:

`V = (4pi*(D^3)/8)/3`

`V = (pi*(D^3)/2)/3 =gt V = (pi*(D^3))/6`

Notice that D = a, hence, you need to substitute a for D in formula of volume of sphere such that:

`V = (pi*(a^3))/6`

You need to remember the formula of volume of the cube such that:

`v = a^3`

You need to evaluate the ratio `v/V`  such that:

`v/V = (a^3)/((pi*(a^3))/6)`

Reducing by `a^3`  yields:

`v/V = 6/pi`

Hence, evaluating the ratio of volumes under given conditions yields `v/V = 6/pi.`

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