Given a tetrahedron KABC where all its edges have equal length x, we take points A1 on KA, such as KA1=1/2 KA, point B1 on KB such as KB1=2/3 KB and point C1 on KC such as KC1=3/4 KC.

Find the volume of tetrahedron KA1B1C1 in relation to a.

I know that the volume of KABC = a^3/6*sqrt(2).

Also the volume is 1/3*(area of a base)*h.

In the second tetrahedron the sides of the base can be calculated (don't know how - I guess with trigonometry??) since we know the angle of each triangle being 60 degrees and the two sides (proportions of a). Then we can calculate the area by Heron's formula. But how do we calculate the height of the new tetrahedron?

Also we know that the volume of the tetrahedron is one third of the cube that contains it.

Any ideas?