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Pythagoras for a Tetrahedron

Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

coordinate diagram
A natural generalisation of Pythagoras' theorem is to consider a right-angled tetrahedron with four faces, three in mutually perpendicular planes and one in the sloping plane. Then ask "what corresponds to the squares of the lengths of the sides?"

The answer must be "the squares of the areas of the faces". If these areas are $P$, $Q$ , $R$ and $S$ respectively then prove that:

$P^2+ Q^2+ R^2= S^2$

Equivalently: (area $OBC$)$^2 + $(area $OCA$)$^2 + $(area $OAB$)$^2 = $(area $ABC$)$^2$.