Question 237540
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If the cylinder with radius *[tex \Large r] fits snugly inside the cube, the cube must have an edge length equal to the diameter of the cylinder, or *[tex \Large 2r].


The lateral surface area of a cube is the area of 4 of the sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L.A._{cube}\ =\ 4e^2]


So for this cube:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L.A._{cube}\ =\ 4(2r)^2\ =\ 16r^2]


The lateral surface area of a cylinder is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L.A._{cyl}\ =\ 2\pi rh]


But for this cylinder, the height is equal to the edge of the cube, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ L.A._{cyl}\ =\ 4\pi r^2]


Compute the difference:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16r^2\ -\ 4\pi r^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4r^2(4\ - \pi)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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