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 CuboWhat is that in the sky? A cube? A tetrahedron? No it’s a cubo-octahedron! The Cubo-Octahedron is made up of six square faces and eight triangular faces. Some people describe the Cubo-Octahedron as looking like a cube with its corners chopped off. For our example in this case the cube had a side length of 12 cm. To find the total surface area of the Cubo-Octahedron you have to break down the different types of faces and find the surface area of each. There is not only one way to find the volume of a cubo-octahedron, but three! The first way to find the volume would be to find the volume of the original cube, then subtract the volumes of the eight corners. The second way to find the volume would be to divide the cubo-octahedron into four rectangular pyramids and a rectangular prism. The third way to find the volume would be to divide the cubo-octahedron into eight tetrahedrons and six square pyramids.Figure 1. Measurements of the Cubo-Octahedron Figure 1 shows that when the side length of the cube is given, you can find the measurements of the other faces of the cubo-octahedron. The 45-45-90 triangle formula can be used to side length of the square face on the cubo. Figure 2: Measurement of a Square FaceFigure 2 shows that when the side length of the cube is 12 cm. you can find one side of the square face is 6√2 cm. The measurement of the side length of the cube was found by plugging in 12 for x and solving for the side length. In Figure 1, it shows the fact that one side length of the square face is equal to ?x√2. After plugging in 12 for x, each side of the square face equaled 6√2 cm.Figure 3. Measurement of a Triangle Face Figure 3 shows that when the side length of the cube is 12 cm. you can find the length of the base is 6√2 cm. and the height of the triangular face is 3√6.The measurement of the base and the height of the triangle was found by plugging in 12 for x and solving for the measurements. In Figure 1, it shows the fact that one side length of the triangular face is ?x√2 just like the side length of the square face. This would mean that the base of the triangular face would be 6√2. With the knowledge of finding the heights of equilateral triangles, the hypotenuse of the triangle can be taken and divided by two to find one half of the base. Then having 3√2 as the base of the triangle you would multiply 3√2 by √3 to find the height of the pyramid which would be 3√6.A=bh(?)A=(6√2)(3√6)(?)A=(18√12)(?)A=(9√12)A=18√3 sq. cm.Figure 4. Surface Area of One Triangle FaceFigure 4 shows how to find the surface area of a triangular face of a Cubo-Octahedron. When given the dimensions of one side of the equilateral triangle you can find the surface area of the triangle. A=bhA=(6√2)(6√2)A=36√4A=72 sq. cm.Figure 5. Surface Area of One Square FaceFigure 5 shows how to find the surface area of a triangular face of a Cubo-Octahedron. When given the dimensions of one side of the square, using the knowledge that all edges of a square are equal, you can find the surface area of the square. A= Area of one triangle face(8)+Area of one square face(6)A=18√3(8)+72(6)A=144√3+432A=144√3+432 sq. cm.Figure 6. Surface Area of the Cubo-OctahedronFigure 6 shows how to find the Total Surface Area of the Cubo-Octahedron. As stated before there is eight triangular faces and six square faces to the cubo-octahedron. After finding the dimensions of one square face and one triangular face, I multiplied the dimension of the face by the number of faces found on the cubo. To find the surface area, the cubo-octahedron is broken into six square faces and eight triangular faces. First, when finding the surface area of one square face the base and the height (Figure 2) should be and multiplied. As in Figure 4, when multiplying the base and the height of the square face ends up with 18√3 sq. cm. for the final surface area. Next, to find the surface area of one triangular face, multiply the base and the height of the triangle together (Figure 3) then divide that number by two because the area of a triangle is being found. As in Figure 5, when multiplying the base and height of the triangular face and dividing it by two the answer should be 72 sq. cm. After finding the area of each of the faces you have to find the surface area of the whole Cubo-Octahedron. To find the Cubo-Octahedron multiply the number of square faces times six plus the number of triangle faces times eight. As shown in Figure 6, the total surface area of the cubo-octahedron is 144√3+432 sq. cm. Case #1Figure 7. Isosceles Right Triangle PyramidFigure 7 shows a visual representation of what an isosceles right triangle pyramid and its dimensions look like. V=(AoB)(h)(?)V=(18)(6)(?)V=(108)(?)V=36 cu. cm.Figure 8. Volume of One Isosceles Right Triangle PyramidFigure 8 shows how to find the volume of one of the eight isosceles right triangle pyramids used to make up the cubo-octahedron. To find the volume of the isosceles right pyramid, the area of the base and the height of the pyramid must be known. V=(AoB)(h)V=(144)12V=1728 cu. cm.Figure 9. Volume Of CubeFigure 9 shows how to find the area of the cube finding the area of the base multiplied by the height of the cube.Volume of Cubo=Area of Cube-8 