properties of regular tetrahedron
It’s apparent that a plane passing through the midpoints of three parallel edges of the cube cuts the regular tetrahedron into two congruent pentahedrons and that the intersection figure is a square, the midpoint of which is the centroid of the tetrahedron.
The angles between the four half-lines from the centroid of the regular tetrahedron to the vertices (http://planetmath.org/Polyhedron) are (), which is equal the angle between the four covalent bonds of a carbon . A half of this angle, , can be found from the right triangle in the below figure, where the catheti are and .
One can consider the regular tetrahedron as a cone. Let its edge be and its height . Because of symmetry, a height line intersects the corresponding base triangle in the centroid of this equilateral triangle. Thus we have (see the below ) the rectangular triangle with hypotenuse , one cathetus and the other cathetus (http://planetmath.org/Cathetus) (i.e. of the median (http://planetmath.org/Median) of the equilateral triangle — see the common point of triangle medians). The Pythagorean theorem then gives
Consequently, the height of the regular tetrahedron is .
Since the area of the base triangle (http://planetmath.org/EquilateralTriangle) is , the volume (one third of the product of the base and the height) of the regular tetrahedron is .
|Title||properties of regular tetrahedron|
|Date of creation||2013-03-22 18:29:39|
|Last modified on||2013-03-22 18:29:39|
|Last modified by||pahio (2872)|