An Inside Look at the
Cuboctahedron
Katlyn Johns
9B Geometry/Algebra with Transformations
Tuesday, March 5, 2013
Johns, Katlyn
9B GAT
5 March, 2013
With any three-dimensional figure, it can easily be chopped into pieces. Such as a cube,
the corners can be cut off at the edge’s midpoints to create a cuboctahedron. A cuboctahedron is
a three-dimensional shape with six, equal square faces and eight regular triangular faces. Since
the cuboctahedron is formed from a cube, the measurements of it can be found just by knowing
the cube’s edge length.
As any three-dimensional figure will, the cuboctahedron has surface area and volume.
Both of these measurements can be found by using the cube’s edge length. This can be done in
remembering a key point in the structure of the cuboctahedron: all of its vertices are the
midpoints of the original cube. Knowing this, the surface area can be found by first finding
where the faces of the cuboctahedron are located. Next, find the area of those faces and add them
up to find the total surface area.
The volume of the cuboctahedron is a little harder to find. A regular figure is defined as
equal sides and angles. The cuboctahedron is made up of regular three-dimensional figures, but
there are so many that the volume can be found in three different ways. This goes back to being
able to cut a three-dimensional shape into fragments; the cuboctahedron can be broken up many
ways or the area surrounding it can be broken up. Various pieces that can be made up of the
cuboctahedron and the surrounding area are pyramids and prisms.
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The cuboctahedron is made up by connecting all the midpoints of a cube and these
midpoints become the vertices to the cuboctahedron. To discover the edge length to the
cuboctahedron, the edge length of the cube it comes from must be known. The edge of a
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cuboctahedron is from one midpoint of a cube’s edge to the midpoint of an adjacent edge. The
edge length of this particular cube is 31.8 centimeters.
Figure 1. This is a face to the cube that the cuboctahedron comes from. The line connecting the
midpoints is an edge to the cuboctahedron.
A midpoint, as the name implies, is the point on a line segment halfway between the
endpoints. To find the distance from an endpoint to the midpoint, divide the edge length by two
(because the midpoint bisects the segment into two equal parts). With the edge length of this
cube being 31.8 centimeters, half would be 15.9 centimeters.
Figure 2. When the cube’s edge length is divided by two, 15.9 centimeters is the length of half
the segment.
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9B GAT
5 March, 2013
When two midpoints, an adjoining vertex, and a segment connecting the midpoints are
put together, a triangle is formed. This particular triangle is actually an isosceles right triangle.
The right angle from the cube’s face makes it a right triangle while the endpoint and two
midpoints form equal legs. By definition, a right isosceles triangle has two equal legs therefore
making the two leftover angles 45° each.
Figure 3. A right isosceles triangle is formed using two midpoints, an adjoining vertex, and a
segment connecting the midpoints (an edge to the cuboctahedron).
In an isosceles triangle, the hypotenuse is found by multiplying a leg by the square root
of two. The hypotenuse is this triangle, however, is an edge to the cuboctahedron. To find its
edge length, the leg (15.9 centimeters) must be multiplied by the square root of two. This can be
done using the formula h= l (√2), h being hypotenuse and l being leg.
Figure 4. The formula used to find the hypotenuse of an isosceles triangle, h being hypotenuse
and l being leg.
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9B GAT
5 March, 2013
When calculated, the edge to the cuboctahedron is 15.9√2 centimeters or 22.486
centimeters exactly.
Surface area is the area of all two-dimensional faces combined. The cuboctahedron has
eight equilateral triangle faces and six square faces. All the edges have the same length. Since all
the edges are the same length, each face is exactly the same as any other face that is the same
shape as itself.
The triangular faces of the cuboctahedron are equilateral triangles. This is because all of
the edges are the same length, each edge of it being 15.9√2 centimeters.
Figure 5. The triangular faces of the cuboctahedron are equilateral triangles with a side length of
15.9√2 centimeters.
In an equilateral triangle, all of the angles are also congruent making it also an
equiangular triangle. All of the angles are 60° in measure.
Figure 6. The triangular faces are also equiangular with each angle measuring 60°.
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9B GAT
5 March, 2013
To find the area of the triangular faces, the base length and height are needed. The base
length is already calculated because the base is any one of the sides. The height, however, is not
present. The height in an equilateral triangle can be found by dropping an altitude from an angle
(which cuts the angle’s measure in half) and making it perpendicular to the opposite side (which
also cuts the base length in half).
