The Pythagorean theorem is a celebrity: if an equation can make it into the Simpsons, I'd say its well-known.

But most of us think the formula only applies to triangles and geometry. Think again. The Pythagorean Theorem can be used with **any shape** and for **any formula that squares a number**.

Read on to see how this 2500-year-old idea can help us understand computer science, physics, even the value of Web 2.0 social networks.

## Understanding How Area Works

I love seeing old topics in a new light and discovering the depth there. For example, I realize I didn't have a **deep grasp** of area until writing this article. Yes, we can rattle off equations, but do we really *understand* the nature of area? This fact may surprise you:

The area of **any shape** can be computed from **any line segment squared**. In a square, our "line segment" is usually a side, and the area is that side squared (side 5, area 25). In a circle, the line segment is often the radius, and the area is pi * r^2 (radius 5, area 25 pi). Easy enough.

**We can pick any line segment** and figure out area from it: every line segment has an "area factor" in this universal equation:

Shape |
Line Segment |
Area |
Area Factor |
---|---|---|---|

Square |
Side [s] | s^{2} |
1 |

Square |
Perimeter [p] | 1/16 p^{2} |
1/16 |

Square |
Diagonal [d] | 1/2 d^{2} |
1/2 |

Circle |
Radius [r] | pi r^{2} |
pi (3.14159...) |

For example, look at the diagonal of a square ("d"). A regular side is d/sqrt(2), so the area becomes 1/2 d^{2}. Our "area constant" 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 p^{2}/16. The area factor is 1/16 if we want to use p^{2}.

## Can we pick any line segment?

You bet. There is always **some** relationship between the "traditional" line segment (the side of a square), and the one you pick (the perimeter, which happens to be 4 times a side). Since we can convert between the "traditional" and "new" segment, it doesn't matter which one we use -- there'll just be a different area factor when we multiply it out.

## Can we pick any shape?

Sort of. A given area formula works for all **similar** shapes, where "similar" means "zoomed versions of each other". For example:

- All squares are similar (area always s
^{2}) - All circles are similar, too (area always pi r
^{2}) - Triangles are
**not similar**: Some are fat and others skinny -- every "type" of triangle has its own area factor based on the line segment you are using. Change the shape of the triangle and the equation changes.

Yes, every triangle follows the rule "area = 1/2 base * height". But the relationship between base and height depends on the type of triangle (base = 2 * height, base = 3 * height, etc.), so even then the area factor will be different.

Why do we need similar shapes to keep the same area equation? Intuitively, when you zoom (scale) a shape, you're changing the absolute size but not the relative ratios within the shape. A square, no matter how zoomed, has a perimeter = 4 * side.

Because the "area factor" is based on ratios inside the shape, any shapes with the same "ratios" will follow the same formula. It's a bit like saying everyone's armspan is about equal to their height. No matter if you're a NBA basketball player or child, the equation holds because it's all relative. (This intuitive argument may not satisfy a mathematical mind -- in that case, take up your concerns with Euclid).

I hope these high-level concepts make sense:

- Area can be be found from
**any line segment squared**, not just the "side" or "radius" - Each line segment has a different "area factor"
- The same area equation works for similar shapes

## Intuitive Look at The Pythagorean Theorem

We can all agree the Pythagorean Theorem is true (here's 75 proofs). But most proofs offer a mechanical understanding: re-arrange the shapes, and voila, the equation holds. But is it really clear, intuitively, that it **must** be a^{2} + b^{2} = c^{2} and not 2a^{2} + b^{2} = c^{2}? No? Well, let's build some intuition.

There's one killer concept we need: **Any right triangle can be split into two similar right triangles.**

Cool, huh? Drawing a perpendicular line through the point splits a right triangle into two smaller ones. Geometry lovers, try the proof yourself: use angle-angle-angle similarity.

This diagram also makes something very clear:

- Area (Big) = Area (Medium) + Area (Small)

Makes sense, right? The smaller triangles were cut from the big one, so the areas must add up. And the kicker: because the **triangles are similar, they have the same area equation.**

Let's call the long side c (5), the middle side b (4), and the small side a (3). Our area equation for these triangles is:

where F is some area factor (6/25 or .24 in this case; the exact number doesn't matter). Now let's play with the equation:

Divide by F on both sides and you get:

Which is our famous theorem! You knew it was true, but now you **know why**:

- A triangle can be split into two smaller, similar ones
- Since the areas must add up, the squared hypotenuses (which determine area) must add up as well.

This takes a bit of time to see, but I hope the result is clear. How could the small triangles **not** add to the larger one?

