The Cuboctahedron
The Volume and Surface Area of a Cubo
Brian Nackoud
Omar Wajeeh
February 28, 2013
Nackoud-Wajeeh 1
So you just finished your cube inside a cube, but you suddenly get intrigued by these 3
dimensional figures called the platonic solids. There are five of them, the cube, the
tetrahedron, the octahedron, the icosahedron, and the dodecahedron. Your favorite is
the cube and the octahedron. Your teacher introduces you to a concept named semiplatonic solids, and one of these is the cuboctohedron (cubo for short). You’re suddenly
excited for the project until you realize something: how will you measure it? You were
given the length of your cube, 12.6 inches, and 3 cases, but not the length of the cubo.
You’ve gotten frustrated, and beating yourself up, but don’t worry. I’m here to help.
Together we’ll find the surface area, volume and dimensions for every case.
The first step to solving the cases is actually finding the measurements of the
cubo. Each of the faces of the cubo is either equilateral triangles or squares. This
makes finding the lengths easier, since finding one length will find them all. To get one
square face of the cubo, draw 4 points on the midpoint of each side of the cube’s
square face. That makes a square with a vertex at 6.3 inches of your square face of the
cube. This creates four 45, 45, 90 triangles on each original vertex of the cube. This
gives you 2 legs of the triangle, 6.3 inches. Knowing 45, 45, 90 triangle properties you
can deduce that the hypotenuse is 6.3√2 inches. Finding the hypotenuse of the triangle
gives us all the sides of your square face: 6.3√2 inches.
Now, it’s time to find the dimensions of the equilateral triangle face. Since each
length should be consistent in a semi-platonic solid, you can confidently say that each
side of the triangular face is 6.3√2 inches. However to find area of a triangle, you must
find the height. The height would make it two 30, 60, 90 triangles. The side would be
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3.15√2 inches, and knowing properties of 30, 60, 90 triangles, you know the big leg is
3.15√6 inches. The formula for the area of all triangle faces is as goes:
1
× 6.3√2 × 3.15√6 = 9.9225√12.
2
9.9225√12 × 8 = 79.38√12
Now, it’s time for the area of the squares:
2
6.3√2 = 79.38
79.38 × 6 = 476.28
This, if you add up the two areas equals 476.28+79.38√12 inches2.
Case 1:
Figure 1: A Cube with Corners Cut Off at the Midpoints of the Edges
Figure 1 shows the layout of case one. You can see that it is just a cube that has
8 triangles that are cut off of the corners. All of these triangles are known to be equal
since they are cut off by the midpoints of the edges.
A cube with the corners cut off on the midpoints is a cuboctahedron itself. This
creates 6 right isosceles pyramids. To easily find the volume, you could simply find the
volume of the cube and subtract it from the 8 pyramids. The midpoint of the cube is 6.3
inches, meaning the 2 legs are 6.3 inches. Knowing properties of 45, 45, 90 triangles,
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you know that the hypotenuse is 6.3√2 inches. The hypotenuses together create an
equilateral triangle. Because of this, every side on the equilateral triangle is 6.3√2
inches.
Now, the volume should be easier, since you already found the dimensions.
1
3
×
6.3×6.3
2
× 6.3 = 41.6745
41.6745 × 6 = 333.396
This means all of the 8 right pyramids combined equal 333.396 inches3
12.63 = 2000.376
2000.376 − 333.396 = 1666.98
This series of equations show the volume of the cut-off pieces (333.396 inches3),
being subtracted from the volume of the cube (2000.376 inches3) to get the volume of
the cubo as 1666.98 inches3.
Case 2:
A right square prism with four rectangular pyramids attached on the lateral faces.
Figure 2: Rectangular Prism and Rectangular pyramid.
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Figure 2 show what we will be using to determine the volume of this case. We
will be using 1 rectangular prism and 4 rectangular pyramids. To do this, you get the
volume of the prism then find the volume of the 4 pyramids and add them all together.
