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The Tetrahedron.
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| Submitted: Mon Dec 22 2003
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THE TETRAHEDRON Figure 1 Let each side of the tetrahedron (in blue) = s. The tetrahedron has 6 sides, 4 faces and 4 vertices. Here the base is marked out in gray: the triangle ABC. From The Equilateral Triangle we know that: Area ABC = (\/¯3 / 4)s². Now we need to get the height of the tetra = ED. E |\ | \ | \ ----\ D A figure 2 From The Equilateral Triangle we know that: DA = (1 / \/¯3)s. Now that we have DA, we can find ED, the height of the tetrahedron. We will call that h. h² = ED = EA² - DA² = s² - (1/3)s² = (2/3)s² h = ED = (\/¯2 / \/¯3)s So V(tetra) = 1/3 * area ABC * height(tetra) = 1/3 * (\/¯3 / 4)s² * (\/¯2 / \/¯3)s, V(tetra) = (\/¯2 / 12) s³ = (1...
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