Question 182723
<font face="Times New Roman" size="+2">


The the weights of two solid masses of similar material are proportional to their volumes.


The volume of your cube, of edge length <i>s</i> is given by:


*[tex \LARGE \text{          }\math V_{you} = s^3]


Since the side length of the larger cube is 2<i>s</i>, the volume of the larger cube must be:


*[tex \LARGE \text{          }\math V_{friend} = (2s)^3 = 8s^3]


So your friend's cube has 8 times the volume and therefore weighs 8 times as much as yours.


The surface area of a cube of edge length <i>s</i> is found by considering that a cube has 6 faces, each of which has an area *[tex \Large A_f=s^2], therefore the total surface area of your cube is:


*[tex \LARGE \text{          }\math A_{you} = 6s^2]


And the total surface area of your friend's cube is:


*[tex \LARGE \text{          }\math A_{friend} = 6(2s)^2= 24s^2]


And your friend's cube has 4 times the surface area as yours.


The problem with the sphere is done in the same fashion.  The formulas for volume and surface area for a sphere are:


*[tex \LARGE \text{          }\math V = \frac{4}{3}\pi r^3]



*[tex \LARGE \text{          }\math A_s = 4 \pi r^2]


Hint:  If your friend's sphere has twice the diameter, it also has twice the radius.


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