Pythagoras for a Tetrahedron

Thanh from Delta Global School in Vietnam, and Ashil and Hadi from St Dominics 6th Form College, Joshua from Bohunt Sixth Form and Lisa, all from the UK, found the sides of triangle $ABC$ in terms of $a, b$ and $c.$ Joshua, Hadi and Ashil then used the cosine rule, and the formula $Area=\frac12ab\sin C.$ This is Ashil's work (click to enlarge):

Thanh began in the same way, but then used a perpendicular 'height' to find the area $S$ (instead of the formula):

Lisa used Heron's formula to find the area $S,$ which gave some very neat algebra. Heron's formula works like this: $$\begin{align}&\text{If a triangle has sides }a,b,c\\ &\text{Let }s=\tfrac12\left(a+b+c\right)\\ &\text{then the area of the triangle is equal to }\\ &\sqrt{s(s-a)(s-b)(s-c)}\end{align}$$Here is Lisa's work (click to enlarge):

Yash from Tanglin Trust School in Singapore and Leo from Stowe School in the UK both took a more algebraic approach. Most of ther work is exactly the same, but they started in subtly different ways. Yash began:

Note that Yash's equation is not true: for the area $P$ to be equal to $\frac12\times AB \times DO,$ $AB$ and $DO$ need to be perpendicular. However, they may not be. This means that Yash has made an unjustified assumption.

Leo defined point $D$ (or point $X$) in a different way, which means that Leo didn't need to make this assumption:

Yash and Leo then proceeded in the same way as each other. Here is the rest of Leo's work: