EASA / CAR Part-66 · Module 1 · 1.3 Geometry · Cat B1/B2

Pythagoras in Two and Three Dimensions

Objective: the theorem you know for a flat triangle works just as well through a solid. Find the diagonal across the floor first, then use that as one side of a second triangle to reach the far top corner.

The box, with both diagonalsdrag to rotate

The two right triangles, flattened

Floor diagonal
0m
Space diagonal
0m
a² + b² + c²
0
Angle to floor
0°
floor d = √(a² + b²)  ·  space D = √(d² + c²) = √(a² + b² + c²)

Do it in two steps, not one. First find the diagonal across the floor using a and b. That diagonal then becomes the base of a second right triangle standing upright, with the height c as the other side, and its hypotenuse is the space diagonal.

The two steps collapse into one formula. Because the floor diagonal squared is already a² + b², substituting it into the second triangle gives √(a² + b² + c²) directly. Knowing where it comes from means you can rebuild it if you forget it.

This is how cable runs and bracing lengths are worked out. A wire running corner to corner through a bay, or a strut across a box structure, is exactly this calculation, which is why it appears in Module 1 and again in structures.