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3 Responses
Ooh, 3D Pythagoras time-my favorite geometric rabbit hole! I always go via a face diagonal first because it feels more step-by-step: if the cuboid has side lengths a, b, and c, then the diagonal across the ab-face should just be a + b (you’re literally going corner to opposite corner along that rectangle), and then the full space diagonal would be sqrt((a + b)^2 + c^2). That way you’re building one right triangle on top of another, which is a nice pattern. I’m a bit unsure though-part of me thinks I might be supposed to square only one of them or maybe even add c at the end, like sqrt(a^2 + b^2) + c, but that seems too linear for a diagonal. I’ve also seen people jump straight to something like sqrt(a^2 + b^2 + c^2), but I can’t tell if that’s just a shortcut or if it’s skipping the geometric reasoning. Does this stepwise “face first, then space” approach make sense to you, or do you prefer going straight for a single formula?
I like to think of the space diagonal as coming from two right triangles glued together: if a cuboid has side lengths a, b, and c, first take a face diagonal d on, say, the a-by-b face: d = sqrt(a^2 + b^2). That diagonal lies flat on the face. Now treat that as one leg of a second right triangle whose other leg is the perpendicular edge c; applying Pythagoras again gives the space diagonal D = sqrt(d^2 + c^2) = sqrt(a^2 + b^2 + c^2). So you can do it in two clear steps (via a face diagonal), or remember the one-step formula directly: the square of the space diagonal equals the sum of the squares of the three mutually perpendicular edges. It doesn’t matter which face you start with, because the roles of a, b, c are symmetric. A short explanation with diagrams is here: https://www.khanacademy.org/math/geometry/hs-geo-analytic-geometry/hs-geo-distance-in-3d/a/three-dimensional-pythagorean-theorem
For a cuboid with edge lengths a, b, and c, the space diagonal d (from one corner to the opposite corner) is d = sqrt(a^2 + b^2 + c^2). You can get this in two steps or one: first find a face diagonal f = sqrt(a^2 + b^2) on any face, then apply Pythagoras again with the third edge to get d = sqrt(f^2 + c^2) = sqrt(a^2 + b^2 + c^2). Personally, I think the two-step picture helps keep the right triangles straight, but doing it in one step is perfectly fine since it’s just the 3D distance formula. I’m pretty sure that’s the cleanest approach; just double-check you’re using three mutually perpendicular edges. A short refresher on the 3D distance idea (same formula) is here: https://www.khanacademy.org/math/geometry-home/geometry-three-dimensions/intro-to-three-dimensions/a/distance-in-three-dimensions.