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'Pythagoras for a Tetrahedron' printed from https://nrich.maths.org/
There are a few different approaches you can try.
- You could use Heron's formula for the area of a triangle.
- You could use $\tfrac 1 2 ab \sin C$ for the area of a triangle. It might be helpful to remember that $\cos^2 \theta+\sin^2 \theta = 1$.
- You could try and find a point $X$ on $AB$ such that $CX$ is perpendicular to $AB$, and then use $\tfrac 1 2 bh$ for the area of a triangle. Vectors might be useful here, as well as the fact that ${\bf p}\cdot {\bf q}=0$ if and only if the vectors ${\bf p}$ and ${\bf q}$ are perpendicular. There are quite a lot of perpendicular vectors in this question!