InterMath | Geometry | 3-D Objects
Title
Volume and Surface Area of a Cube
Problem Statement
Suppose you build a cube whose length, width, and height are each 2 cm. What is the
volume of this cube? What is its surface area? Repeat this procedure for a 3 x 3 x 3 cube
and then for a 4 x 4 x 4 cube. What patterns do you notice? Can you predict the volume and
surface area of a 17 x 17 x 17 cube? How about an s x s x s cube?
Problem setup
What is the volume and surface area of a cube with the dimensions of 2 x 2 x 2, 3 x 3 x 3, and 4
x 4 x 4? Is there a pattern? By predicting, what would be the surface area and volume of a 17 x
17 x 17 cube and an s x s x s cube?
Plans to Solve/Investigate the Problem
First I will construct a cube in GSP and label the dimensions of the cube. Then using the
formula for surface area and volume I will determine surface area and volume of a cube with
dimensions of 2 x 2 x 2. Next I will find the surface area and volume the same way with cubes
of dimensions 3 x 3 x 3 and 4 x 4 x 4. After determining these measurements I will compare
them to try to determine a pattern. Using this pattern I will predict what the surface area and
volume would be of a cube with the dimensions of 17 x 17 x 17 and s x s x s.
Investigation/Exploration of the Problem
1. Construct a square
2. Construct another square and connect them to form a cube
a
c
b
The dimensions of this cube are 2 x 2 x 2. Therefore, a = 2, b = 2, and c = 2. The formula to
determine the volume of a shape is V = side3. Surface area is a measurement of the area of all
the faces of an object. A cube has 6 faces of equal area. Therefore, to find the surface area of a
figure you add up the area of the six faces. The formula for the area of a square is equal to s2 so
the formula for surface area would be s 2  s 2  s 2  s 2  s 2  s 2 . This could be rewritten as SA =
6 s2 .
For a cube with dimensions 2 x 2 x 2:
V  side
SA  6 s 2
3
SA  6(2cm) 2
V  (2cm)3
SA  6(4cm 2 )
V  8cm3
SA  24cm 2
In terms of volume, this means that for a cube of dimension 2 x 2 x 2 it would be filled up with 8
1 cm cubes.
For a cube with dimensions 3 x 3 x 3
3 cm
a
c
b
3cm
3cm
V  side
SA  6 s 2
3
SA  6(3cm) 2
V  (3cm)3
SA  6(9cm 2 )
V  27cm 3
SA  54cm 2
For s cube with dimensions 4 x 4 x 4
4 cm
a
b
c
4cm
4cm
SA  6s 2
V  side3
V  (4cm)
3
V  64cm3
SA  6(4cm) 2
SA  6(16cm2 )
SA  96cm2
The volumes of the cubes have all been cubes roots because the formula for the volume of a cube
cubes the length of the side. The volume of a cube with the dimensions of 17 x 17 x 17 should
be 173 which equals 4,913 cm3. The surface areas for the cubes are all multiples of 6. The
surface area if equal to a square times six. Therefore, for 17 x 17 x 17, surface area would equal
172 * 6. This equals 1734.
For a cube with dimensions s x s x s,
s
a
b
c
s
s
V = s3
SA = 6s2
Extensions of the Problem
Find the volume of an irregularly shaped large object. First, find a small plastic model of
that object - one that you wouldn't mind getting wet (like a model of a dinosaur). Next,
record the volume of the model using displacement. That is, fill a graduated cylinder about
half-way with water and record the volume; then, completely submerge the model into the
water and record the new volume. The volume of the model can be determined by
subtracting the original volume from the new volume. Finally, find a scale factor between
the model and the large object, and then use this scale factor and the model's volume to
compute the volume of the original object.
I am going to find the volume of a flip-flop. To do this I will use displacement to find the
volume of a smaller model of the object.
The volume of the water before adding the object – 5cm
The volume of the water after adding the object – 6.5 cm
Volume of the object found by subtracting the volume of the water before the object is added
from the volume of the water after adding the object: 6.5 cm – 5 cm = 1.5 cm3
Scale factor – the amount of the smaller objects inside the flip-flop.
The scale factor between these two lines would be approximately 3:1 (three for every
one).
The scale factor of the flip-flop to the smaller object is approximately 70. It is an approximation
because you cannot find an exact scale factor from looking and estimating between the sizes of
the flip-flops.
To determine the volume of the larger flip-flop you would multiply the scale factor by the
volume of the smaller flip flop.
Therefore, we are multiplying 70 by 1.5 cm3
The volume of the larger flip-flop is 105 cm3
Author & Contact
Tabitha Davis – middle grades cohort mathematics concentration
tabbydavis240@yahoo.com