Tetrahedron volume

Tetrahedron volume

Postby harry.petrone » Sun Jul 30, 2017 5:13 pm

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?





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Re: Tetrahedron volume

Postby Guest » Mon Jul 31, 2017 7:07 am

The height is 1/2 the heigth of the big tetrahedron. That can be proved with similar triangles.

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Re: Tetrahedron volume

Postby harry.petrone » Mon Jul 31, 2017 12:42 pm

How?

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Re: Tetrahedron volume

Postby Guest » Tue Aug 01, 2017 4:37 am

Could you please attach an image?

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