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Pythagorean theorem as sweeping area – Better expressed



The pythagorean position is often taken as a fact about the right triangles.

Let's try a broader interpretation: The Pythagorean chair explains how the 2D area can be combined.

Here is what I mean. Suppose we have two lines lying down (the creative name Line A and Line B ). We can spin them to create area:

  soplinje

Okay, funny enough. Where is the mystery?

Well, what if we combine the line segments before they spin them?

 sweep-combined-line

Whoa. The area swept out seems to change. Should only move the lines, not extend them, change area?

Running the numbers

The above eyeballing chart seems to be growing. Let's work out the details.

As an example, assume $ a = 6 $ and $ b = 8 $. When swept in circles ($ text {area} in r ^ 2 $) we get:

 displaystyle {text {Circle A} = pi a ^ 2 = pi 6 ^ 2 ) = 36 pi} "align =" absmiddle "class =" tex

 displaystyle {text {Circle B} = pi a ^ 2 = pi (8 ^ 2) = 64 pi} "align =" absmiddle "class =" tex

For a total of $ 36 pi + 64 pi = 100 pi $.

The combined segment has length $ c = a + b = $ 14, and when we spin we get:

 displaystyle {text {Circle C} = pi c ^ 2 = pi ( 14 ^ 2) = 196 pi} "align =" absmiddle "=" tex

Uh oh. It is much more than before.

The problem

What happened? Well, Circle A did not change. But Circle B is much smaller than Ring B (just look at it!).

The problem: When Line B spins itself, "it just knocks out" 8 units. When we attach line B to line A, it reaches 6 + 8 = 14 units. Now, the circular sweep covers more area, which means that Circle B is smaller than Ring B.

 sweep-near-parallel

Mathematically here is what happened. Underbrace {b ^ 2} _ {text {Circle B}}} "alt =" displaystyle {underbrace {[a + b] ^ 2} _ {text {Circle C}} = underbrace {a ^ 2} _ {text {Circle A}} + subbrace {2ab + b ^ 2} _ {text {Ring B}}> underbrace {a ^ 2} _ {text {Circle A} } + underbrace {b ^ 2} _ {text {Circle B}}} "align =" absmiddle "class =" tex “/>

Ignore $ in $ for a moment because it is a common term. When you expand $ c ^ 2 = (a + b) ^ 2 = a ^ 2 + 2ab + b ^ 2 $ there is a new $ 2ab $ term that must go somewhere. Since Circle A does not change, this extra area must be displayed in Ring B.

Making things up

It … makes sense that the area changes, but I do not like it. Just moving things around shouldn't have this effect! Can the site ever be the same?

Sure, if we remove the $ 2ab $ term. The simple solution is to set $ a = 0 $, but it's cheating and you know it.

Let's find a smart solution. Intuitively, the question is: How can line A length not help line B when spinning?

Tilt it! When we rotate line B, it has less benefit from line A length. Ladders are useless when lying on the floor, right?

 sweep-when-parallel

When we go Full Perpendicular ™, $ 2ab $ term and Circle B = Ring B. (In vector terms, the dot product is zero: $ a cdot b = 0 $) .

Ah – that's the meaning of the Pythagorean theorem. When the line segments are perpendicular, the same area is swept if the lines are combined or separated.

Checking the Math

It's not a bad idea to make sure the numbers are right. Since the segments are now perpendicular, we know $ c ^ 2 = a ^ 2 + b ^ 2 $, so:

 displaystyle {text {Full distance} = c = srt {a ^ 2 + b ^ 2}} "align =" absmiddle "class =" tex

 displaystyle {text {Width of the ring} = text {c - a} = sqrt {a ^ 2 + b ^ 2} - a} "align =" absmiddle "class =" tex

Now we can calculate:

 displaystyle {text {Ring B} = text {Circle C} - text {Circle Pi c ^ 2 - pi a ^ 2 = pi (a ^ 2 + b ^ 2) - pi a ^ 2 = pi b ^ 2 = text {Circle B}} "align =" absmiddle "class =" tex

Tada! Call and circle sweep the same area.

In our example, we have Circle A = $ 36 in $, Circle B = $ 64 in $, $ c = sq {36 + 64} = $ 10. The ring width is $ 10-6 = 4 $.

Summary

Don't think about the pythagorean theorem just about the area. When parts are perpendicular, the area they do is independent of how they are arranged.

Appendix: Mixed Thoughts

  • Cosin's law expressly expresses the $ 2ab $ term assumed to be zero in Pythagorean theorem. The area in Ring B can even be "negative" if we tilt line B to point inside.
  • We can combine area from multiple dimensions ($ x ^ 2 + y ^ 2 + z ^ 2 + … $). As long as they are mutually perpendicular, the area swept by each dimension is swept by the sum.
  • The Pythagorean rate is a ratio in the 2D domain domain ($ c ^ 2 = a ^ 2 + b ^ 2 $). We start here and convert it to a ratio in the 1D domain ($ c = sqrt {a ^ 2 + b ^ 2} $). The transformation happens as often as we forget where it started.
  • Additional reading in sweeping area: https://www.cut-the-knot.org/Curriculum/Geometry/PythFromRing.shtml

Happy math.


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