Then, by the triangular prism volume formula above. Let $V_A$ be the volume of the truncated triangular prism over right-triangular base $\triangle BCD$ likewise, $V_B$, $V_C$, $V_D$. So, let's explore the subdivided prism scenario:Īs above, our base $\square ABCD$ has side $s$, and the depths to the vertices are $a$, $b$, $c$, $d$. OP comments below that the top isn't necessarily flat, and notes elsewhere that only an approximation is expected. The volume of that figure $s^2h$ is twice as big as we want, because the figure contains two copies of our target.Įdit. In the triangular prism calculator, you can easily find out the volume of that solid. This follows from the triangular formula, but also from the fact that you can fit such a prism together with its mirror image to make a complete (non-truncated) right prism with parallel square bases. Let the base $\square ABCD$ have edge length $s$, and let the depths to the vertices be $a$, $b$, $c$, $d$ let $h$ be the common sum of opposite depths: $h := a c=b d$. The least mentally taxing way to do this is the calculate the volume of the first sphere, then multiply by 2.5, and then apply the formula in reverse to find the radius and then diameter of the second sphere. In order to calculate the c side, use the Pythagorean theorem. If the table-top really is supposed to be flat. The right triangular prism formula looks as follows: Volume Length ( (a b) / 2) Where: a and b are the sides of the triangle that touch the right angle and Length means the length of the entire prism, i.e., distance between two faces. 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. But you still have to solve the height h 1. Where $A$ is the volume of the triangular base, and $a$, $b$, $c$ are depths to each vertex of the base. Theres a formula in terms of h 1 and A 1, A 2 (the areas of the base triangles) V 1 3 h 1 ( A 1 A 1 A 2 A 2). ("Depths" to opposite vertices must sum to the same value, but $30 80 \neq 0 120$.) If we allow the table-top to have one or more creases, then OP can subdivide the square prism into triangular ones and use the formula The question statement suggests that OP wants the formula for the volume of a truncated right-rectangular (actually -square) prism however, the sample data doesn't fit this situation.
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