Pythagoras for a Tetrahedron
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Consider a right-angled tetrahedron with vertices at $O(0,0,0)$, $A(a, 0, 0)$, $B(0, b, 0)$ and $C(0, 0, c)$.
Let the area of face $AOB$ be $P$, the area of $BOC$ be $Q$ and the area of $COA$ be $R$. Also let the slanted face $ABC$ have area $S$.
($S$ is not shown on the diagram above!). Can you prove that $P^2+ Q^2+ R^2= S^2$?
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Equivalently: (area $OBC$)$^2 + $(area $OCA$)$^2 + $(area $OAB$)$^2 = $(area $ABC$)$^2$.
Extension
If you enjoyed this question, you might like explore STEP Support Programme Foundation Assignment 5 which asks a question about the perpendicular distance of face $ABC$ from the origin.
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