Do I have enough iMac boxes to make a full circle? My work has a bunch of iMac boxes and because of their slightly wedged shape we are curious how many it would take to make a complete circle.
We already did some calculations and also laid enough out to make 1/4 of a circle so we know how many it would take, but I'm curious to see how others would approach this problem mathematically and see if you came up with the same answer.
Our stock consists of 12 of the 27" iMac boxes and 16 of the 21.5" iMac boxes.
The 21.5" box has the following dimensions

top: 5"
height: 21.25"
bottom: 8.75"

and the 27" box is

top: 5.75"
height: 23.75"
bottom: 9.5"

21.5" iMac box picture for reference

 A: Trying to avoid trigonometric functions: The outer circle must be longer than the inner circle by $2\pi$ times the height, so compute $$\frac{2\pi\text{height}}{(\text{bottom}-\text{top})} $$
as a good approximation.
A: Suppose we have $n$ wedge-shaped boxes wrapped around to make a circle, with an inner length (per box) of $C_1$, an outer length of $C_2$, and a 'diametric' length of $D$.  Then putting the $n$ boxes together will yield an inner circle with circumference $n\times C_1$, and an outer circle with circumference $n\times C_2$.  By the formula for the area of a circle, the diameter of the inner circle is then $\frac{nC_1}{\pi}$, and similarly for the outer circle, so the difference in diameters is $\frac{n}{\pi} (C_2-C_1)$ - but this has to be exactly equal to the twice the diametric length, $2\times D$ (since there we add an additional distance of $D$ on either side of the diameter).  In other words, we have $\frac{n}{\pi}(C_2-C_1)=2\times D$, or solving for $n$, $n=\dfrac{2\pi D}{C_2-C_1}$.
For instance, for your first case, we have $C_1=5$ (inches), $C_2=8\frac34$, and $D=21\frac14$, so we get $n=\dfrac{2\cdot\frac{85}{4}\cdot\pi}{\frac{35}{4}-\frac{20}{4}} = \frac{34}{3}\pi\approx 35.6$ boxes.  Unfortunately, since this isn't very close to an integer, any circle you make will be either a little tight or a little loose.  I'll let you solve the second case yourself...
