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How do I arrive at a formula to calculate the volume of the following 3D shape? Does this shape have a proper name?

enter image description here

It kind of looks like an irregular triangular prism with 2 similar triangles as bases. The base edf is bigger in this example, but could be smaller, too.

All edges are different sizes and their lengths are known. Angles α and β are known too. Bases are parallel to each other. ab, df, gf, ga are all right angles.

Couldn't find a formula for a shape like this. I tried subdividing ot into other 3D shapes with known volume formulas, but I wasn't able to infer the sizes of all the necessary dimensions to plug into their volume formulas. Any ideas?

Can / should this be solved using integral calculus or algorithmically, since this can be viewed as the abc triangle riding on the g edge and linearly increasing / decreasing in size until it becomes bfe?..

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    $\begingroup$ In your diagram it looks like sides a,g,f,i are all on one plane. Is that the case? If so you could make slices parallel to the base, determine their areas, and finally integrate that (variable) area as it sweeps out the shape. $\endgroup$ Commented May 10, 2022 at 6:09
  • $\begingroup$ @coffeemath yep, they're on one plane. I think you're right. Also the two similar triangles are parallel each other so I could take triangular slices and integrate their variable area, right? $\endgroup$ Commented May 10, 2022 at 6:16
  • $\begingroup$ After staring at the drawing for one more hour, I realized this is simply a truncated irregular triangular pyramid. So volume(desiredShape) = volume(entireImaginaryPyramid) - volume(truncatedTip). And pyramidVolume = base area * height / 3. $\endgroup$ Commented May 10, 2022 at 6:54
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    $\begingroup$ Assuming abc and def are similar triangles, then it is Indeed a truncated pyramid, also more generally called a frustum. $\endgroup$ Commented May 10, 2022 at 7:10

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The solid is a cone (or a pyramid since the base is a polygon) with top cut off. Let $h_1$ be the distance between the planes of the triangles, i.e. the height of the cut pyramid. There's a formula in terms of $h_1$ and $A_1, A_2$ (the areas of the base triangles)

$$V=\frac{1}{3}h_1 \left(A_1 +\sqrt{A_1A_2}+A_2 \right).$$

But you still have to solve the height $h_1$. This can be done by setting the figure into coordinate space by setting the right angle of the bigger triangle to origin and giving the two other points the coordinates $(d,0,0)$ and $(0,f,0)$. Now you want to know the $z$-coordinate of the point that is the right angle point of the smaller triangle. You will know its distance to three points: $g$ to origin and other two can be solved from the diagonals on the sides (they are trapezoids). From there, I think we are able to solve the coordinates.

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  • $\begingroup$ I think that's perfect, thanks! Even easier in my case I think, because the distance between the two planes h1 = g on my drawing, which is given. g is perpendicular to both triangle planes by definition, the fact that I did not clearly show on my drawing. $\endgroup$ Commented May 10, 2022 at 8:28

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