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Mario Nona/Merna Sana

Acre

GAT 9A

28 February 2014

Cuboctahedron

The cuboctahedron is a very unique 3 dimensional shape. It is a polyhedron made up of

six squares and eight triangles, creating a total of 14 faces. It has a 12 vertices with two squares

and two triangles meeting at each and 24 identical edges, separating a square from a triangle. The

dual of the cuboctahedron is the rhombic dodecahedron. Apart from Platonic Solids, it is an

Archimedean Solid because it’s faces use two shapes. There are many different procedures that

can be used to find the side lengths, surface area, and volume of the cuboctahedron, mostly

because it can be looked at and thought of in several different ways.

The side length, surface area, and volume can all be found using just the side length of

the cube that encases the cuboctahedron. Remembering that surface area is the total sum of the

areas of all the faces on a (or net of a) 3 dimensional shape and volume is the amount of space a

3 dimensional shape takes up. The simplest process to use to find the side length of the

cuboctahedron is with the base of the actual cube. Put the midpoints of a face of the cube on each

side of the square and draw lines to and from consecutive sides to create another square within

the original. Using the known side lengths and the pythagorean theorem, the side length of one

side of the inner square can be found, which is the side length of the cuboctahedron. For this

problem, the side length of the cube is 13.2 cm.

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Figure 1. Finding the Side Length of the Cuboctahedron

Figure 1 shows how the side length of the cuboctahedron was found. The midpoints of

the sides of the square of the original cube were drawn out to start. Leaving the side length cut in

half into two pieces that are 6.6 cm. After, the midpoints were connected to make a square (face

of octahedron) and the hypotenuse was calculated. The corner triangle is a 45°-45°-90° regular

right triangle (where both legs are congruent so the hypotenuse is leg length times root 2, making

the side length of the cuboctahedron 6.6√2 cm.

This measurement is the same for every edge on the cuboctahedron, 6.6√2 cm.

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Figure 2. Finding the Area of a Triangle Face on the Cuboctahedron

Figure 2 shows how the area of the triangle faces of the cuboctahedron were found,

already have discussed that one edge of the cuboctahedron is 6.6√2 cm. The midpoint of one of

the edges (that will be called the base based on how it is being looked at) is drawn and connected

with a line segment to the opposite vertex to cut the triangle in half. This new triangle made is a

30-60-90 triangle. In a 30-60-90 triangle, the side opposite of the 30° angle is half of the

hypotenuse. The other side that connects the 90° angle and 30° angle is half of the hypotenuse

times root 3. The hypotenuse is already known, which is also the cuboctahedron edge, 6.6√2.

The base of the new triangle is half of the original, so it is 3.3√2 cm long. The last leg of the

right triangle, also the height, is (3.3√2)(√3), which can be simplified to 3.3√6 cm.

Table 1. Surface Area of a Triangle on the Cuboctahedron.

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SA = ½*b*h

SA= ½*(6.6√2)*(3.3√6)

SA= 10.89√12

SA= 21.78√3 cm^2

Table 1 shows the surface area and how it is calculated of just one triangle of the

cuboctahedron.

Table 2. Surface Area of a Square on the Cuboctahedron.

SA= (½x√2)^2

SA= (6.6√2)^2

SA= 87.12 cm^2

Table 2 displays how the surface area of a square on the cuboctahedron was calculated,

which ends up being 87.12 cm^2.

Table 3. Total Surface Area of all the Triangles on the Cuboctahedron.

All Triangles 21.78√3 * 8

All Triangles 174.24√3 cm^2

Table 3 shows the area of one triangle on the cuboctahedron being multiplied by 8,

because there are eight on the cuboctahedron. To get a total of 174.24√3 cm^2.

Table 4. Total Surface Area of all the Squares on the Cuboctahedron.

All Squares 87.12 * 6

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All Squares 522.72 cm^2

Table 4 shows the total surface area of all the squares on the cuboctahedron. The area for

one square is 87.12 cm, and multiply that by 6 because there are six squares on the

cuboctahedron, to get a total of 522.72 cm^2.

Table 5. Total Surface Area of the Cuboctahedron.

Total Surface Area 522.72 + 174.24√3 cm^2

Table 5 shows the total surface area of the cuboctahedron, which is 522.72+174.24√3

cm^2.

To sum it up, the area of one square on the cuboctahedron is 87.12 cm^2, the area of a

triangle on the cuboctahedron is 21.78√3 cm^2, and the total surface area of the cuboctahedron is

522.72 + 174.24√3 cm^2.

Volume

Case #1

The cuboctahedron can be thought of in many ways because of how unique it is. The

effect of this is that there are also many ways to find the volume. The first method that will be

used is to cut off a cubes corners at the midpoints of the edges.

Figure 3. An Image of a Cuboctahedron within its Cube.

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Figure 3 shows a picture of the cube with the corners being shown to be cut off, making

an image of the cuboctahedron within the cube. Notice there are eight corner pyramids.

Figure 4. Net of a Corner Pyramid of the Cube.

Figure 4 shows a new of a corner pyramid from the cube. It also displays all the

dimensions and side length of the pyramid. The equilateral triangle is a face on the

cuboctahedron, and shares an edge that are all 6.6√2 cm. The right triangles have leg (line

segments connected to the 90° angle) lengths of 6.6 cm because they are half of the cubes edge.

Also since it is a 45-45-90 triangle, the hypotenuse is 6.6√2 cm long.

