SOLUTION: Amber plans to ship a mini-basketball she bought for her nephew. The circumference of the ball is 24 inches and the package she wants to ship it in is a rectangular box that measur

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Question 961854: Amber plans to ship a mini-basketball she bought for her nephew. The circumference of the ball is 24 inches and the package she wants to ship it in is a rectangular box that measures 8 inches X 8 inches X 9 inches. Will the basketball fit in the box?
I tied to do it, and I'm just not sure if I getting the right answer.
The Circle/ Basketball
C= 24
2πr= 24
R=12π
V=4/3π(12π)^3
V=1558.5 in^3
The Box
V=8x8x9
V=576 in^3
If this is correct the basketball would not fit in the box, correct?

Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
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 space diagonal 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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If you need more one-on-one help, email me at jim_thompson5910@hotmail.com. You can ask me a few more questions for free, but afterwards, I would charge you ($2 a problem to have steps shown or $1 a problem for answer only).

Alternatively, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Any amount is greatly appreciated as it helps me a lot. This donation is to support free tutoring. Thank you.

Jim
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