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  Pythagoras in 3D (Geometry)


The Pythagorean theorem in three dimensions

Earlier I said that everything is triangles. That applies even to 3-D shapes.

But to save you some trouble, let’s just agree that the Pythagorean Theorem works just fine in three dimensions. An example will make this clearer:

If I have a box that’s 2 by 3 by 6, what’s the length of its body diagonal? (Note: a body diagonal is a line segment connecting two opposite vertices of a box, for example from upper right near corner to lower left far corner.)

In this case, we use the 3-D Pythagorean Theorem:

a2 + b2 + c2 = d2

...where a, b, and c are our three edges, and d is the body diagonal.

By the way, this formula is not necessarily something to memorize, because it follows from the regular Pythagorean Theorem. I’ll show you. Take a look at the picture of the glass brick below. To find the first diagonal -- the one across the bottom -- use the Pythagorean Theorem and get √61. Then, to find the second diagonal -- the one that actually goes through the brick -- notice that it’s the hypotenuse of the triangle whose base is the first diagonal and whose height is three.

So, using the Pythagorean Theorem again, we take the base, √61, and square it. That’s 61. Then we take the height, 3, and square it. That’s 9. Add them together, 70. Take the square root, √70, which is the thing we were looking for.

So now that you’ve seen it as a formula and as a process that only requires the Pythagorean Theorem, I’ll leave it to you whether you’d prefer to memorize the 3-D version, or just work it out from the regular version whenever you need it. Both are good approaches; it just depends on whether you prefer to memorize or work things out.

Either way, for our original problem, we get d = √49 = 7. The answer is 7.

Anyhow, this problem is pretty deep into the specifics of this test. The rest of this section will talk about the general math strategies that are especially helpful for this test.