Hello, Zellator!

If we are allowed to use Trig, I have a solution.

The diagonals of a parallelogram are 16 and 20.

One side of the parallelogram is 12.

Find the area of the parallelogram. Code:

D C
o * * * * * * * o
* * * *
* * 8 * *
* * E * 10 *
* ♥ *
* 10 * * *
* * 8 * *
* * * *
o * * * * * * * o
A 12 B

We have parallelogram $\displaystyle ABCD$

. . with diagonals $\displaystyle BD = 16,\:AC = 20$

. . which are bisected at $\displaystyle E.$

Consider triangle $\displaystyle EAB.$

Code:

E
o
10 * : * 8
* :h *
* @ : *
A o * * * * * * * * * * * o B
12

Let $\displaystyle h$ be the altitude from $\displaystyle E.$

Let $\displaystyle \theta = \angle EAB.$

$\displaystyle \text{Law of Cosines: }\:\cos\theta \:=\:\frac{10^2 + 12^2 - 8^2}{2\!\cdot\!10\!\cdot\!12} \:=\:\frac{3}{4}$

$\displaystyle \sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \sin^2\!\theta + \left(\tfrac{3}{4}\right)^2 \:=\:1 $

. . $\displaystyle \sin^2\!\theta \:=\:\tfrac{7}{16} \quad\Rightarrow\quad \sin\theta \:=\:\tfrac{\sqrt{7}}{4}$

$\displaystyle \sin\theta \:=\:\frac{h}{10} \quad\Rightarrow\quad h \:=\:10\sin\theta \:=\:10\left(\tfrac{\sqrt{7}}{4}\right) \:=\:\tfrac{5\sqrt{7}}{2}$

$\displaystyle \text{Hence, the height of the parallelogram is: }\:H \:=\:2h \:=\:2\left(\tfrac{5\sqrt{7}}{2}\right) \:=\:5\sqrt{7}$

$\displaystyle \text{The base of the parallelogram is: }\:B \:=\:12$

$\displaystyle \text{Therefore, its area is: }\:BH \:=\:(12)(5\sqrt{7}) \:=\:60\sqrt{7}$