Actually, it turns out the Pythagorean Theorem depends on the assumptions of Euclidean geometry and doesn't work on spheres or globes, for example. But we'll save that discussion for another time.

## Useful Application: Try Any Shape

We used triangles in our diagram, the simplest 2-D shape. But the line segment can belong to **any** shape. Take circles, for example:

Now what happens when we add them together?

You guessed it: Circle of radius 5 = Circle of radius 4 + Circle of radius 3.

Pretty wild, eh? We can multiply the Pythagorean Theorem by our area factor (pi, in this case) and come up with a relationship for any shape.

Remember, the line segment can be **any portion of the shape**. We could have picked the circle's radius, diameter, or circumference -- there would be a different area factor, but the 3-4-5 relationship would still hold.

So, whether you're adding up pizzas or Richard Nixon masks, the Pythagorean theorem helps you relate the areas of any similar shapes. Now that's something they didn't teach you in grade school.

## Useful Application: Conservation of Squares

The Pythagorean Theorem applies to **any** equation that has a square. The triangle-splitting means you can split any amount (c^{2}) into two smaller amounts (a^{2} + b^{2}) based on the sides of a right triangle. In reality, the "length" of a side can be distance, energy, work, time, or even people in a social network:

**Social Networks.**

Metcalfe's Law (if you believe it) says the value of a network is about n^{2} (the number of relationships). In terms of value,

- Network of 50M = Network of 40M + Network of 30M.

Pretty amazing -- the 2nd and 3rd networks have 70M people total, but they aren't a coherent whole. The network with 50 million people is as valuable as the others combined.

**Computer Science**

Some programs with n inputs take n^{2} time to run (bubble sort, for example). In terms of processing time:

- 50 inputs = 40 inputs + 30 inputs

Pretty interesting. 70 elements spread among two groups can be sorted as fast as 50 items in one group. (Yeah, there may be constant overhead/start up time, just work with me here).

Given this relationship, it makes sense to partition elements into separate groups and then sort the subgroups. Indeed, that's the approach used in quicksort, one of the best general-purpose sorting methods. The Pythagorean theorem helps show how sorting 50 combined elements can be as slow as sorting 30 and 40 separate ones.

**Surface Area**

The surface area of a sphere is 4 pi r^{2}. So, in terms of surface area of spheres:

- Area of radius 50 = area of radius 40 + area of radius 30

We don't often have spheres lying around, but boat hulls may have the same relationship (they're like deformed spheres, right?). Assuming the boats are similarly shaped, the paint needed to coat one 50 foot yacht could instead paint a 40 and 30-footer. Yowza.

**Physics**

If you remember your old physics classes, the kinetic energy of an object with mass m and velocity v is **1/2 m v ^{2}**. In terms of energy,

- Energy at 500 mph = Energy at 400 mph + Energy at 300 mph

With the energy used to accelerate one bullet to 500 mph, we could accelerate two others to 400 and 300 mph.

## Try Any Number

You can use any set of numbers that make a right triangle. For example, enter a total amount (50) and one subportion (30), and the remainder will appear below:

Suppose you want to see if a large pizza (16 inches) is bigger than two mediums (12 inches). Plug in 16 for C, and 12 for A. It looks like the large pizza can be split into a 12-inch and 10.5-inch pizza, so two-mediums are in fact bigger.

## Enjoy Your New Insight

Throughout our school life we think the Pythagorean Theorem is about triangles and geometry. **It's not.**

When you see a right triangle, realize the sides can represent the lengths of any portion of a shape, and the sides can represent variables in **any equation** that has a square. Maybe it's just me, but I find this pretty surprising.

There's much, much more to this beautiful theorem, such as measuring any distance. Enjoy.

Wow, thats really cool.

So if I’m getting this right, when a car is traveling at 100 mph, the energy that is required to take it to 110 mph (or the difference in its speed vector) is the same as that of two other cars travelling at 100 and 45,8 mph.

So, if they travel for 1 hour, they would have covered 100 miles (car A) and 110 miles (car C) respectively, but car A would have enough fuel to go for another 45 miles.

Makes you think about fuel economy/efficiency..

Einstein found this method of proof for Pythagorean Theorem when he was 12 or so; though he probably wasn’t the first (or last) to have done so.

@Spyros: I think that’s right from a pure energy point of view: the energy needed to go from 100-110mph equals the energy to go from 0-45.8 mph.

I’m not sure how well the pure energy analogy works (drag, efficiency of engine, etc.), but there is an ideal speed for fuel efficiency cars, and it’s not around 110mph :).

@Kaizyn: Thanks for the info. Yep, I like this particular proof compared to others because it focuses on the larger concepts, not just re-arrangement of area.

