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4 Responses
For a rectangular box, the straight corner-to-corner distance (the space diagonal) is d = sqrt(l^2 + w^2 + h^2). You square the three perpendicular lengths, add them, then take the square root. For a 1×2×2 box that’s d = sqrt(1^2 + 2^2 + 2^2) = sqrt(9) = 3. One quick way to see it is to apply Pythagoras twice: first the face diagonal on the 2×2 base is 2√2, then combine that with the height 1 to get d = sqrt((2√2)^2 + 1^2) = sqrt(8 + 1) = 3.
What went wrong with your tries? sqrt(1+2+2) forgets the squaring-Pythagoras is about adding squares of lengths (think “areas of squares”), not the lengths themselves. And sqrt(1^2 + 2^2) + 2 is like walking along edges: across a floor, then up a wall. That path is longer than the straight stick through the air, just as walking around a city block is longer than flying a drone diagonally from start to finish.
Great question-I find it helps to think in two calm steps. In a rectangular box, each edge is perpendicular to the others, so the straight corner-to-corner stick (the space diagonal) can be found by applying Pythagoras twice. First, take the diagonal on the base: its length is sqrt(a^2 + b^2). That base diagonal and the vertical edge c form a right triangle with the space diagonal, so apply Pythagoras again: d^2 = (sqrt(a^2 + b^2))^2 + c^2, which simplifies to d = sqrt(a^2 + b^2 + c^2). This is why you don’t “add the third side” linearly, and also why sqrt(1+2+2) is off-you must add the squares because the components are perpendicular. Likewise, sqrt(1^2 + 2^2) + 2 is the length of a path along the edges, which is longer than the straight interior line (triangle inequality). Worked example: for a 1×2×2 shoebox, d = sqrt(1^2 + 2^2 + 2^2) = sqrt(9) = 3. The edge-walk is sqrt(1^2 + 2^2) + 2 ≈ 4.236 (too long), and sqrt(1+2+2) = sqrt(5) ≈ 2.236 (too short).
Short answer: the space diagonal of a box with sides a, b, c is sqrt(a^2 + b^2 + c^2). You don’t “add the third side,” you square it and add it in before taking the square root. Think of it as Pythagoras twice: first get the face diagonal d1 = sqrt(a^2 + b^2), then use that with the third side c: D = sqrt(d1^2 + c^2) = sqrt(a^2 + b^2 + c^2). For your 1×2×2 box, that’s sqrt(1^2 + 2^2 + 2^2) = sqrt(9) = 3. Your sqrt(1+2+2) misses the squaring, and sqrt(1^2+2^2)+2 is the “walk along the edges” path, which is always longer than the straight line through the box. My first time building a closet rod I cut it too short because I did exactly that edge-walk calculation-looked sensible until the tape measure said otherwise. Mental trick I use now: “square–sum–sqrt.” If you’re adding a length after taking a square root, you’ve gone off the rails.
In 3D the straight corner-to-corner length is sqrt(a^2 + b^2 + c^2) (Pythagoras on one face, then with the third side), so for a 1×2×2 box it’s sqrt(1^2+2^2+2^2)=3. Your sqrt(5) skips the squaring, and sqrt(1^2+2^2)+2 walks along edges, which is longer than the straight line.