Investigations | Geometry | 3-D Objects
Recommended Investigations | Same but Different
Can two solids with the same volume have different surface areas? If so, provide an
example. If not, explain why not.
Extensions
If one solid has a greater volume than another, must it also have a greater surface area?
If you begin with a rectangular solid which has a given length(l), width(w) and height(h),
you can assign a numerical value to each and then find it’s volume(V) and surface
area(SA).
I will use the following formulas:
Volume (V) = l*w*h
Surface Area (SA) = 2(lw) + 2(lh) + 2(wh)
The problem states that I need to find two solids with the same volume yet different
surface areas.
If I use the number 16 to represent the volume, I can assign various lengths, widths and
heights to the solid using the factors of 16.
1 x 1 x 16
V = 16
SA = 2(1*1) + 2(1*16) +2(1*16) = 66
1x2x8
V = 16
SA = 2(1*2) + 2(1*8) + 2(2*8) = 52
1x4x4
V = 16
SA = 2(1*4) + 2 (1* 4) + 2(4*4) =48
THUS, I have found three solids with the same volume yet different surface areas.
An extension of this problem is to find if the volume is larger, must the surface area be
larger too.
Again, I assigned various values to the length, width and height of a solid, giving one a
volume just larger than the other.
EXAMPLE
1x1x9
V=9
SA = 2(1*1) + 2(1*9) + 2(1*9) = 38
1x2x5
V = 10
SA = 2(1*2) + 2(1*5) + 2(2*5) = 34
In the example above, the solid with the greater volume (V) did NOT have the greater
surface area (SA).
A further extension would be to compare the problem using a CUBE.
This extension fails immediately due to the nature of a cube having all sides congruent.
EXAMPLE
Cube with a volume of 8 has ONE AND ONLY ONE way to assign length, width, and
height:
2x2x2
V=8
SA = 2(2*2) + 2(2*2) + 2(2*2) =24
AGAIN, there are no other ways to assign measures to a cube with a volume of 8.
Thus, there does not exist another “cube” with the same volume yet different surface
area.