Isosceles right trianglesV=1728-(36*8)V=1728-288V=1440 cu. cm.Figure 10. Volume of the Cubo-OctahedronFigure 10 shows to find the volume of the cubo-octahedron the volume of the cube must be found and then the volume of a isosceles right triangle pyramid times eight must be subtracted.The first way to find the volume of the cubo-octahedron would be to find the volume of the cube and subtract the volume of eight isosceles right triangle pyramids. The first thing that should be done while finding the volume of the cubo-octahedron would be to find the volume of the cube. To find the volume of the cube the area of the base must be multiplied by the height of the cube. The length of the cube is 12 cm. is known because the information was given. As shown in Figure 9, the area of the base (144 cm.) is multiplied by the height of the cube which is 12 to find the total volume of the cube which is 1728 cu. cm. The next step to finding the volume of the cubo-octahedron would be to find the volume of one of the isosceles right triangle pyramids. The area of the base has to be known to find the volume of one isosceles right triangle. The area of the base would be found by multiplying the length of the pyramid by the width of the pyramid, which is 9√4 (knowing both the width and the length of the pyramid are 3√2 in Figure 7). After simplifying the root to 18 proceed with multiplying 18 by the height of the pyramid which is six (Figure 7). After multiplying those numbers, divide the product by three, since it is a pyramid to find the volume of one isosceles right triangle which is 36 cu. cm. After knowing the volumes of both the cube and one isosceles triangle right pyramid the volume of the cubo-octahedron can be found by subtracting the volume of the isosceles right triangle pyramid by the volume of the cube (Figure 10). Figure 10 shows when you take the volume of the cube, which is 1728 cu. cm., and subtract the volume of eight isosceles right triangle pyramids, which is 288 cu. cm. because 36 needed to be multiplied 8, to get the final answer of 1440 cu. cm. Case #2Figure 11. Rectangular Prism Inside the CubeFigure 11 shows where the Rectangular prism is located inside the cube.Figure 12. Tetrahedron inside the CubeFigure 12 shows where one of the four tetrahedron is located inside the cube. Figure 13. Dimensions of the Rectangular PrismFigure 13 shows the dimensions of the rectangular prism. The height of the prism is the same as the height of the cube which is 12. The side length of base is the same as the square face of the cubo, to find the length see Figure 1.Figure 14. Dimensions of the Rectangular PyramidFigure 14 shows the dimensions of the rectangular pyramid. The side lengths of the base are equal to one lateral side of the prism. The height of the pyramid can be found by using the Pythagorean Theorem. Rectangular Prism:V=(AoB)hV=(6√2*6√2) V=(72)12V=864 cu. cm.Figure 15. Volume of Rectangular PrismFigure 15 shows how to find the volume of the rectangular prism by multiplying the area of the base and the height of the prism. Rectangular Pyramid:a^2+b^2=c^2a^2+6^2=(6√2)^2a^2+36=36√4a^2+36=72√a^2=√36a=6 cm.Figure 16. Slant Height of Triangle Figure 16 shows how to find the slant height of one of the lateral sides of the pyramid by using the Pythagorean Theorem. In this equation “b” equals half of the height of the base and “c” equals the length of one edge. a^2+b^2=c^2a^2+(3√2)^2=6^2a^2+9√4=36a^2+18=36a=√18a=3√2 cm.Figure 17. Height of PyramidFigure 17 shows how to find the height of the pyramid by using the Pythagorean Theorem. When finding the height “b” equals half of the width of the base and “c” equals the slant height (Figure 16). The height is simplified from √18cm. to 3√2 cm.AoB=bhAoB=(12)(6√2)AoB=72√2 sq. cm.Figure 18. Area of BaseFigure 18 shows how to find the area of the base by using the area formula which is base times height.V=bh(?) V=(72√2)(3√2)(?)V=(216√4)(?)V=(216(2))(?)V=(432)(?)V=144 cu. cm.Figure 19. Volume of Pyramid Figure 19 shows that using volume formula for a pyramid the volume of the rectangular pyramid. The volume formula is base times height times ?.V=Volume of Rectangular Prism + (Volume of Rectangular Pyramid*4)V=864+(144*4)V=864+576V=1440 cu. cm.Figure 20. Volume of CuboFigure 20 shows how by first multiplying the volume of the rectangular pyramid by four, to account for the 4 pyramids needed to make the volume of cubo, then adding the volume of the rectangular prism.The second way to find the volume of the cubo-octahedron is to find the volume of the of the inner prism and the volume of the four outer pyramids. To start off, the dimensions of the prism and pyramids must found (Figure 13 and Figure 14). The next step is to find the volume of the rectangular prism (Figure 15); in this case the volume is 864 cu. cm. After the volume of one of the four rectangular pyramids must found; however unlike the prism the pyramid takes several steps. The first part is finding the slant height of the lateral side (Figure 16). This is achieved by using the Pythagorean Theorem; here the slant height is six cm. Next by again using the Pythagorean Theorem the height of the pyramid is found (Figure 17); here the height is 3√2 cm. Finding the area of the base of the pyramid