Figure 7. The altitude of an equilateral triangle cuts the angle and opposite edge in half.
Once the altitude is placed inside of the equilateral triangle, the triangle is cleanly cut in
half. Also, a 30°-60°-90° triangle is formed which makes it easier to find the height of the
cuboctahedron’s triangular face. Knowing that the altitude cuts the base in half, the original base
length can be divided by two to get the length of the short leg (leg opposite the 30° angle) in the
30°-60°-90° triangle.
Figure 8. The base length (15.9√2 centimeters) cut in half (7.95√2 centimeters) gives the length
of the short leg in the 30°-60°-90° triangle.
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9B GAT
5 March, 2013
Now that the length of the short leg has been found, the height of the equilateral triangle
can be found. The short leg is needed because the rule for a 30°-60°-90° triangle is that the
length of the long leg (the leg opposite the 60° angle) is the quantity of the short leg and the
square root of three. The formula that can be used for this is b=a√3, where b is the long leg and a
is the short leg.
Figure 9. The formula to find the long leg of a 30°-60°-90° triangle, b being the long leg and a
being the short leg.
The long leg has now been calculated which is also the height of the whole triangle. The
face’s area, being triangular, can be found using the formula 𝐴𝑇 = .5𝑏𝑇 ℎ𝑇 . 𝐴𝑇 being area of one
triangular face, 𝑏𝑇 being base of a triangular face, and ℎ𝑇 being height of one triangular face.
Figure 10. The formula used to find the area of a triangle. 𝐴𝑇 being area of one triangular face,
𝑏𝑇 being base of a triangular face, and ℎ𝑇 being the height of one triangular face.
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9B GAT
5 March, 2013
Now that the area of a triangular face is calculated, the area of a square face needs to be
determined. In a square, all of the edges are the same length.
Figure 11. A square face of the cuboctahedron with the side lengths labeled.
As with most quadrilaterals, the area formula is the quantity of the base and the height.
With squares though, the area can be calculated by doing an edge length squared. The formula
for this will be 𝐴𝑆 = 𝑒 2 . 𝐴𝑆 being area of one square face and e being edge of the square face.
Figure 12. Formula used to find the area of a square. 𝐴𝑆 being area of a square face and e being
edge length.
The area of a square face has now been found. Now that the areas of a single square face
and a single triangular face have been found, the surface area of the cuboctahedron can be found
using the formula 𝑇𝑆𝐴 = 6𝐴𝑆 + 8𝐴𝑇 , where TSA is total surface area, 𝐴𝑆 being area of a square
face, and 𝐴𝑇 being area of a triangular face. The area of the square face is multiplied by six
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because there are six square faces and the triangular faces are multiplied by eight because there
are eight triangular faces.
Figure 13. Formula used to find the total surface area of the cuboctahedron. TSA is for total
surface area, 𝐴𝑆 being area of one square face, and 𝐴𝑇 being area of a triangular face.
When calculated, the surface area for the entire cuboctahedron is 3,033.72+1,011.24√3
centimeters squared. The two numbers in the solution are separated because a number and a
number with a square root attached cannot be combined as like terms.
The cuboctahedron, capable of being made in many different formats, can have its
volume found in multiple ways. One way the volume can be found is to slice off a pyramid at
each corner of the original cube where the midpoints are. This will give the cuboctahedron its
shape.
Figure 14. Pyramids of the original cube (the darkened lines) being sliced off at the midpoints (a)
to create the cuboctahedron (b).
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9B GAT
5 March, 2013
When these pyramids are located, the volume can be found by subtracting the sum of all
eight of the pyramid’s volume from the original cube’s volume. The first component that needs
to be known is the volume of the original cube. This can be calculated by knowing a single edge
length because, by definition, all edges on a cube are congruent. The volume is found with the
formula 𝑉𝐶 = 𝑒 3 , where 𝑉𝐶 is volume of the cube and e is edge length. The edge length is cubed
because the volume of the cube is extended out an equal distance along the x, y, and z axes.
Figure 15. The formula used to find the volume of a cube. 𝑉𝐶 being volume of the cube and
being edge length.
When calculated, the volume of the cube that the cuboctahedron is contained within is
32,157.432 centimeters cubed.
