Steven Prascius
9A/GAT
A cuboctahedron, or cubo, is a polyhedron with six square faces and eight triangle faces.
The cubo has 12 vertices, 24 edges, and can fit inside of a cube. The cubo is basically a cube
with the corners cut off. To find an edge of the cubo you must connect the midpoints of each
edge of the cube, forming the cubo. There are many ways to find the volume for the
cuboctahedron.
The length of one edge of my cube was 18cm. Since one vertex of the cubo bisects one
edge of the cube, it creates two equal segments on each edge of the cube. Each of the bisected
segments are 9cm because the original edge was 18cm and the new segment is half of that, or
9cm. Two of these equal segments forms a 45-45-90 triangle at a corner of the cube where the
two equal legs are x and the hypotenuse is x 2 .I found the measurements of the legs of this
triangle by knowing that each leg was exactly half of the cube’s edge, which is 18 cm, meaning
the legs were both 9 cm. I then used the formula H=L ( 2 ), or hypotenuse= leg (square root of
2), to find the hypotenuse of the 45-45-90 triangle. By using this formula, H=9( 2 ), I found that
the length of the hypotenuse was 9 2 cm. The hypotenuse for this triangle is the edge of the
cubo which means that one edge of my cubo is 9 2 cm.
9
9 cm
2 cm
9 cm
To find the total surface area of the cubo, you must find the area of both the eight triangular
faces and the six square faces of the cubo. In order to find the area of the square face, I used
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Steven Prascius
9A/GAT
the formula A= (B) (H), or Area= (Base) (Height). Since each edge of my cubo was 9 2 cm, the
formula was A= (9 2 ) (9 2 ), which comes out to equal 81 4 , or 162 cm2. The area for one
of the square faces of my cubo is 162 cm2. To find the area of all the square faces, I then
multiplied the area of one square face, 162 cm2, by six because of the total six square faces.
This came out to 972 cm2. The total surface area of the square faces is 972 cm2. I then found
the area of the triangular face of my cubo. In order to do this I had to find the height of the
triangular face. The triangle face is an equilateral triangle, which means it can be divided into
two 30-60-90 triangles. The base of the 30-60-90 triangle is x, the hypotenuse is 2x and the
other leg, or height, is x 3 . Since the hypotenuse is 9 2 cm, the base of the triangle is 9 2
divided by two, or 4.5 2 cm. Therefore the height of the triangle is 4.5 2 times 3 , or 4.5 6
cm. I then used the triangle area formula A= 1/2(B) (H), or Area=one half (Base) (Height), to find
the area of the triangle. This comes out to be A= 1/2(9 2 ) (4.5 6 ) which equals 81 3 cm2.
From here I multiplied the area of one triangular face, 81 3 cm2, by eight because of the eight
triangular faces. This came out to equal 648 3 cm2. The total surface area for the triangular
faces is 648 3 cm2.
9 2 cm
4.5 6 cm
9 2 cm
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Steven Prascius
9A/GAT
The total surface area of the cubo is the sum of all the faces, which is the total surface area of
the square faces added to the total surface area of the triangular faces, which equals 972 cm 2 +
648 3 cm2. The total surface area for the whole cubo is 972 cm2 + 648 3 cm2.
In Case 1 I had to find the dimensions and volume for the corner pyramid of the cubo. I
found the dimensions of the corner pyramid by using the formula for a 45-45-90 triangle where
the legs are x and the hypotenuse is x 2 . Since the sides of the base are the edges of the cubo,
I knew they each were 9 2 cm. The three triangles around the base are 45-45-90 triangles so
each leg is 9 cm because the hypotenuse is the edge of the cubo, which is 9 2 cm, and the
formula for a 45-45-90 triangle says that legs are x while hypotenuse is x 2 , meaning the two
legs of the 45-45-90 triangle are each 9 cm.
