Responses to PUZZLE #33: Surface and volume

It is interesting to compare this puzzle with #31.

If you calculate the values of the surface area and the volume of each solid and compare the results it will be clear what is going on.

SURFACE AREA:

The box has 2 ends of area 1x2 = 2in.², a front and back of area 1x4 = 4in², & a top and bottom area of 2x4 = 8in². So the total surface area is 2(2 + 4 + 8) = 28in².

Each side of the cube has area 2x2 = 4in² and there are 6 sides, so the total surface area is 6x4 = 24in².

Using the equation for surface area of a sphere, the sphere of radius r = 1.382in. has surface area 4pi(1.382)² = 24.000812in².

VOLUME:

The box has volume 1x2x4 = 8in³.

The cube has volume 2x2x2 = 8in³.

Using the equation for the volume of a sphere, the sphere of radius r = 1.382in. has volume 4pi(1.382)³/3 = 11.056374in³.

CONCLUSION:

So, we have 2 solids with surface area of 24in², the cube and the sphere (assuming that the sphere is almost 24in²), but the volumes of each are very different.

On the other hand, we have 2 solids with the same volume of 8in³, the box and the cube, but the surface areas of each are very different.