The next part that needs to be determined in order to find the cuboctahedron’s volume is
the measurements and volume of a corner pyramid. The corner pyramid is completely made of
triangles; one of the triangles is a face to the cuboctahedron. This is true because when the
pyramid is separated from the cube by the midpoints, the midpoints of three adjacent edges
create an equilateral triangle.
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9B GAT
5 March, 2013
Figure 16. One face to the corner pyramid is an equilateral triangle formed by connecting three
adjacent midpoints. The darkened edges are the original cube.
The other three faces of the corner pyramid are right isosceles triangles. The right angle
comes from the cube’s face because all of the angles in a square are right angles. The two
congruent legs come from the edges to the cuboctahedron. All of its edges are congruent because
the midpoints are always the center of the edges making the distance between two midpoints
equidistant between any two midpoints. The last two angles would be 45° because a right
isosceles triangle has two 45° angles. These faces are made by connecting the midpoints of two
adjacent sides on the cube (an edge to the cuboctahedron) and linking these two same midpoints
to a vertex of the cube. There are three of these faces because three adjacent faces on a cube
share one vertex.
Figure 17. Three midpoints and a vertex of the cube form the vertices of the corner pyramid.
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9B GAT
5 March, 2013
The three isosceles triangles measurements can be found using the cube’s midpoints. A
midpoint is halfway between a segment’s endpoints so half of 31.8 (the cube’s edge length) is
15.9. The legs of the isosceles triangles are now calculated as 15.9 centimeters.
Figure 18. An unfolded version of the corner pyramid called a net. The legs of the isosceles
triangle are half of the cube’s edge length making it 15.9 centimeters.
The hypotenuses of these isosceles triangles are also the sides to the equilateral triangle
piece. The hypotenuse of an isosceles triangle is the quantity of a leg and the square root of two.
This gives the hypotenuse plus the sides to the equilateral triangle a length of 15.9√2 centimeters
(refer to Figure 4).
Figure 19. A net of the corner pyramid with all the lengths labeled.
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5 March, 2013
Now that the measurements of the corner pyramid have been determined, the volume can
now be found. Since all of the faces are triangles, anyone can be used as the base. Considering
the fact that three of the faces are isosceles triangles, the volume would be easiest to calculate
with one of these as a base. This is because the height of a pyramid is perpendicular to the base
and one of the angles in the isosceles triangle is 90°.
Figure 20. Two adjacent right isosceles triangles are perpendicular to another one used as a base.
The darkened segment would be the height.
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To find the volume of a pyramid, the formula would be 𝑉𝐶𝑃 = 3 𝐴𝑏 ℎ𝐶𝑃 . 𝑉𝐶𝑃 is volume of
a corner pyramid, 𝐴𝑏 is area of the base, and ℎ𝐶𝑃 is height of the corner pyramid. The area of the
base and the height are multiplied by one third because three pyramids put together make a prism
and the volume of a prism is found by multiplying the area of the base and its height.
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9B GAT
5 March, 2013
Figure 21. The formula used to find the volume of a pyramid. 𝑉𝐶𝑃 is volume of a corner pyramid,
𝐴𝑏 is area of the base, and ℎ𝐶𝑃 is height of the pyramid.
The area of the base is 126.405 centimeters squared because the base is a triangle and the
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area formula for a triangle is 𝐴𝑡 = 2 𝑏𝑡 ℎ𝑡 . 𝐴𝑡 is area of the triangle, 𝑏𝑡 is base length of the
triangle, and ℎ𝑡 is height length of the triangle.
Figure 22. The formula used to find the area of a triangle. 𝐴𝑡 is area of the triangle, 𝑏𝑡 is base
length of the triangle, and ℎ𝑡 is height.
The height of the corner pyramid is 15.9 centimeters because the height of the pyramid is
a leg to one of the isosceles triangles. The volume of a corner pyramid is now calculated as
669.9465 centimeters cubed.
Johns, Katlyn
9B GAT
5 March, 2013
As the volumes of the cube and a corner pyramid have now been confirmed, the volume
of the cuboctahedron can now be calculated. To find the volume, the sum of all the volumes of
all the corner pyramids needs to be subtracted from the volume of the cube. There are going to
be eight corner pyramids because there are eight corners in a cube. The volume formula will be
𝑉𝐶𝐵 = 𝑉𝐶 − 8𝑉𝐶𝑃 . 𝑉𝐶𝐵 is volume of the cuboctahedron, 𝑉𝐶 is volume of the cube, and 𝑉𝐶𝑃 is
volume of a corner pyramid.