Spyros,

Thats the energy required to accelerate the vehicle to that speed, not the energy required to maintain the speed, nearly all of which is required to overcome wind resistance at high speeds.

The coolest thing I’ve ever discovered about Pythagorean’s Theorem is an alternate way to calculate it. 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.

Hi Mccoyn, thanks for the great info! The note about precision is especially useful, I hadn’t thought about the impact of order-of-operations on the calculation. I’m sure that comes in useful in graphics programming, etc.

Wow nice article. That puts the ‘ol Pythagorean theorem into a whole new perspective.

What program do you use to draw pictures in your articles?

Thanks, glad you liked the article. Yep, I’ve discovered that there are many gems to be found when we revisit concepts we “know” :).

I used PowerPoint 2007 to make the diagrams.

“The Pythagorean theorem is a celebrity: if an equation can make it into the Simpsons, I’d say its well-known.”

Stopped reading right there.

Ah, that’s too bad — you may have found it interesting! [Both the article and the Simpsons episode ]

Your site is triggering a virus warning. Some java class it’s loading has a worm.

http://s3.amazonaws.com/instacalc/release/build516/javascripts/library-internal.js.packed

Hi, thanks for the info. Avast appears to be have a false positive (many AV companies mis-detect javascript as Feebs, see more here: http://isc.sans.org/diary.html?storyid=2319&rss).

I split my files into smaller chunks and it should be ok now. Appreciate the tip.

Your proof of the Pythagorean theorem is very appealing. However, to be complete, you’d need to prove your supposition about similar triangle areas: concretely, the special case that the area of a right triangle can be computed as constant * hypotenuse^2. I tried a few different approaches, but they all end up having to apply Pythagoras in the end, which makes the whole exercise circular and thus ultimately pointless. If you try to establish the result by concrete computation it seems you are bound to run into this same difficulty–how do you get around it?

Ignore that; I’m an idiot. If you treat it abstractly, it’s almost blindingly obvious.

You just have to prove that L^2 / A is constant within a similarity class. Take two members of the same similarity class, of areas A and A’ and lengths L and L’. Let F be the factor of the dilation that maps the first figure onto the other. Then A = F^2 * A’ and L = F * L. Squaring the length equation gives L^2 = F^2 * L’^2. Dividing the area equation by this, the F^2 factors cancel, yielding A / L^2 = A’ / L’^2. So the area to squared length ratio is indeed constant.

Hi Per, glad you wrote it out :). No, you’ve saved me a bit of trouble as I wanted to discuss that aspect too!

If a “proof” of the Pythagorean Theorem does not bring in the Euclidean nature of the space under consideration, in what sense can it be considered a proof at all?

According to this wikipedia article, the “Euclidean metric” — which could also be called the “Pythagorean metric” — is actually one of the axioms of a properly-specified Euclidean geometry.

More interesting (to me, at least) would be a discussion of how the Pythagorean Metric leads to our concept of geometric area.

Sorry, I just can’t see how doing it the other way around makes sense.

Hi Ralph, I think discussions of other geometries would be pretty interesting and a good follow-up (I need to read up on them). This wasn’t meant to be a rigorous proof from first principles, more a new intuition onto an old result.

Hi

i m school kid

and dis seems to be terrific

Hi Harshil, glad you liked it.

i need a specific answer why circles are similar

hello, im a school student.. and i have a project on pythagorean theorem.. and i cant find out in concise the uses/applications of the same in daily life.. please help..

how does one calculate the height of i.e. a waterfall using the pathagorean

excellent , mind blowing work, i got a very big help from u for my maths project

@shashank: Glad it was helpful!

I’m doing a research paper and was wondering what your first and last name is so I can site the paper correctly. Thanks!

Hi Ryan, my name is Kalid Azad — good luck with your paper!

this give an complete information on pythagoraus !

I now love maths more.

Kalid,

You might want to make it clear that you’re restricting your discussion to squares, since the statement

is not true for rectangles.

Actually, I think I see what you mean. (Sorry, I’m commenting as I read. Maybe I should write down my comments and not say them until I’m finished reading, huh?)

I would like to see a derivation of mccoyn’s result of c = a * sqrt(1 + b^2 / a^2) being equivalent to c = a^2 + b^2, if anyone has one.

Hi John, thanks for the comments! No worries about the inline comments, it’s interesting seeing the thought process. Yep, rectangles can still follow the rule constant * side squared, but that constant will be different for each shape. In the case of a square, the constant is 1 (it is a different way to look at it).

