I’m prepping for a test and I keep getting tangled up on Pythagoras in 3D. In 2D I’m fine: I can spot the right triangle and life is good. But as soon as the problem jumps into a box/room shape, my brain does that loading wheel thing.
For example, say I’ve got a rectangular box with lengths 8, 6, and 3, and they ask for the straight-line distance from one corner to the opposite corner through the inside. I feel like that should be straightforward (like the longest straw you could fit in the box), but I keep second-guessing whether I’m supposed to do it in one go or in two stages. Do I find a diagonal on a face first, and then somehow use that with the third dimension? Or is there a simpler “one-and-done” way I’m supposed to recognize?
I get even more stuck when it’s not the opposite corner. Like: bottom-front-left corner to the center of the top face of a 10 by 12 by 5 box. I try to picture which right triangle that line actually belongs to, and then my sketch turns into a potato and I lose where the right angle is supposed to be. Same thing if it’s from one corner to a point halfway up the back edge. I don’t know which lengths I’m allowed to pair together without accidentally making a weird diagonal that doesn’t sit on a right triangle.
Is there a simple way to decide which edges or distances to combine in 3D? Do I always “flatten” it mentally onto a net and do Pythagoras twice, or is there a reliable shortcut rule for when that makes sense? And if the problem gives coordinates (like a point at one corner and another point somewhere inside), is there a one-step approach I should be recognizing, or should I still be thinking in terms of two right triangles glued together?
If it helps, the way I’m picturing it is like pulling a tight string inside a shoebox from one point to another. Sometimes the string lies along two sides and then cuts through the air, and sometimes I feel like it should just zip straight across. Kind of like finding a shortcut in Minecraft-you’ve got x, y, and z, and I’m not sure when I’m allowed to combine them all at once versus taking them one plane at a time.
I’m not looking for full working, I just want a clear way to know which triangle is “the” right triangle in 3D problems like these. Any tips for spotting it quickly during a test so I don’t panic and start guessing?
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3 Responses
The reliable rule is: take the differences along three mutually perpendicular directions and use d = sqrt((Δx)^2 + (Δy)^2 + (Δz)^2). That is the 3D Pythagoras, and it is exactly the same as “do a face diagonal, then combine with the third dimension” because (Δx^2 + Δy^2) gives the face diagonal squared, and adding Δz^2 completes it. So for a box 8 by 6 by 3, opposite corners are sqrt(8^2 + 6^2 + 3^2). For “bottom‑front‑left” to the center of the top face in a 10 by 12 by 5 box, set coordinates: (0,0,0) to (5,6,5), giving sqrt(5^2 + 6^2 + 5^2). If it’s to a point halfway up a back vertical edge, I’m assuming the back edge is at x = 10, y = 12, so (0,0,0) to (10,12,2.5) gives sqrt(10^2 + 12^2 + 2.5^2); if you mean a different edge, adjust the zeroed coordinates accordingly. The “right triangle” you’re after is the one whose legs are exactly those perpendicular component differences; never mix lengths that aren’t perpendicular as legs. In practice: assign axes (length, width, height), write the three deltas, square–sum–root. If both points lie in the same plane, one delta is zero and it collapses to the usual 2D Pythagoras; if they lie on the same line, two deltas are zero. If you prefer a visual check, project the segment onto a face to get the face diagonal first, then combine with the remaining perpendicular component-the algebra matches the one‑and‑done formula. I might be misreading one of your edge descriptions, but this component rule will steer you right under exam pressure.
I love the “longest straw in a shoebox” picture-that’s exactly right, and here’s the clean rule that stops the brain’s loading wheel: in a rectangular box, break the move from A to B into how far you go in three mutually perpendicular directions (call them x, y, z), then the straight-line distance is one-and-done: distance = sqrt((dx)^2 + (dy)^2 + (dz)^2). That’s just Pythagoras stacked once more, and it’s the same as doing it in two stages (face diagonal first, then combine with the third direction), because squaring that face diagonal just gives you the sum of squares anyway. Think Minecraft: how many blocks east-west, how many north-south, how many up-down; only add squares of moves that are at right angles. So the “right triangle” you’re after is always the one whose legs are exactly those three perpendicular components; the segment from A to B lies in some plane, and in that plane those three components combine into a single hypotenuse. Worked example: bottom-front-left to the center of the top face of a 10 by 12 by 5 box-start at (0,0,0); the top-face center is (5,6,5), so dx=5, dy=6, dz=5, and the distance is sqrt(5^2+6^2+5^2)=sqrt(86)≈9.27. For the “longest straw” in an 8 by 6 by 3 box, it’s the same idea with dx=8, dy=6, dz=3, so distance = sqrt(8^2+6^2+3^2)=sqrt(109) (doing the face first gives the same result). If the target is on a face, one of dx, dy, dz is zero (back to 2D Pythagoras); if it’s on an edge, two are zero (just a straight edge length). Quick test tip: sketch a tiny L-shaped path along the box edges from start to finish-read off those three perpendicular moves, square-add-sqrt, and you’ve got the right triangle without guessing.
Quick rule: line up x, y, z with the box edges and use the perpendicular changes as the legs-distance = sqrt((Δx)^2+(Δy)^2+(Δz)^2); so opposite corners of an 8×6×3 box give sqrt(8^2+6^2+3^2), and doing it via a face diagonal then the third side is the same thing. I used to draw wobbly potato‑boxes until someone told me “list the three moves your string makes-along x, y, z-square, add, square‑root,” which makes, say, bottom‑front‑left to the top‑face center in a 10×12×5 box just Δx=10/2, Δy=12/2, Δz=5.