First, you have to find the dimensions of the rectangular prism. The rectangular
prism is the same height as the cube, so the long length is 12.6 inches. The vertices
making the square base can be put on the midpoint of each side of the corner, making a
45, 45, 90 triangle. Since half of 12.6 is 6.3, and that you know properties of special
right triangles, you know that the sides making the square are 6.3√2 inches. So, each
square base is 6.3√2 by 6.3√2, and each rectangle face is 12.6 by 6.3√2.
(6.3√2)2 = 79.38
79.38 × 12.6 = 1000.188
Now to find the volume of the 4 triangular pyramids. The sides of the base of the
pyramid line up with the prism meaning the shorter side of the rectangle is 6.3√2 in and
the longer sides are 12.6 in. To get the length of the slant height of triangle on the
shorter side, and since this is an equilateral triangle, all of the angles are congruent
meaning they all equal 60 degrees. When going for the height of the triangle make a
line from the top vertex down to the opposite side such that it is perpendicular to the
opposite side. This made a 30-60-90 triangle. You can then use the 30-60-90 triangle
method to get the height. The shortest length on the bottom would be half of the side
length making it 3.15√2 in. You then multiply that by √3 in to get the slant height. This
leaves you with 3.15√6 in. Now all that is left to find is the height of the pyramid. To find
this you need the slant height of one of the lateral triangles, and half of the length of the
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long side of the base, which is 6.3 in. If you use the Pythagorean Theorem to get the
height it would look like this:
𝑐 2 = 𝑎2 + 𝑏 2
3.15√62 = 6.32 + 𝑏 2
59.535 = 39.69 + 𝑏 2
√19.845 = 𝑏
Now that we know that the height is √19.845 in we can find the volume of the
pyramids. First we should find the area of the base, which is solved like this:
12.6 × 6.3√2 = 79.38√2
If we multiply this by the height we should get the volume of just one of the
pyramids.
79.38√2 × √19.845
= 166.698
3
So now multiply it by 4 to get the volume of all of the pyramids.
166.698 × 4 = 666.792
And when we add this to the volume of the prism, we would get this:
666.798 + 1000.188 = 1666.98
So this shows us that the volume of all of the shapes combined is equal to
1666.98 in3.
Case 3:
Now for the third and final method of finding the volume of the cubo. This case
includes the use of eight tetrahedrons and 6 regular square pyramids. To find the
volume of the tetrahedron use the formula:
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𝑉=
1
×𝑏×ℎ
3
The sides of the tetrahedron are just the side of the cubo, since all sides of a
tetrahedron are congruent. This means all sides are 6.3√2 in. To find the height of a
triangle on the tetrahedron is simple. Since this is an equilateral triangle, all of the
angles are congruent meaning they all equal 60 degrees. When going for the height of
the triangle make a line from the top vertex down to the opposite side such that it is
perpendicular to the side. This made a 30-60-90 triangle. You can then use the 30-6090 triangle method to get the height. The shortest length on the bottom would be half of
the side length making it 3.15√2 in. You then multiply that by √3 in to get the slant
height. This leaves you with 3.15√6 in. The formula for the base of the tetrahedron is
shown as:
𝐴=
1
×𝑏×ℎ
2
Now plug in the side length for the “b” and the slant height for the “h”. You should have
this:
𝐴=
1
× 6.3√2 × 3.15√6
2
When this is solved out you should get 9.9225√12 in2.
Now that you have the “b” in the volume equation for the tetrahedron it’s time to
find the height of it. You can find it by using the Pythagorean Theorem which looks like
this:
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𝑐 2 = 𝑎2 + 𝑏 2
The “c” is the hypotenuse of a triangle. The hypotenuse in this case would be the slant
height. The “a” would be represented as a third of the slant height since the height of
the tetrahedron touches the centroid of the base. Here is how to solve the equation:
2
3.15√6 = (
3.15√6 2
)
3
+ 𝑏2
59.535 = 6.615 + 𝑏 2
√52.92 = √𝑏
2
√52.92 = 𝑏
Here you can see that the height of the tetrahedron is √52.92 in. This is what the
lengths should end up being:
Height=
√52.92cm
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Figure 3:
Figure 1 shows what the length is for each measurement. You can see that the
slant height is 3.15√6 in, the height of the tetrahedron is √52.92 in, and all of the sides
are equal to 6.3√2 in. The bottom side of the triangle that is on the base of the
tetrahedron is 3.15√6/3 in.
