How to find the height of an irregular triangular pyramid.

Question

I have an irregular triangular pyramid. In this drawing, I have use arbitrary side lengths for simplicity. How do I find the altitude from the base (A,B,C) to the peak D?

The answer in a post made no sense to me about the height of an oblique pyramid here but one comment suggests (to me) that I set up the following equations:

$$X=(AD^2 - BD^2)/2AB + AB/2$$ $$Y=(AD^2 - CD^2)/2AC + AC/2$$ $$H=\sqrt{AD^2 - X^2 - Y^2}$$ Is this correct? I have no idea what I'm doing.

P.S. Somebody noticed the 6,8,10 in my figure. Please disregard the coincidence of a Pythagorean triplet in the diagram. Treat it as though the triplet were 6,8,11 or something else not a $right$ triangle.

Answer 1

Here is a general solution that is easy enough to understand. As others have done, let A=(0,0,0) and C=(7,0,0). From your example, $a=5$; $b=6$; $c=7$; Use the law of cosines to get angle $\alpha$ at point A in the xy plane. $a^2+c^2-2ac\,\cos(\alpha)=b^2$. Doing that, point $B=A + a\cdot(cos(\alpha), sin(\alpha),0)=(2.71429,4.19913,0)$ Next we will find point $D$, and having done so, the z-coordinate of $D$ is your height answer. To find $D$, we simply write the equation for 3 spheres with radii being the lengths of the 3 tetrahedron legs going out of the plane. Let's start with the hardest one, being the sphere centered at point $B$. $$(x-2.71429)^2 + (y-4.19913)^2 + z^2= 8^2$$ Now we write spheres for the two easy ones centered at $A=(0,0,0)$ and $C=(7,0,0)$. $$x^2 + y^2 + z^2 = 9^2$$ $$(x-7)^2 +y^2 +z^2=10^2$$ Finally, for the last part (which could be tedious or easy depending on your use of CAS), solve the three spheres simultaneously for $x,y,z$ and you are done. $$D=(2.14286,3.61591,7.95822)$$

Answer 2

We know that the volume of a tetrahedron is one-third the area of the base times the height, so if we can find the are of $\triangle ABC$ and the volume of the tetrahedron, we are home free. We can find the area of a triangle, given the lengths of the sides, by Heron's formula. At the bottom of the same Wikipedia page, there is a Heron-type formula for the volume of a tetrahedron, due to David Robbins. It doesn't look pleasant, but it should do the job.

I found another discussion of a tetrahedral volume formula (due to the renaissance painter Pierro della Francesca!) in Kevin Brown's Math Pages. Some care is necessary in using the formula, because as explained in the article, the formula depends on the pairings of opposite sides of the tetrahedron, and there are actually two tetrahedra, with different volumes, for a given pairing of six lengths. This is a complicated formula also, but it doesn't have all those square roots!

EDIT

Alternatively, just solve it with coordinate geometry. Here's a python script

from sympy.core.symbol import symbols
from sympy.solvers.solveset import nonlinsolve

# A(0,0,0)
# B(u,v,0)
# C(0,7,0)
# D(x,y,z)
# AB = 5
# AC = 7
# BC = 6
# AD = 9
# BD = 8
# CD = 10

u,v,x,y,z = symbols('u,v,x,y,z', real=True)
equations = (u**2+v**2-25, 
             u**2+(7-v)**2-36,
             x**2+y**2+z**2-81,
             x**2+(y-7)**2+z**2-100,
             (x-u)**2+(v-y)**2+z**2-64) 
solns = nonlinsolve(equations, [u,v,x,y,z])
for soln in solns:
    print(soln)

This produces the output

(-12*sqrt(6)/7, 19/7, -31*sqrt(6)/21, 15/7, -sqrt(570)/3)
(-12*sqrt(6)/7, 19/7, -31*sqrt(6)/21, 15/7, sqrt(570)/3)
(12*sqrt(6)/7, 19/7, 31*sqrt(6)/21, 15/7, -sqrt(570)/3)
(12*sqrt(6)/7, 19/7, 31*sqrt(6)/21, 15/7, sqrt(570)/3)

so that the height is $\boxed{{\sqrt{570}\over3}}\approx7.95822425754$

You should check that the solutions actually satisfy the given conditions, because I haven't done that.

Answer 3

The angle at $B$ (DBC) made by the $6, 8, 10$ triangle is $90$ degrees. The angle at $B$ (ABC) made by the $5, 6, 7$ triangle is determined from the law of cosines $7^2 = 5^2 + 6^2 -2(5)(6)\cos(B)$. So $B = 78.4630$ degrees.

Looking end on along the edge of length 6, we see a foreshortened AB as a length of:

AB' $= 5\sin(78.463) = 4.89898$.

With length $AA' = \sqrt{5^2 - 4.89898^2} = 1$, we see a foreshortened length AD as:

$AD' = \sqrt{9^2-1^2} = 8.94427$.

This gives us an apparent $8, 8.94427, 4.89898$ triangle from which to calculate the elevation h from the $4.89898$ side.

Apparent angle $ABD' = \cos^{-1}\frac{8}{16\sqrt{24}} = 84.1421$ degrees.

$h = 8\sin(84.1421) = 7.9582$