Figure 23. One of the options that can be used to find the volume of the cuboctahedron where
𝑉𝐶𝐵 is volume of a cuboctahedron, 𝑉𝐶 is volume of the cube, and 𝑉𝐶𝑃 is volume of a corner
pyramid.
The volume of the cube is calculated by cubing the cube’s edge length (refer to Figure
15) which gives the cube a volume of 32,157.432 centimeters cubed. The volume formula of a
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corner pyramid is 𝑉𝐶𝑃 = 3 𝐴𝑏 ℎ𝐶𝑃 (refer to Figure 21) which gives a corner pyramid a volume of
669.9465 centimeters cubed. The volume of the cuboctahedron is calculated to be 26,797.86
centimeters cubed.
A second way to find the volume of a cuboctahedron is to take a rectangular prism and
attach rectangular pyramids to the lateral faces of the prism to create a cuboctahedron. The
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rectangular prism would be as tall as the cube and its bases vertices are the midpoints to a cube’s
face. The rectangular pyramid’s base is exactly the same as the prism’s lateral face and its apex
is a midpoint to one of the cube’s edges.
Figure 24. A rectangular prism and a rectangular pyramid placed inside constructing part of the
cuboctahedron within the cube where the darkened lines are the original cube.
The first part that needs to be found is the measurements and volume of the rectangular
prism. The height of the prism is already known. This is due to the fact that the prism is as tall as
the cube giving it a height of 31.8 centimeters. The side lengths of the base now need to be found
knowing its vertices lie on the midpoints of one of the cube’s faces. The edge length of the cube
is 31.8 centimeters and the midpoint is halfway so half of the segment is 15.9 centimeters long.
A single angle of a square face is 90°. Since there are two equal sides attached to the right angle
(these are equal because all of the edges on a cube are congruent making half a segment on each
edge equidistant) and the midpoints of two adjacent sides are connected by a side of the prism’s
base, it is a right isosceles triangle. The hypotenuse (the base edge) length is the quantity of a leg
and the square root of two making the base edge length 15.9√2 centimeters (see Figure 4). The
area of the base can now be found. The formula for this will be 𝐴𝑆 = 𝑏𝑆 ℎ𝑆 where 𝐴𝑆 represents
area of the square, 𝑏𝑆 is base length of the square, and ℎ𝑆 being height of the square.
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5 March, 2013
A simpler formula that can be used as well is 𝐴𝑆 = 𝑒 2 , 𝐴𝑆 is area of the square and e is
edge length. When all put together, the area of the prism’s base is 505.62 centimeters squared
(see Figure 12). The volume of the prism can now be determined. The formula to calculate the
volume of a prism is 𝑉𝑃 = 𝐴𝑏 ℎ𝑃 . 𝑉𝑃 is volume of the prism, 𝐴𝑏 is area of the base, and ℎ𝑃 is
height of the prism.
Figure 25. The volume used to find the volume of a rectangular prism. The representations being
as follows: 𝑉𝑃 being volume of the prism, 𝐴𝑏 being area of the base, and ℎ𝑃 being height of the
prism.
When calculated, the volume of the prism is 16,078.716 centimeters cubed. Now, the
volume and measurements of a rectangular pyramid are needed. As it has already been found
out, the base is a lateral face to the prism. Therefore, the base edge lengths to the pyramid are the
same as a lateral face on the rectangular prism.
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5 March, 2013
Figure 26. An unfolded version of the rectangular pyramid called a net with the base edge
lengths labeled.
As for the rest of the edges, they are edges to the cuboctahedron. These edges connect the
midpoints of two adjacent sides on the cube. So, after calculating, the edges are 15.9√2
centimeters (refer to Figure 4). This makes the lateral edges to the rectangular pyramid 15.9√2
centimeters as well.
Figure 27. A net of the rectangular pyramid with all edge lengths labeled.
