Question 961854
The mistake you made was when you got the radius to be 12pi when it should be 12/pi 


Volume of Ball:


V = (4/3)*pi*r^3


V = (4/3)*pi*(12/pi)^3


V = 2304/(pi^2) ... exact volume


V = 233.444007111947 ... approximate volume


The volume of the basketball is roughly 233.444007111947 cubic inches


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Volume of box = L*W*H = 8*8*9 = 64*9 = 576 cubic inches


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Since the volume of the ball (233.444007111947 cubic inches) is less than the volume of the box (576 cubic inches), this means the ball will fit with plenty of room leftover.


Side note: 233.444007111947/576 = 0.40528 = 40.528% of the box is filled with the ball. The other 100% - 40.528% = 59.472% of the volume is other stuff like air or packing material.


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Now we need to check if the dimensions will work. Imagine working with a box and a very long stick. Surely the box has a much much greater volume than the small stick, but will the stick fit in the box without breaking? The answer is "no" if the stick is very long (especially if it exceeds the <a href="http://www.emathematics.net/triangrectangulo.php?a=3&pita=6">space diagonal</a> of the box)


So this is why we must find the diameter and make sure it is smaller than all of the dimensions of the box.


Circumference = pi*diameter


C = pi*d


24 = pi*d


d = 24/pi ... exact diameter


d = 7.63943726841098 ... approximate diameter


The diameter of the ball is approximately 7.63943726841098 inches. This is less than 8 inches, which is the smallest dimension of the box. By extension, the diameter is also less than the other dimension of 9 inches.


Therefore, this ball will fit in the box just fine.


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