## Monday, August 3, 2015

### Unizor - Geometry3D - Prisms - Triangular Prism

Unizor - Creative Minds through Art of Mathematics - Math4Teens

A prism with a triangle as its directrix is called a triangular prism.

According to a mini-theorem proven about all prisms, the upper and lower bases of a triangular prism are congruent triangles and the side faces are parallelograms.

Obviously, we can talk about right triangular prism if all the side edges are perpendicular to bases.

Of particular interest is the volume of a triangular prism. To approach this problem, we will perform the following transformation of a triangular prism.

Assume we have a triangular prism with a lower base ABC and an upper base A'B'C'. All side faces are parallelograms, including BCC'B'. In that parallelogram we choose a point of intersection of its diagonals
P=BC'∩B'C.

Now let's use this point P as a center of symmetry and reflect each vertex of our triangular prism relatively to this center. That means, we connect each vertex of a prism (say, A) with this center P by a segment (AP in this case) and extend that segment further behind a center P by a length of this segment, getting a centrally symmetrical point. In this case a new point D' will be symmetrical to point A and a new point D will be symmetrical to point A', while all other vertices will be symmetrical among themselves: B - to C' and C - to B'.

Prisms ABCA'B'C' and BCDB'C'D' are symmetrical and, therefore, congruent and have the same volume. But, considered together as an object ABCDA'B'C'D', they form a parallelepiped. It follows from the properties of central symmetry of transforming lines to lines, planes to planes, segments to segments of equal length and angles to angles of equal measure. We leave the details of this proof to students and will be happy to include it into this writing with all the credentials if submitted.

Now, since our original triangular prism is a half of parallelepiped ABCDA'B'C'D', its volume VTprism is also a half of the volume of this parallelepiped VPped, which is equal to a product of the area of its base SPped by its altitude H:
VPped = SPped·H
Therefore,
VTprism = (1/2)·SPped·H
But, since area of the base of our triangular prism equals to a half of the area of parallelogram ABCD,
STprism = (1/2)·SPped
and, therefore, we came to the same formula for a volume of a triangular prism, that is it's equal to a product of the area of the base by the altitude:
VTprism = STprism·H