interesting geometry question...

[snakeeyes1000]

i have a question... i just came across a figure and i couldn't figure how to mathematically find its volume? let me describe it...

-it has a base of a regular hexagon, with sides 2 inches.
-it has a base of a regular rectangle(aka a square) with sides 3 inches.
-one pair of opposite angles of the square base connect to two angles each on the hexagonal base; the other pair of opposite angles of the square connect to only one angle each on the hexagonal base
-the vertical(not slant) height is known to be 15 inches.

Now I was just wondering how to measure the volume of this? I originally thought to just get the average of two different prisms; one with the square base as both bases, and the other with the hexagonal base as both bases. would that work? just curious. any help is much appreciated(and water displacement is a no-no! it's made out of paper...)

[DiGiTaIErRoR]

If this is an actual object. See how much water it displaces.

That's the easy way.

[sql_lall]

I'm not sure exactly of what it would look like, but is it possiblt to combine two of theses (either square-to-square or hex-to-hex) to make one larger object, find it's volume, then divide by two??

[snakeeyes1000]

sql, that's what i thought... i did it and it worked. and water displacement s a no-no, it is paper and would dissolve.(i measured it by sand displacement)

[prog_tom]

the surface area of this object is:

S = 16.96 in^2 * 6
= 101.76

Because V = s^3

101.76 in^2 = 6s^2
s = 4.118252056394800126318947785679 in (note s in this case is the lengths of each side on a cube)

V = 69.845554876455810142369354445115 in^3

[sql_lall]

Wow.
Is V=s^3 always the case, or just in this example?