Now we can calculate the volume of the tetrahedron. Using the equation for the
volume above, we can plug in √52.92 in for the “h” and 19.845√12 in2 for the “b”. Here is
how to solve it out:
𝑉=
1
× 9.9225√12 × √52.92
3
𝑉=
1
× 9.9225√635.04
3
𝑉=
1
× 250.047
3
𝑉 = 83.349
So now we know that the volume of the tetrahedron we should multiply it by eight since
there are 8 tetrahedrons.
𝑉 = 83.349 × 8
𝑉 = 666.792
This means that the sum of all of the tetrahedrons combined is equal to 666.792 in3.
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Now that is only the volume for the 8 tetrahedrons. It is now time to get the
volume of the 6 square pyramids. The lengths of all of the sides in this pyramid are also
equivalent to the sides of the cubo. Using the same method as the tetrahedron to find
the slant height, you can conclude that the slant height is 3.15√6 in. To find the height of
the pyramid you just use the Pythagorean Theorem again. This time, since the base is a
square, the “a” would be substituted for 3.15√2 in, which is half of the side length. Here
is how to solve this one out:
𝑐 2 = 𝑎2 + 𝑏 2
2
(3.15√6) = (3.15√2)2 + 𝑏 2
59.535 = 19.845 + 𝑏 2
√39.69 = √𝑏 2
√39.69 = 𝑏
At this point we have all of our lengths. Here is how it should look like so far:
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cm
Figure 4: Square pyramid
Figure 4 shows what the length is for each measurement. You can see that the
slant height is 3.15√6 in, the height of the square pyramid is √39.69 in, and all of the
sides are equal to 6.3√2 in. The bottom side of the triangle that is on the base of the
square pyramid is 3.15√6 in.
Now that we have the height of the square pyramid, all we need is the area of the base.
To find the area of the base, which is a square, all you do is square the side length. In
this case it is 6.3√2 in. So if we square 6.3√2 in, you should get 79.38 in. Now to find the
volume of just one of the square pyramids, use the same equation as before. Here is
how to solve it:
𝑉=
1
× 79.38 × √39.69
3
𝑉=
1
× 500.094
3
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𝑉 = 166.698
So now we know that the volume of the square pyramid we should multiply it by 6 since
there are 6 square pyramids.
𝑉 = 166.698 × 6
𝑉 = 1000.188
This means that the sum of all of the square pyramids combined is equal to
1000.188 in3.To get the final volume for the cubo, add the volume of the 8 tetrahedrons
to the volume of the 6 square pyramids. It would look like this:
𝑉 = 1000.188 + 666.792
𝑉 = 1666.98
This means that the final volume for this case is 1666.98 in3.
In conclusion the volume of a cuboctahedron or “cubo” for short can be found in
three ways. The first way of cutting 8 corners off of the cube, the volume was 1666.98
in3. The second way of finding the volume of a square pyramid and then adding the
volume of 4 rectangular pyramids was also 1666.98 in3. The third way is to find the
volume of 6 square pyramids and 8 tetrahedrons this also turned out to be 1666.98 in3.
The reason they were all the same is because all of the cases were used to figure out
the same thing except with different methods. The Total Surface Area was
476.28+79.38√12 cm² which was found by finding the individual surface areas of the
cubes and triangles adding those together.
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Sources
http://www.math.wichita.edu/history/Images/tetrahedron2.gif
http://www.applet-magic.com/cuboctahedron.gif
http://kevin6p.edublogs.org/files/2012/10/square_pyramid-z38fqv.gif