Finding volume of a tetrahedron

Question

The edge is 12 and the height is 12 for the tetrahedron. I know that tetrahedron consists of 4 equilateral triangles.

I have a problem with finding the volume because i had two answers. first I did the pyramid formula. Found the base's area and plugged it in 1/3 bh and got 144 square root of 3. 

I wanted to check my work I went around online searching a formula for the volume of tetrahedron. Found one that was : (squareroot of 2 / 12) a^3. (a = edge). plugged info in and got 144 squareroot of 2. 

I got different squareroots so now I am confused. Which of the answer is correct? or show me how you would find the volume.

Thank you very much

Answer 1

Here is another way.

A regular tetrahedron of edge length s, can be inscribed in a cube of edge length s/√2.

This tetrahedron's edges are the face diagonals of the cube, and cutting it out from the cube leaves four congruent right-tetrahedra.

Each right tetrahedron has 1/6 the area of the cube, if you use the formula A = bh/2 for triangles and V = bh/3 for tetrahedra.

So the total of the right tetrahedra is 2/3 of the cube, so that the regular tetrahedron is 1/3 of the cube and you get V = (s/√2)³ / 3 = s³ / (6√2) = (√2 / 12) s³.

Your tetrahedron has volume (√2 / 12) * 12³ = 144√2

 

Answer 2

The correct answer is 144√2.

Here is my work: The base area = (1/2)(12)(6√3) = 33√3.

The height can be calculated from √[12² - (4√3)²] = 4√6

So, the volume of the tetrahedron = (1/3) base area x height = 144√2.