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# Volume of Tetrahedrons

Is the volume formula V=(1/3){base area}[height] valid for irregular tetrahedrons, or is it for regular tetrahedrons only?

If it is then can some of you prove it. ( Please don't use higher mathematics unless required, I'm just a 12th grader.)

Note by Vraj Mistry
11 months, 4 weeks ago

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It works for any tetrahedron, regular or not. Let's say that the cross section area is a square function from the apex

$$A(z) = k {z}^{2}$$

where $$k$$ is some constant, and $$z$$ is the distance from the apex to the cross section. Then we integrate this

$$\displaystyle \int _{ 0 }^{ h }{ k{ z }^{ 2 } } dz=\frac { 1 }{ 3 } k { h }^{ 3 }=\frac { 1 }{ 3 } \left( k{ h }^{ 2 } \right)h =\frac { 1 }{ 3 } Bh$$

where $$B=A(h)$$ is the area at distance $$z=h$$ from the apex. You should recognize the familiar pyramid formula, and you should realize that this works for any kind of base, provided that all the cross sections are similar in shape following the square function of the distance from the apex.

- 11 months, 4 weeks ago

But, how is the base area a square function, couldn't it be anything else like dependent on two different variables rathere than one variable squared. For example: A(z1,z2)=k(z1)(z2)

- 11 months, 4 weeks ago

If you have any 2D shape with a definite area, then similar shapes have areas equal to the first times the square of the ratio of size. For example, a given polygon may have area B, and the same thing but 1/3 the size will have area 1/9 B. The pyramid formula doesn't compute the constant $$k$$ itself, that depends on the particular shape that the base and all the cross sections have, which is supposed to be all similar.

- 11 months, 4 weeks ago

Thank you

- 11 months, 4 weeks ago

cool

- 11 months, 4 weeks ago