| Solution The base area of the given tetrahedron is the area of the equilateral triangle with the side measure x. So, the base area is equal to Next, let us find the measure of the height of the given tetrahedron (Figure 1b). The height OP of the given tetrahedron drops to the center O of its base, which is the intersection point of the base altitudes, medians and angle bisectors. It is well known fact of Planimetry that the intersection point of medians of a triangle divides them in proportion 2:1 counting from the vertices (see the lesson Medians of a triangle are concurrent in this site). |
Figure 1a. |
| Thus the segment OA in Figure 1b has the length of two third of the altitude of the base triangle, i.e. |OA| = Now, the height of the pyramid is and the volume of our tetrahedron is Answer. The volume of the regular tetrahedron with the edge length x is |
Figure 1b. |