Let S be the area of triangle T in three-dimensional space. Let A, B, and C be area of the projections of T to the xy, yz, and xz planes respectively. Then

S² = A² + B² + C².

There’s an elegant proof of this theorem here using differential forms. Below I’ll sketch a less elegant but more elementary proof.

You could prove the identity above by using the fact that the area of a triangle spanned by two vectors is half the length of their cross product. Suppose a, b, and c are the locations of the three corners of T. Then

S² = v²/2,

where

v = (a – b) × (c – b)

and by v² we mean the dot product of v with itself.

Write out the components of v² and you get three squared terms. Notice that when you set the x components to zero, i.e. project onto the yz plane, the first of the three terms is unchanged and the other two are zero. In other words, the first of the three terms of v² is A². A similar argument shows that the other two terms are B² and C².