The second result

c^2 = a^2 + b^2

c^2 = a^2 * (1 + b^2/a^2) [rearrange right side]

c = a * sqrt(1 + b^2/a^2) [square root of both sides]

Now, the physical meaning of this is interesting. It basically gives you a constant [sqrt(1 + b^2/a^2)] that maps you distance in the “a” direction to relative distance in the “c” direction.

There’s more details here if you like:

http://betterexplained.com/articles/rescaling-the-pythagorean-theorem/

@ritesh: Glad you enjoyed it!

Thanks Kalid,Superb work,its crystal clear to understand.

Thanks Naresh!

why do we use the pythagorean theorem??? (plz answer)

@Teresa: Hi, there’s several uses of the Pythagorean theorem mentioned in the article.

Can anyone explain the uses of Pythagorus theurom?(please can you answer this question)

In the section on proving the Pythagorean Theorem (Intuitive Look at the Pythagorean Theorem), I am unsure why the Area = F*hypotenuse^2. Can you please explain? I tried to use the A= 1/2bh equation and substitute one of the sides with h = sqrt(hyp^2 – B^2) but could not come back to the F * hyp^2.

Thanks,

Ashley

I Think This Is Good . But I Need Something That Could Be More/Better Explained !

@Ashley: Hi, that equation

Area = F*hypotenuse^2

refers to the fact that any triangle can have an equation formula like this. The amount of F will change on the shape though.

Area = 1/2 b * h

is a more useful equation because it works for every type of triangle. But the first one gets at the idea that the area of any shape is essentially based on the hypotenuse squared (or any side squared, for a different F). For squares, area is 1 * side^2, or 1/16 * perimeter^2. In both cases, it’s “some number time a measurement squared”.

Hope this helps!

thats all bakvass………

Thank you very much ! It’s the best thing I’ve found about Pythagoras…

Could u please explain me the concept of Area factor. Im really not getting it. With respect to various shapes how can i associate it with the given figure.

Could u please explain me the concept of Area factor. Im really not getting it. With respect to various shapes how can i associate it with the given figure.

got something new to know about pythagoras

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.).

Hi Kalid

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 :).

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 a*Sqrt(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=a*Sqrt(n) long. That’s because k*x^2=k*n*a^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 k*l^2. We can paint as many smaller boats as we want (all the same or different), as long as k*l^2=k*a^2+k*b^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=n*x^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

LOVE it!

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!

@Yash: You’re welcome!

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.

@Swati: Thanks, glad it helped!

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!!!

and one more point..since we can wrap any shape in a square there is always going to be a (line segment)^2 in the area..

one more observation..for a circle, if we find the best enclosing square, the total area of that square is 4*(radius^2)..but the actual area has to be lesser than this..so we have pi instead of 4.but can you tell me how this pi was reached??

darn you, Kalid! i’m in 3rd year calculus, but now all I want to do is go back to geometry again!

seriously, thanks for putting this up. this makes math fun again.

@Anonymous: Awesome, glad it helped! Yeah, it’s so easy to get distracted when math becomes enjoyable again :).

For those of you that wonders if if maths are useful. Understandig Pythagoras helped me building my home. How to make 90 degrees corners in a house? Knowing Pythagoras says that marking 3 feets from the corner of one wall and 4 feets of the adjacent wall and then using the marks adjusting the sides till you get your 5 feet diagonal and voila.. the sides forms a perfect 90 degree angle.

@Eyvind: That’s a great example of going backwards — yes, you can make a perfect right angle at home by making a triangle of sides 3-4-5!

Wow those are great examples. I found some other site that actually has circle, triangle, rectangle, and many other calculators.

http://www.thexorb.com/Algebra/Pythagorean/Pythagorean-Theorem-Calculation.aspx

I was confused by the Area Factor bit as well. So I did a little formulating.

Starting with: Area = F * hypotenuse^2

Replace Area: 1/2 Base * Height = F * hypotenuse^2

Rearrange: (Base * Height) / (2 * hypotenuse^2) = F

There is your area factor.

Making this useful, if you only knew you had two _similar_ triangles with hypotenuses 3 and 4 respectively (as in the example) simply go back to Pythagoreas and solve for the missing hypotenuse of the combined triangles, knowing of course that the respective smaller hypotenuses become the sides of the larger triangle.

As Kalid said the Area Factor doesn’t really matter, but I still wanted to find a formula for it…voila.

vision sort of locked 3d sighting “Topologie ia branche mat who you presenting ability see it plus de sides is the key of a triangles example bending Arcsin(x):

sides is statement the many mat formats

krob

Brilliant. Exactly what I was looking for.

this was so interesting……..

can’t be better than this

@tahira: Thanks!

