Diagonal calculation in a 3D square

Question

I have the information as shown in the following image: i.e.: all sides (AB, BC CD and DA) and one diagonal (BD) and the height difference between the point A and the points B/C/D (855.35mm).

How do I calculate the length of diagonal AC (red line)?

Answer 1

It is not a square but a tetrahedron with base $BCD$ and height $h = 855.35$.

Label the edge lengths as $$\begin{cases} b &= |CD| = 2578,\\ c &= |BD| = 2828,\\ d &= |BC| = 2128,\\ b_1 &= |AB| = 2060,\\ c_1 &= |AC|\\ d_1 &= |AD| = 2045\\ \end{cases}$$

$c_1$ will be the length we seek.

You can compute the volume of this tetrahedron in two different manners.

• $V = \frac13 \mathcal{A} h$ where $\mathcal{A}$ is area of $\triangle BCD$. You can use Heron's formula to get $\mathcal{A}$. $$16\mathcal{A}^2 = (b+c+d)(-b+c+d)(b-c+d)(b+c-d)$$

• You can also compute the volume using Cayley Menger determinant.

$$288V^2 = \left|\begin{matrix} 0 & 1 & 1 & 1 & 1\\ 1 & 0 & b_1^2 & c_1^2 & d_1^2\\ 1 & b_1^2 & 0 & d^2 & c^2\\ 1 & c_1^2 & d^2 & 0 & b^2\\ 1 & d_1^2 & c^2 & b^2 & 0 \end{matrix}\right|$$

Expanding the CM determinant and combine it with first result, one obtain following equation which is quadratic in $c_1^2$:

$$\begin{align} 16\mathcal{A}^2h^2 = 144V^2 =& \phantom{+}\; b^2 b_1^2(-b^2-b_1^2 + c^2 + \color{red}{c_1^2} + d^2 + d_1^2)\\ & + c^2 \color{red}{c_1^2}(\;b^2+b_1^2 - c^2 - \color{red}{c_1^2} + d^2 + d_1^2)\\ & + d^2 d_1^2(\;b^2+b_1^2 + c^2 + \color{red}{c_1^2} - d^2 - d_1^2)\\ & - (b^2c^2d^2 + b^2 \color{red}{c_1^2} d_1^2 + b_1^2 c^2 d_1^2 + b_1^2\color{red}{c_1^2} d^2) \end{align}$$

With help of a CAS, we can substitute back the numerical values of $b,c,d,b_1,d_1$ into above equation and simplify. The end result is

$$\begin{align}c_1 &= \sqrt{\frac{116153047144695\pm\sqrt{8168336037304042557755678133}}{19993960}}\\ &\approx 1135.385089196282 \;\text{ or }\; 3213.987289241557 \end{align} $$ There are two possible solutions for $c_1$. The '+' solution corresponds to the case where the dihedral angle between the planes holding $\triangle ABD$ and $\triangle BCD$ is obtuse ( $> 90^\circ$). For the '-' solution, the dihedral angle is acute ($< 90^\circ$).

Judging from your picture, the dihedral angle at edge $BD$ is obtuse. The length you seek is the one $\approx 3213.987289241557$.

Answer 2

Judging from the fact that A = (?, ?, 855.35), I think the figure is a tetrahedron instead of a square. It has BCD as the base and AA’ (= 855.35) as its vertical height. Therefore, all arrow-marked angles shown are $90^0$. Also, AC is a slant edge instead of a diagonal.

Apply Pythagoras to [$\triangle ABA’$ with AB and AA’ known] to get BA’.

Similarly, DA’ can be found from $\triangle ADA'$.

Apply Cosine law to [$\triangle BDC$ with all 3 sides known] to get $\angle BDC$.

Similarly, apply Cosine law to [$\triangle BDA’$ with all 3 sides known] to get $\angle BDA’$.

$\angle A’DC = \angle BDC - \angle BDA’$

Then, apply Cosine law [$\triangle A’DC$ with DA’, DC and $\angle A’DC$ known to get A’C.

Finally, apply Pythagoras to [$\triangle AA’C$] to get AC.