If looked at more closely, the two lateral faces on the shorter sides of the base are
equilateral triangles because all of the sides equal 15.9√2 centimeters. Then, if the two lateral
faces of the longer sides of the base are observed, they are isosceles triangles because two of the
sides equal 15.9√2 centimeters. To find the volume of the rectangular pyramid, the formula
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1
would be 𝑉𝑟𝑝 = 3 𝐴𝑏 ℎ𝑟𝑝 . 𝑉𝑟𝑝 is volume of the rectangular pyramid, 𝐴𝑏 being area of the base, and
ℎ𝑟𝑝 is height of the rectangular pyramid. The area of the base and the height is multiplied by one
third because tree pyramids pieced together create a prism, where the volume is area of the base
times height.
First, the area of the base needs to be found. This can be done by multiplying the base
length by the base height.
Figure 28. Formula used to determine the area of a rectangle. 𝐴𝑟𝑝𝑏 is base area of the rectangular
pyramid, 𝑏𝑙 is base length, and 𝑏ℎ is base height.
When this is completed, the area of the base is 505.62√2 centimeters squared. Now the
other component needed to find the volume of the rectangular pyramid is its height. This can be
found by forming a right triangle within the pyramid where a leg is the height, another leg is half
the base length, and the hypotenuse is the slant height of the pyramid.
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5 March, 2013
Figure 29. A triangle can be found within the pyramid where the height of the pyramid is
included as a leg.
Now that a triangle has been formed, the Pythagorean Theorem (the theorem that states
that the sum of each leg squared is equal to the hypotenuse squared) can be used to find the
height of the pyramid. All that needs to be done beforehand is finding the slant height of the
pyramid (the height of a triangular face) and half the length of the base edge. For the leg of the
triangle that is half the base length, it is running parallel to the longer base edge so half of the
longer leg is 15.9 centimeters. One leg of the triangle has now been found. The slant height that
will be used is that of the equilateral triangle face.
Figure 30. An equilateral triangular face of the pyramid where the face’s slant height (x) will be
the hypotenuse of the triangle within the pyramid.
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9B GAT
5 March, 2013
To find the height, the triangle can be cut in half along its height. This will create two
30°-60°-90° triangles because an equilateral triangle is also equiangular making each angle 60°.
When the height is placed in, it cuts an angle and the angles opposite side in half at a 90° angle.
The angle that is cut in half becomes 30° and the 90° angle comes from the height being
perpendicular to the triangles edge (refer to Figure 7). The side that is cut in half becomes 7.95√2
centimeters (refer to Figure 8). In a 30°-60°-90° triangle, the long leg is the quantity of the short
leg and the square root of three. This makes the height of the triangle 7.95√6 centimeters (see
Figure 9). The slant height of the pyramid and the hypotenuse of the internal triangle is now
found to be 7.95√6 centimeters long.
Figure 31. The triangle formed within the rectangular pyramid. The height of the pyramid is
included as a leg.
Now that two of the triangles sides have been determined, the Pythagorean Theorem can
now be applied to find the height of the rectangular pyramid. The formula for this theorem would
be 𝑎2 + 𝑏 2 = 𝑐 2 , where a and b are legs of the right triangle and c is the hypotenuse. The
Pythagorean Theorem can only be used with right triangles. The triangle within the pyramid is a
right triangle because the height of a pyramid is always perpendicular to its base.
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5 March, 2013
Figure 32. The Pythagorean Theorem is used to find the height of the pyramid. a and b are legs
of a right triangle while c is the hypotenuse.
When the Pythagorean Theorem is through with, the height of this pyramid is calculated
to be 7.95√2 centimeters. The height of the pyramid and the area of the base have now been
calculated allowing the volume of the rectangular pyramid to be found. The volume formula will
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be 𝑉𝑟𝑝 = 3 𝐴𝑏 ℎ𝑟𝑝 where 𝑉𝑟𝑝 is volume of a rectangular pyramid, 𝐴𝑏 is area of the base, and ℎ𝑟𝑝 is
height of the rectangular pyramid. The area of the base and the height are multiplied by one third
because three pyramids together form a prism where the volume is the quantity of the area of the
base and height.
Figure 33. Formula used to determine the volume for a rectangular pyramid. 𝑉𝑟𝑝 is volume of a
rectangular pyramid, 𝐴𝑏 is area of the base, and ℎ𝑟𝑝 is height of the rectangular pyramid.
