how can in incorporate Pythagoras into this theory
Sorry, I am still gropling with the question - how the Pythagorean theory is helpful in our life?
@Amrit: This is a tough question since the theorem is everywhere – it’s in the structure of the universe :). It’s a bit like asking how circles are useful in our life. Most people don’t “make” circles that often, but it’s a concept which is everywhere once you start looking for it.
If you ever need to find the distance between two things (driving, parts in a machine, diagrams on a piece of paper) the Pythagorean theorem was used. Most people don’t use it directly (most of us aren’t involved in computing things) but it’s one of the most useful results ever. Anything that involves a distance measurement likely involves the Pythagorean theorem; that’s just the starting point (it can define equations for circles, etc.).
Thanks for the well written article and great diagrams! I really enjoyed it and learned a lot. Your examples and humour were great and made for an enjoyable read - yes, Im learning that math can be enjoyable!
@Deb: You’re welcome, really glad you enjoyed it! I think any subject can be made enjoyable if looked at the right way :).
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can v use pythagoros theorem in the field of medicine. ( please answer )
I really like the boat thing. You can also put it this way: if with a bucket of paint I can paint the hull of an “a” feet boat, how long a boat can I paint (assuming the boats are similar) if I have 2 buckets? Answer: I can paint a boat which is aSqrt(2) long (and not twice as long!).
More generally, if with one bucket of paint I can do an “a” long boat, with “n” buckets of paint, I can do a boat which is x=aSqrt(n) long. That’s because kx^2=kna^2, where k is the form coefficient: the two “k” cancel out, and we solve for “x”, discarding, of course, the negative solution. This “k” thing is the reason why Pythagorean theorem holds for any shape: “k” cancels out when we solve.
But what if we want to paint many small boats with the same bucket that allows us to do an “l” long boat? As you stated, any triangle can be split in 2 similar triangles. But then, any form, e.g. the hull of a boat, can be split in 2 similar forms and so on and on. Let the surface of the boat be kl^2. We can paint as many smaller boats as we want (all the same or different), as long as kl^2=ka^2+kb^2+k^c^2… or, after “k” cancels out, l^2=a^2+b^2+c^2…
If the smaller boats are all the same, and we have “n” of them, that becomes l^2=nx^2 and, if we solve for x, x=(l^2)/(Sqrt(n)), discarding the negative solution.
If feel that Pythagorean theorem can indeed be extanded beyond triangles, but then, we are left with plain algebra to work with. Pythagorean theorem also can’t easily solve the “n” boats problem: we must use algebra for that.
wow you have helped me sooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo(etc.)
much on my project!
thhanks, i’m impressed.
@antonia: Awesome, glad you liked it!
you really helped out in my school project. thanks!
This dint help at all
@Anon: Sorry it didn’t help, feel free to leave a comment about what parts were confusing.
Its good! I always find maths difficult to understand so this too was a bit difficult.
Thanks. My maths project has become easy.
i have a maths question : investigate wheather the theorem holds if equilaterial triangkes are drawn instead of squares?
Quick one on the SQRT computer formulation of
Pythagoras - "If you write a program that uses the distance form c = sqrt(a^2 + b^2) you will suffer from the lose of half of your available precision because the square root operation is last. A more accurate calculation is c = a * sqrt(1 + b^2 / a^2). If a is less than b, you should swap them and of course handle the special case of a = 0.
mccoyn — October 29, 2007 @ 6:24 am
I can follow the logic with one exception the ‘discovery’ that where a<b that the formula requires b to be switched for a. Its true as
1 + a^2/b^2 does not equal 1 + b^2/a^2 numerically
but what defines which of the relevant 'sides’
in the original theorem should come first since
any order in their squaring will hold for
a^2 + b^2 = c^2 … thanks
hey…your articles are simply superb…while reading this one i started thinking as to what s the intuition behind area of a square being side^2 etc…and then came the "aha"moment…if we place these line segments one above the other there will be total of s(side) line segments…so the total area is side*area of one line segment…since lines are having unit thickness area of 1 line segment is simply its length…and thus area of square=side^2…
in fact if we consider the diagonal of a square, we can move it to the corners and then we span a square twice the size of the original square…so actual area=1/2 * d^2…
same way we can think of other areas too…its amazing!!!