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3 Responses
Yup, you’ve got the right idea! In a rectangular box, the long space diagonal is the hypotenuse of a right triangle whose legs are the base diagonal and the vertical edge. Since the base is perpendicular to the height, any diagonal drawn on the base is also perpendicular to that vertical edge-so you can safely do the base first and then combine with the height. That’s exactly the 3D version of Pythagoras: d = sqrt(3^2 + 4^2 + 12^2).
Even better, this one has a lovely pattern. The base diagonal from the 3-by-4 face is 5 (because 3-4-5), and then the space diagonal is sqrt(5^2 + 12^2) = sqrt(25 + 144) = sqrt(169) = 13. So the whole box secretly hides a 5-12-13 triple in 3D! Unless I’m misunderstanding the orientation (which I don’t think I am), the long diagonal is 13.
You’ve got it right: the long corner-to-corner diagonal of a 3×4×12 box is sqrt(3^2 + 4^2 + 12^2), and doing it in two steps is exactly why that works. Think of it like this: first, look at the base rectangle (3 by 4). Its diagonal is sqrt(3^2 + 4^2) = 5. Now picture the triangle whose one leg is that base diagonal (length 5), the second leg is the vertical edge of the box (length 12), and the hypotenuse is the space diagonal you want. Those two legs are perpendicular because the vertical edge is perpendicular to the whole base plane, so by Pythagoras again the space diagonal is sqrt(5^2 + 12^2) = sqrt(25 + 144) = sqrt(169) = 13. Doing it all at once just combines those steps: sqrt(3^2 + 4^2 + 12^2) gives the same result. I remember the first time this clicked for me was while helping a friend measure if a long box would fit in a car-someone said “try the floor diagonal first, then up,” and I thought we were cheating until we got a perfect 13 from the 3-4-12-felt like a magic trick I should’ve already known!
You’ve got it right! In a rectangular box, the long corner-to-corner diagonal is √(a² + b² + c²), and doing it in two Pythagoras steps is exactly why that formula works. Picture the bottom face: its diagonal is √(3² + 4²) = 5. Now that bottom-face diagonal lies flat, and the 12-edge sticks straight up, perpendicular to the whole face, so those two are legs of a right triangle whose hypotenuse is the space diagonal. So the space diagonal is √(5² + 12²) = √(25 + 144) = √169 = 13. If you do it “all at once,” you’re just substituting 5² = 3² + 4² into that last step, which gives √(3² + 4² + 12²) anyway. I used to worry I was double-counting something, but it’s legit because each pair of edges you combine are perpendicular. A tiny worked example to show the idea: in a 1×2×2 box, the diagonal is √(1² + 2² + 2²) = √9 = 3; or do it in steps: bottom diagonal √(1² + 2²) = √5, then √(5 + 4) = 3.