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9B GAT
5 March, 2013
After the formula is put into effect, the volume of a single rectangular pyramid is
2,679.786 centimeters cubed. The volume of the cuboctahedron can again be found by this time
attaching a rectangular pyramid to each lateral face of the rectangular prism. There will be four
rectangular pyramids because a rectangular prism has four lateral faces. The volume formula for
this cuboctahedron will be 𝑉𝐶𝐵 = 𝑉𝑃 + 4𝑉𝑟𝑝 . 𝑉𝐶𝐵 is volume of the cuboctahedron, 𝑉𝑃 is volume
of the prism, and 𝑉𝑟𝑝 is volume of a rectangular pyramid.
Figure 34. The formula that was used to find the volume of the cuboctahedron, this time
determining the volume with a different method, 𝑉𝐶𝐵 is volume of the cuboctahedron, 𝑉𝑃 is
volume of the prism, and 𝑉𝑟𝑝 is volume of a rectangular pyramid.
When all values are plugged in, the volume of the cuboctahedron is still 26,797.86
centimeters cubed. This is due to the fact that the same edge length of the original cube is being
used. The volume of the cuboctahedron will always be the same if the original cube is the same
size every time.
The last, but not least, way to find the volume of a cuboctahedron is to construct it out of
eight tetrahedrons and six square pyramids. There are eight tetrahedrons because the base is a
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single triangular face to the cuboctahedron. The same goes for the square pyramids except they
are the square faces of the cuboctahedron.
Figure 35. Tetrahedrons and square pyramids forming a cuboctahedron. Only one of each is
shown.
To find the volume of the cuboctahedron, the volume of all the pyramids would be added
together. So the volume of only one tetrahedron and one square pyramid are needed for now. The
volume and dimensions will be found for the square pyramid first. The vertices of the base lie on
the midpoints of the cube’s edges of one face. The edges of the base connect two midpoints
together. Since the midpoints are the middle of each edge and each angle of a cube’s face is 90°,
an isosceles right triangle is formed. Half of each edge is 15.9 centimeters making the
hypotenuse of each isosceles triangle 15.9√2 centimeters. This is because the hypotenuse is an
isosceles triangle is the quantity of a leg and the square root of two (refer to Figure 4). So each
edge to the square pyramid’s base is 15.9√2 centimeters. The area of the base can now be found
using the formula 𝐴𝑆 = 𝑒 2 . 𝐴𝑆 is area of a square face and e is edge length. The area of the base
will be 505.62 centimeters squared (see Figure 12). To find the volume of the square pyramid,
the height of the figure is still needed. In this cuboctahedron, however, all of the apexes of the
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square pyramids meet up in the very center of the cube. The heights of opposite square pyramids
align to form a straight line running through the middle of the cube.
Figure 36. Opposite square pyramids meet at the apex with the heights aligning in a straight line.
Since the square pyramids meet halfway, the height of the pyramid is half the height of
the cube, which is 31.8 centimeters. The height of the square pyramid is 15.9 centimeters. The
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volume of the square pyramid can now be found. The formula for this will be 𝑉𝑆𝑃 = 𝐴𝑏 ℎ𝑆𝑃
3
where 𝑉𝑆𝑃 is volume of the square pyramid, 𝐴𝑏 is area of the base, and ℎ𝑆𝑃 is height of the
square pyramid. The area of the base and the height is again multiplied by one third because
three pyramids form a prism.
Figure 37. Volume formula for the square pyramid where represents 𝑉𝑆𝑃 is volume of the square
pyramid, 𝐴𝑏 is area of the base, and ℎ𝑆𝑃 is height of the square pyramid.
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9B GAT
5 March, 2013
When all values are plugged in, the volume of the square pyramid is determined to be
2,979.786 centimeters cubed. The edge lengths and volume must now be calculated for the
tetrahedron so the cuboctahedron volume can be found. By definition, all of a tetrahedron’s
edges are equal in length; so all four of its faces are equilateral triangles. Its edges also align with
those of the square pyramid so its edge length is 15.9√2 centimeters. To find the area of the base,
the height of the base needs to be found. To do this, the triangle can be cut in half to form two
30°-60°-90° triangles. The base gets cut in half in the process (refer to Figure 8) making the short
leg 7.95√2 centimeters. The long leg of the 30°-60°-90° triangle is the quantity of the short leg
and the square root of three giving the long leg and the height of the triangle 7.95√6 centimeters
(see Figure 9). The area of the base can now be calculated using the formula 𝐴𝑇 = .5𝑏𝑇 ℎ𝑇 (refer
to Figure 10) to get a base area of 126.405√3 centimeters squared. To find the volume of the
tetrahedron, the height of the figure is still needed. The height of a tetrahedron can be found
1
using the formula ℎ𝑡 = 3 𝑒√6. In the formula, ℎ𝑡 is height of a tetrahedron and e is edge length.