s
s
s
9 2 cm
9 cm
The volume of the pyramid can be found by using the formula V=1/3 (area of base) (Height). In
order to use this formula, I had to find the area of the base. I used the 45-45-90 triangle
formula where hypotenuse is x 2 , and legs are x. Both of the legs are 9 cm because the
hypotenuse was 9 2 cm. The formula for area is A=1/2 (B) (H), or Area= one half (base)
(height), which was 1/2 (9) (9), or 40.5 cm2. Then I found the height of the pyramid was 9 cm
because the corner pyramid is a right triangle, making the slant height of 9cm the height. I then
plugged in the area and the height into the volume formula V=1/3 (area of base) (Height) to get
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Steven Prascius
9A/GAT
V=1/3 (40.5) (9 cm), or 121.5 cm2. 121.5 cm2 is the volume of the corner pyramid. Then to get
the the total volume of the cubo, I had to find the total volume of the cube. To get this volume
I used the formula V=Ab (H), or Volume=Area of the base (Height). The area of the base of the
cube was 18 cm multiplied by 18 cm because each edge of the cube was 18 cm. This came out
to equal 324 cm2, which then is put into the volume formula to get v= 324(18), or 5832 cm3.
Then to get the volume of the cubo I had to subtract 8 times the volume of the corner pyramid
by the volume of the entire cube because there are eight triangular corners. So, the volume of
the cubo was 5832 cm3-(121.5 cm3)8, which equals 4860 cm3. The total volume of my cubo is
4860 cm3.
Corner Pyramid
In Case 2 I had to find the dimensions of the square prism and rectangular pyramid. The
square prism is the center piece of the cubo, meaning it has a height of 18 cm because it
stretches from the top of the cube down to the bottom. This means it is the same length as one
of the cube’s edges. The base of the prism is 9 2 cm because it is the square face of the cubo,
meaning it has edges of 9 2 cm. The base is a square so, I used the formula A=B (H), or Area=
Base (Height) to find the area. Since all sides of the square are the same, the area was A= 9 2
(9 2 ), or 162 cm2. The height of the prism was 18 cm so I used the formula V= (Ab) (H) to find
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Steven Prascius
9A/GAT
volume. This came out to be V= (162) (18), or 2916 cm3. The volume of the square prism is 2916
cm3.
Square
Prism
18 cm
9 2 cm
The rectangular pyramid’s base is connected to the lateral sides of the square prism. Therefore
the base has a height 18 cm and a base of 9 2 cm. To find the area of the base I used the formula
A=B (H), or Area= Base (Height). I plugged my height of 18 cm and my base of 9 2 cm in the area
formula to get the area of 162 2 cm2. To find the height of the pyramid, I used Pythagoreans
theorem, which is a2+b2=c2. Before I used that, though, I had to find the height of the equilateral
triangles that make up two parts of the rectangular pyramid. Two of the triangles are equilateral
triangles with side lengths of 9 2 cm, and to find the height of these triangles I used the 30-60-90
triangle formula, were the hypotenuse is 2x, base is x, and height is x 3 . When I divided the triangle
into two parts the base was 4.5 2 cm, hypotenuse was 9 2 cm, and height was 4.5 6 cm. The two
other two triangles were 45-45-90 triangles when I divided them in half, where the legs are x and the
hypotenuse is x 2 . This means that both legs are 9 cm because they are half of the base height, which
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Steven Prascius
9A/GAT
is 18 cm; this means that the hypotenuse is 9 2 cm. After I found my height and dimensions, I used the
Pythagorean Theorem, where a2and b2 are the legs, and c2 is the hypotenuse.
C2
2
B
Rectangular pyramid
A2
The hypotenuse of the triangle would be the height of the equilateral triangle. One of the legs would be
half the measure of the height of the base, and the other leg would be the height of the whole pyramid.
My hypotenuse was 4.5 6 cm, my one leg was 9 cm, and the other leg is b2 because it is the height I am
trying to find. These values make 92+b2=4.5 6 2, which simplifies to 81+b2=121.5, then to b2=40.5, then
b= 20.25 * 2 , and finally to b=4.5 2 . The height of the pyramid is 4.5 2 cm. After finding the height
I was able to use the volume formula v= 1/3(Ab) (h), or Volume=one third (Area of base) (height of the
pyramid), to find the volume of the pyramid, which is V=1/3(162 2 ) (4.5 2 ), or 486 cm3. Since there
are four rectangular pyramids that make up the cubo in case 2, I had to multiply the volume of one
pyramid by four, which is 1944 cm3. The volume for all four rectangular pyramids is 1944 cm3. To find
the volume of the entire cubo, I had to add the volume of the square prism to the total volume of all the
rectangular prisms. This would be 2916 cm3+ 1944 cm3, which equals 4860 cm3. The volume of the cubo
is 4860 cm3.