Table 6. Volume of the Corner Pyramid

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V= ⅓*A(base)*h(pyramid)

V= ⅓*(21.78)*(6.6)

V= 47.916 cm^3

Table 6 shows the volume of one corner pyramid from the cube. The volume of the figure

is 47.916 cm^3. The area of the base is 21.78 cm^2, which the math is shown in table 1.

Table 7. Volume of Cubo

V= V(pyramid)*8

V= 47.916*8

V= 383.328 cm^3

V(cubo)= V(cube) - V

V(cubo)= 2299.97 - 383.328

V(cubo)= 1916.64 cm^3

Table 7 shows the method that was used for finding the volume. The volume of the cube

(13.2 squared) was needed to be found. The corner pyramid volume was also needed, and

multiplied by eight to account for all of the pyramids. Finally, the volume of the pyramids were

subtracted from the cube to get the volume of the cuboctahedron. The volume of the

cuboctahedron resulted in being 1916.64 cm^3.

Case #2

The second method here involves a right square prism with four rectangular pyramids

attached on the lateral faces of the prism. The volumes of all five objects will be added, because

they make a cuboctahedron. That volume should be the same as the one found in the first

method.

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Figure 5. Right Square Prism with a Rectangular Pyramid.

Figure 5 shows a right square prism and a rectangular pyramid to help visual how this

method of finding the volume will be done. The volume will be found using one prism and four

of the pyramids.

Table 8. Volume of the Right Square Prism

V= A(base)*h(prism)

V= (6.6√2 * 6.6√2) * 13.2

V= 87.12 * 13.2

V= 1149.98 cm^3

Table 8 shows the steps to finding the volume of the right square prism within the cubo.

The volume of the inner prism is 1149.98 cm^3.

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Figure 6. Net of the Inner Right Square Pyramid of the Cuboctahedron.

Figure 6 shows a net of the inner right square prism of the cuboctahedron. The height is

13.2 cm just like the original cube.

Table 9. Volume of the Rectangular Pyramid

V= ⅓*A(base)*h(prism)

V= ⅓*(87.12√2)*(4.669)

V= ⅓*(406.58√2)

V= 135.527√2 cm^3

Table 9 shows the steps to finding the volume of the rectangular prism. The volume

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equaled out to be 135.527√2 cm^3.

Figure 7. Net of the Rectangular Pyramid of the Cuboctahedron.

Figure 7 displays the net of the rectangular pyramid of the cuboctahedron. The base of

the pyramid is also a face of the right square pyramid.

Table 10. The Volume of the Cuboctahedron.

V = 4 x (V of pyramid) + (V of prism)

V = 4 x 135.527√2 + 1149.98

V = 766.656 + 1149.98

V = 1916.64 cm^3

Table 10 shows the volume of the cuboctahedron, found using the second method

discussed. Since there are four pyramids, its four times the volume of one pyramid plus the

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volume of the prism. The volume came to be the same, just like it should be, 1916.64 cm^3.

Case #3

The third and last method that will be used to find the volume of the cuboctahedron will

include eight tetrahedrons and six square pyramids. These all can be put together to make the

cuboctahedron, also with the eight triangular faces and six squares corresponding with the faces

on the cuboctahedron.

Table 11. Volume of the Pyramid with a Square Base

V= ⅓*A(base)*h(pyramid)

V= ⅓*(87.12)*(6.6)

V= 191.664 cm^3

Table 11 shows the steps to finding the volume of the pyramid. The area of the base that

was used was found earlier as shown in table 2. The volume equaled out to be 191.664 cm^3.

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Figure 8. Net of the Square Pyramid.

Figure 8 shows a net of the square pyramid within the cuboctahedron. Used for the third

method of finding the volume of the cuboctahedron.

Table 12. Height of the Tetrahedron

h= ⅓*(edge)*√6

h= ⅓*(6.6√2)*√6

h= 2.2√12

h= 2.2*(2√3)

h= 4.4√3 cm

Table 12 shows the steps to finding the height of the tetrahedron. The final height

calculated out to be 4.4√3 cm.

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Table 13. Volume of the Tetrahedron

V= ⅓*A(base)*h(pyramid)

V= ⅓*(21.78√3)*(4.4√3)

V= 95.832 cm^3

Table 13 shows the steps to finding the volume of the tetrahedron. The area of the base

that was used was found earlier as shown in table 1. The height that was used can be found on

table 11. The volume calculated came out to be 95.832 cm^3.

Table 14. The Volume of the Cuboctahedron

V= 6*(V of pyramid) + 8*(V of tetrahedron)

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V= 6*(191.664) + 8*(95.832)

V= 1149.98 + 766.656

V= 1916.64 cm^3

Table 14 shows the volume of the cuboctahedron found in the third, and final, method.

The volumes of the eight tetrahedrons and six square pyramids were added up to get the final

volume. Just like before, the volume came to be 1916.64 cm^3.

In conclusion, the cuboctahedron is a very magnificent 3 dimensional figure, that has

many properties to it. Just like how there are many different type of people in the world, the

cuboctahedron can be seen and thought of in many different ways by different people. Finding

the side lengths, surface area, and volume in alternative ways was quite a challenge. This

Archimedean Solid has two different types of faces, triangles and squares. There are eight

triangles and six squares, the variety made finding the results challenging with several decimals,

roots, and squares. The cuboctahedron was a flawless object to have to explore all about.