I’m stuck on when it’s valid to combine Pythagoras in 3D. Suppose I have a rectangular box 10 by 6 by 8. Let M be the midpoint of the vertical edge at the front-left (so halfway up that corner), and I want the straight-line distance from M to the opposite top-back-right corner.
My first instinct was to do it in two steps: base diagonal first, then use the height. I got base diagonal sqrt(10^2 + 6^2) = sqrt(136), and then I used the full height 8, giving sqrt(136 + 8^2) = sqrt(200). But then I realized M is halfway up, so maybe the vertical difference is 4, not 8, which would give sqrt(136 + 4^2) = sqrt(152).
Here’s my confusion: is it even valid to pair “half the height” with the full base diagonal like that? The base diagonal seems to connect two bottom corners, while my start point is halfway up, so I’m not sure that triangle is actually a right triangle with the distance I want as the hypotenuse.
Can someone explain which right triangle I should be using here (and why it’s right-angled)? A quick, clean way to set this up without overcomplicating it would really help.
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3 Responses
Yup-set coordinates so M=(0,0,4) and the opposite corner is (10,6,8), then the distance is sqrt((10)^2+(6)^2+(8−4)^2)=sqrt(152); equivalently, the horizontal separation is sqrt(10^2+6^2) and the vertical difference is 4, and those are perpendicular (vertical vs horizontal), so Pythagoras applies. Hope this helps!
I think the clean way is to use the 3D distance: from M=(0,0,4) to the opposite corner (10,6,8), the horizontal leg is the base diagonal sqrt(10^2+6^2)=sqrt(136) and the vertical change is 4, so the distance is sqrt(136+4^2)=sqrt(152). Picture sliding horizontally to S=(10,6,4) and then straight up-horizontal vs vertical makes a right angle, so Pythagoras applies; quick refresher: https://www.mathsisfun.com/algebra/distance-3d-points.html
Yes-let H be the point directly below the opposite corner but at M’s height (10,6,4): MH is a purely horizontal base diagonal of length sqrt(10^2+6^2), HT is a vertical 4, they’re perpendicular, so MT = sqrt(10^2+6^2+4^2) = sqrt(152). I once kept using the bottom diagonal until a teacher said “slide then climb,” and drawing that little right triangle (horizontal slide, then vertical climb) finally made it click for me.