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Steven Prascius
9A/GAT
9 2 cm
b
a
9 2 cm
a
18 cm
18 cm
9 2 cm
In Case 3 I had to find the dimensions of the tetrahedron, and the regular square pyramid. The
Tetrahedron is one of the triangular faces of the cubo, and is made up of four equilateral triangles.
Since the base is the triangular face of the cubo, its edges are 9 2 cm, and because it is made up of four
equilateral triangles all sides are 9 2 cm. The area of the base can be found by using the 30-60-90
triangle formula where the base is x, hypotenuse is 2x, and the height is x 3 . When I divide the base
triangle in half I get 4.5 2 cm for the base length, 9 2 cm for the hypotenuse, and 4.5
6 cm for the
height.
Side
9 2 cm
H
4.5 6
cm
4.5 2 cm
Then to find the area of the base I used the formula A=1/2 b (h), or Area=one half base (height), which
came out to be A=1/2 (9 2 ) (4.5 6 ), or 40.5
3 cm2. To find the height of the tetrahedron I used the
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Steven Prascius
9A/GAT
formula H=1/3(edge) 6 , or height of tetrahedron=one third (one edge) (square root of six), to get a
height of H=1/3(9 2 ) 6 , or 6 3 cm. Then I plugged in the height of tetrahedron, 6 3 cm, and the
area of the base, 40.5 3 cm2, into the formula v=1/3(Ab) (h), or volume=one third (area of base)
(height of tetrahedron), to get a volume of 243 cm3. The volume of the tetrahedron is 243 cm3. The
regular square pyramid is the square base of the cubo, meaning the base has an edge length of 9 2 cm.
All of its edges are 9 2 cm because the four triangles that are connected to the square base are
equilateral triangles. To find the area of the square base I used the formula A=b (h), or area= base
(height), to get A= 9 2 (9 2 ), or 162 cm2. To find the height I had to use the Pythagorean Theorem,
but first I had to find the height of the triangle sides. I did this by using the 30-60-90 triangle formula,
where the base is x, hypotenuse is 2x, and height is x 3 . When I divided the equilateral triangle in half I
found 4.5 2 cm to be the base, 9 2 cm to be the hypotenuse, and the height to be 4.5 6 cm. I then
put the height and one leg, which is 4.5 2 cm or half of base edge, into the Pythagorean Theorem
which came out to be 4.5 2 2+b2=4.5 6 , which simplifies to 40.5+b2=121.5, then to b= 81 , and
finally to b=9. The height of the regular square pyramid is 9 cm. To find the volume I used the formula
V=1/3(Ab) (h), which came to equal V=1/3(162) (9), or 486 cm3. The volume for the regular square
pyramid is 486 cm3.
a
9 2 cm
a
9 2
cm
Height 9cm
Tetrahedron
Height 6 3 cm
8
Regular square
pyramid
Steven Prascius
9A/GAT
To find the volume of the entire cubo using tetrahedrons and regular square pyramids, I multiplied the
volume of the tetrahedron by eight and the volume of the regular square pyramid by six because the
cubo has six square faces and eight triangular faces. When I did this I got (243)8+486(6), or 4860 cm3.
The total volume of the cubo is 4860 cm3.
Regular square
pyramid
Tetrahedron
A cuboctahedron, or cubo, is a polyhedron with eight triangular faces and six square faces. The
total surface area of the cubo was 972 cm2 + 648 3 cm2, and the volume of the cubo was 4860
cm3. Throughout this project I learned that it does not matter which method you use to find the
volume of the cubo because you will always get the same answer. I came across mathematical
errors while trying to find many of the volumes, which threw off my whole answer. As I have
learned, there are many ways to find the volume of a cuboctahedreon.
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