# Surprising Uses of the Pythagorean Theorem

#101

I would like to add to the application to sorting algorithms. Yes, it is easier to partition the input and sort them separately. But where has the remaining time/effort gone? It is in order to “merge” the two partitioned sets into a single sorted array. However this takes O(n) and not O(n^2). This is an early indication that sorting is possible in lesser than O(n^2) time.

#102

Hi,
Everyday application of Pythogoras theorem(extended to power 3)is in buying coconuts/fruits.
If a fruit of 3"dia costs 10 bucks, a fruit of 4" dia costing less than 23 bucks as well as a fruit of 5"dia costing less than 46 bucks are cheaper.

#103

@NANDEESH: Ah, in this case I think it might be more of a regular area comparison?

3" dia = 1.5" rad = 2.25 * pi area
4" dia = 2" rad = 4 * pi area. As long as it’s cheaper than 10 * (4/2.25) = 17.77 it will be better deal.

#104

I suppose we have to compare volumes rather than areas of cross section.
More precisely, we may have to compare shell volumes (of say 1" thickness) in case of coconuts.

#105

@NANDEESH: Ah! Yes, that makes sense – my mistake!

#106

How does this work for areas under the graph of y = x^2?

#107

[…] Here are some surprising uses of the Pythagorean Theorem […]

#108

The Pythagorean theorem is based on similarity. I found a quite interesting site at: http://www.echteinfach.tv/2011/10/rechtwinklige-dreiecke-satz-pythagoras.html#w
If you look for the paragraph “Das Geheimnis hinter dem Satz des Pythagoras” (the secret behind Pythagoras) you find some interesting stuff:

Conclusion: All 3 triangles are similar, so all 3 squares are similar. This is the foundation of the theorem.

The description that I like as well (‘expanded similarity’?): each square originates from its corresponding triangle, it’s the triangle itself, increased in size, and changed in form! See images on the website to understand that.

#109

love this

#110

[…] also takes credit on measurements. For instance, you can apply the pythagorean theorem to know the shortest way to go to one place. Even angles are given too much focus on this area; so […]

#111

i like this website it help me finish a math report i will definetly use this website more often there is so much i learned just on the first page i didnt believe that you could uses for any shape that can be squared so i tried it out and its true you can that just made math so much easier

#112

@sabrina: Thanks, glad you liked it!

#113

yeah it is helpful and interesting also

#114

I can’t seem to understand this:
"Assuming the boats are similarly shaped, the paint needed to coat one 50 foot yacht could instead paint a 40 and 30-footer."
I don’t understand how physically that’s possible, or the fact that two separate things could equal larger area, but end up equaling a smaller one as the hypotenuse. I’m sorry I’ve read over this a few times over the course of a few days but it’s not clicking. Mark’s comment does not help. (jan 14th 2010)

#115

#116

@CrazyFeetKait: Awesome, glad you enjoyed it! Thanks for the feedback, I would like to clarify that part, probably with a diagram.

I see the area factor as the conversion rate depending on what you want to measure. One way to think about it:

• The average person has their armspan equal to their height (“height factor of 1”)
• The average person has their height equal to 10 times their foot size (“height factor of 10”)

I’m not sure if the 10x is exactly right, but the idea is that depending on what part you measure, you get a different “formula” (armspan * 1, foot size * 10, etc.). The neat thing is the formula works on babies to adults since it’s in term of “armspan” and not absolute height. Hope this helps a bit! I would like to clarify this more.

#117

Woah. Simply mind blowing from what can derive from one simple little theory!
Who knew the Pythagorean Theorum had so much more potential that many didn’t even know about! :o

The only part that had me a bit confuzzled was:

"For example, look at the diagonal of a square (“d”). A regular side is d/sqrt(2), so the area becomes 1/2 d2. Our “area contant” is 1/2 in this case, if we want to use the diagonal as our line segment to be squared.

Now, use the entire perimeter (“p”) as the line segment. A side is p/4, so the area is p2/16. The area factor is 1/16 if we want to use p2."

The whole area factor concept is hard to catch onto…

(the area becomes 1/2 d2)
(A side is p/4, so the area is p2/16. The area factor is 1/16 if we want to use p2.")

But overall, this entire article is Mathtastic!

#118

#119