is next, because the base is a rectangle the formula “A=B*H” can be used (Figure 18). This time the area is 72√2 sq. cm. Now that all the steps are done the volume of pyramid can be calculated (Figure 19); the volume is 144 cu. cm. Now, to find the volume of the whole cubo-octahedron the volume of the pyramid is multiplied by four, then added to the volume of the prism (Figure 20). Just like in case #1 the volume equals 1440 cu. cm.Case #3Figure 21. Dimensions of Square PyramidFigure 21 shows the dimensions of the square pyramid, including the edge lengths, the height of the pyramid, and the slant height. The slant height was found by using the 30-60-90 degree triangle formula and the height was found using the Pythagorean Theorem using the edge length and half of the diagonal of the base.Figure 22. Dimensions of Tetrahedron Figure 22 shows the dimensions of the tetrahedron, including the edge lengths, the height of the pyramid, and the slant height. The slant height was found by using the 30-60-90 degree triangle formula and the height was found using the formula to find the height of tetrahedrons (?(edge)√6).Tetrahedron:H=?(edge)(√6)H=?(6√2)√6H=(2√2)√6H=2√12H=4√3 cm.Figure 23. Height of TetrahedronFigure 23 shows how to find height of the tetrahedron. The height is simplified from 2√12 cm. to 4√3 cm.A=(?)bhA=(?)(6√2)(3√6)A=(?)(18√12)A=9√12A=18√3 sq. cm.Figure 24. Area of BaseFigure 24 shows that using the area formula for a triangle the area of the base, of the tetrahedron can be found. The height is simplified to 18√3 sq. cm.V=(?)(AoB)(H)V=(?)(18√3)(4√3)V=(?)(72√9)V=24√9V=24(3)V=72 cu. cm.Figure 25. Volume of TetrahedronFigure 25 shows how to find the volume of the tetrahedron. “AoB” equals area of base and “H” equals height.Square Pyramid:a^2+b^2=c^26^2+b^2=(6√2)^236+b^2=36√436+b^2=36(2)36+b^2=72√b^2=√36b=6 cm.Figure 26. Height of Square PyramidFigure 26 shows that with using the Pythagorean Theorem the height of the square pyramid can be found. “A” equals half of the diagonal of the base, “b” is the height, and “c” equals length of the edge.A=B*HA=(6√2)(6√2)A=36√4A=36(2)A=72 sq. cm.Figure 27. Area of BaseFigure 27 shows how to find the area of the base by using the area formula. Because the base is a square both the base and the height are the same.V=(?)(AoB)(H)V= (?)(72)(6)V=(?)(432)V=144 cu. cm.Figure 28. Volume of Square PyramidFigure 28 shows how to find the volume of the square pyramid. “AoB” means area of base and “H” equals height of the pyramid. V=(Volume of Square Pyramid*6)+(Volume of Tetrahedron*8)V=(144*6)+(72*8)V=864+576V=1440 cu. cm.Figure 29. Volume of CuboFigure 29 shows that by first multiplying the volume of the square pyramid by six, and the volume of the tetrahedron by eight then adding them together, the volume of the Cubo.In this case the volume of a tetrahedron is found, then multiplied by eight because there is one tetrahedron for every triangular face of the cubo-octahedron. The volumes of six square pyramids are found to account for all six faces. To find the volume of tetrahedron, the first thing to do is to find the height (Figure 23). This is accomplished by using the formula for finding the height of a tetrahedron. Next, the area of the base needs to be found (Figure 24), here it is 18√3 sq. cm. Now both of these are used to find the volume of one tetrahedron (Figure 25). The same steps are used to find the volume of the square pyramid. Like the tetrahedron the height must first be found (Figure 26), however there is not a special formula that can be used. The area of the base is found next (Figure 27), then the volume of the pyramid is found (Figure 28). The volume of the tetrahedron is multiplied by eight and the volume of the square pyramid multiplied by six, then both are added together to find the volume of the cubo. If the volumes of the cubo are the same in all three cases then the math is right, here the volume is equal to the other cases, 1440 cu. cm.After making the cubo-octahedron, it is known that it is not difficult to find the surface area and volume of a cubo-octahedron. Overall, the hardest part of this project would be figuring out the proper way to perform case three. The basic understanding and structure of case three was the confusing part of it. If the project were to be constructed again a few steps would have been done differently. The first thing that would have been done differently would be to create the nets before writing the bulk of the paper because then there would more of a visual while writing. Other than that slight change to the process of this project, nothing else would would have been changed. This project was easier than originally believed it would be from writing the paper to making the 3D figures. In all three cases the volume should be the same, no matter the way of finding the volume it is still the same shape therefore the same volume. Finding volume in three different ways was a way to think outside of the comfort zone and expand ways of looking at the shape. To conclude, by finding the volume three ways the cubo-octahedron was forced to be broken into many different shapes and it will help to think more critically in the future. ................
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