Figure 38. The formula used to determine the height of a tetrahedron.
The height of the tetrahedron is now calculated as 10.6√3 centimeters. The tetrahedron’s
1
volume can now be found using the formula 𝑉𝑡 = 3 𝐴𝑏 ℎ𝑡 . 𝑉𝑡 is volume of the tetrahedron, 𝐴𝑏 is
Johns, Katlyn
9B GAT
5 March, 2013
area of the base, and ℎ𝑡 is height of the tetrahedron. The area of the base and the height is yet
again multiplied by one third because three pyramids form a prism.
Figure 39. Volume formula for a tetrahedron where represents 𝑉𝑡 is volume of the tetrahedron,
𝐴𝑏 is area of the base, and ℎ𝑡 is height of the tetrahedron.
When all the values are inserted, the volume of the tetrahedron is 1,339.893 centimeters
cubed. Actually, this could have been calculated sooner because in the third step, the √9 (which
equals 3) canceled out the one third. What was left over would have been the volume of the
tetrahedron as well. The volumes of the tetrahedron and square pyramid have both been found
now, so the volume of the cuboctahedron can be found once more. This will be done by
multiplying the volume of the tetrahedron by eight and the square pyramid’s volume by six. This
is because there are eight tetrahedrons and six square pyramids. The volume formula will be
𝑉𝐶𝐵 = 8𝑉𝑡 + 6𝑉𝑆𝑃 where 𝑉𝐶𝐵 is volume of the cuboctahedron, 𝑉𝑡 is volume of the tetrahedron,
and 𝑉𝑆𝑃 is volume of the square pyramid.
Johns, Katlyn
9B GAT
5 March, 2013
Figure 40. Volume formula used to calculate the volume of the cuboctahedron using tetrahedrons
and square pyramids to construct it. 𝑉𝐶𝐵 is volume of the cuboctahedron, 𝑉𝑡 is volume of the
tetrahedron, and 𝑉𝑆𝑃 is volume of the square pyramid.
The volume of the cuboctahedron is again calculated as 26,797.86 centimeters cubed.
This again proves that if the same edge length is used, the same volume will be found for the
cuboctahedron, no matter what method is used.
***
The cuboctahedron is a three-dimensional figure with six, equal square faces and eight,
equal triangular faces. All of its vertices are the midpoints of the cube that the cuboctahedron is
contained within. The vertices of the triangular faces on the cuboctahedron are the midpoints on
three adjacent edges of the cube. The vertices of the square faces, however, are the midpoints of
a face on the cube.
The vertices of the cuboctahedron always lie on the cube’s midpoints. The midpoint is
defined as the middle point on a line segment, so unless the cube’s edge length changes, the
midpoint will always be the same. Also, a cube is defined as having all of its edges the same
length, so all of the midpoints are in the same spot on each edge. Therefore, the sides to the
Johns, Katlyn
9B GAT
5 March, 2013
cuboctahedron will stay the same as well because the edges connect the cube’s midpoints. In all
the ways that the volume is found, the pyramids and prisms rely on the midpoints and the faces
of the cuboctahedron to keep their shape. If the midpoints of the cube were to change, all of the
figures used to find the volume would vary and the volume would change every time.
Many times, the height of the pyramids could have been calculated different ways. Issues
occurred when deciding which method was best or most exact. Some ways that could have been
used are the Pythagorean Theorem and the rules of 30°-60°-90° triangles. The problem with the
Pythagorean Theorem is that square roots are not always rational or they are weird decimals. The
30°-60°-90° rules are not always accurate, either, because people cannot always be sure if it even
is a 30°-60°-90° triangle.
Issues will always occur, but no matter what method is used, the volume of a
cuboctahedron will remain the same if the same cube edge length is used with each method.
Different methods are preferred more than others, but the same answer will be found each time.
The end result will never change because the definition of a midpoint does not change. All the
edges in a cube are equal length so a midpoint will not change.