Thanks. Your explanations on the physic were brilliant. It must surely also applies to others like the

mgh,pVg(buoyant force),hpg(pressure)and others.@Hanka: Thank you!

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.

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.

@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.

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.

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

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

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.

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!

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!

@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.

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

@sabrina: Thanks, glad you liked it!

that was the most fantastic idea i ever came across in maths.

@patrick: Thanks!

love this

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)

Hi Mary, the idea is that a yacht is essentially a deformed sphere. Spheres have surface area 4 * pi * r^2.

So we can do

50^2 = 40^2 + 30^2

and multiply both sides by 4 * pi to get

4*pi*50^2 = 4*pi*40^2 + 4*pi*30^2

which says “A ball of radius 50 has the same surface area as a ball of radius 40 and a ball of radius 30″. In our case, we’re using ship hulls, vs. perfect spheres, but the relationship is (theoretically) the same.

man this is awesume. it has helped me a lot in my project

The Pythagorean Theorem as modern art, featuring a squared circle:

http://www.aitnaru.org/images/Pi_Corral.pdf (see Pythagorean Pi design)

Geometers easily comprehend that this new concept of Pi simply complements one ratio (Pi) with another (ASR) … and both ratios include the same mysterious and stimulating essence of irrationality! Such is the nature of squared circles.

How not to square the circle?

Believe that it is impossible.

Just curious on your mention of using perimeter to get area on a square. When you use the formula Area = (1/16)P^2 , there is no way that makes sense. Simply using a 10×10 square, you can figure out this equation doesn’t work out when you wind up with a (100/16) answer instead of 100. What is this?

Hi Justin, great question. In the case of a 10×10 we have Perimeter = 40. So we can plug in P = 40 and get:

Area = (1/16) * (40)^2 = (1/16) * 1600 = 100

Hope that helps!

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Great work!! And a big thank you for confirming some of my own theories . I pretty much used the same reasoning to prove several mathematical conjectures such as Fermat’s Last Theorem. I know. Wiles has already proved it but certainly not like what Fermat would have. Upon seeing Fermat’s equation I immediately said that’s just the Pythagorean theorem. Its almost exactly like you said except the “factor” is N^n-2. But since you have a,b, and c then N^n-2 is a different factor for each variable. But you can get around that by expressing all the variables in terms of c or the hypoteneuse, i.e., a=c*sine and b =c*cos. Gives a very short sweet margin proof more in keeping with what Fermat actually came up with.

I am convinced that if mathematician would actually sit down and think about some of these problems intuitively rather than trying to impress other mathematicians with “rigor” then far more long standing mathematical problems would be solved.

It turns out that

is true even when area is measured by some other shape — when area is measured by something other than tiny squares.

http://math.stackexchange.com/questions/800699/why-square-units

i liked it

wow .thanks men you help me in our science ka look alike

Maybe I’m just being slow, but I don’t see much more than just manipulation of equations here! It’s not incredibly profound.

hi Kalid hope you have time:

so if i see c as a vector that can be split into components (a,b) i can just as well see the area defined by c as a vector that can be split into components of the areas defined by a and b.

so when we split the right triangle into two self-similar triangles, we really split the area into the components of its growth in the two orthogonal dimensions…

what do you think of this? i found this really beautiful but i just wonder how to think of it.

then what we see as the area is really the magnitude of the area vector just as the hypotenuse is just the magnitude of the vector c. and what is magnitude? in fact it is just projection in a different coordinate system..

somehow this is really whirling up quite a bit of my understanding of direction. pretty interesting..

the comment might have become a bit confusing in the end.

i guess what im really find interesting is that in the right triangle the self similar areas defined by the two sides add up to the area defined by the hypotenuse, such that one might say that the area is split into the two dimensional components, but the lenghts of the sides obviously dont add up taxicab-like to the lenght of the hypotenuse. yet in the end there is this beautiful relationship of the area to the euclidean norm/the hypotenuse.

well in the end i think that i should probably stay humble and let those patterns be as awesome as they are, instead of speculating all too much. sure its nice to think about those patterns, but i just looked at a pair and it was of such beautiful complex structure, that i just felt like i should spend my time rather on using the patterns to construct something nice. the patterns are there anyway and for the pythagorean theorem i actually have quite a nice intuition already, thanks to your article

Hi Khalid, I noticed that you applied the relationship for the well known 5×5 = 4×4 + 3×3 Pythagorean to circular areas. Did you know that 6x6x6 = 5x5x5 + 4x4x4 + 3x3x3 would similarly apply to spherical volumes?