What's the intuition behind Pythagoras' theorem?

Question

Today we learned about Pythagoras' theorem. Sadly, I can't understand the logic behind it.

$A^{2} + B^{2} = C^{2}$

$C^{2} = (5 \text{ cm})^2 + (7 \text{ cm})^2$
$C^{2} = 25 \text{ cm}^2 + 49 \text{ cm}^2$
$C^{2} = 74 \text{ cm}^2$

${x} = +\sqrt{74} \text{ cm}$

Why does the area of a square with a side of $5$ cm + the area of a square with a side of $7$ cm always equal to the missing side's length squared?

I asked my teacher but she's clueless and said Pythagoras' theorem had nothing to do with squares.

However, I know it does because this formula has to somehow make sense. Otherwise, it wouldn't exist.

Answer 1

Ponder this image and you will see why your intuition about what is going on is correct:

Answer 2

It is not quite important that the shapes of the figures you put on the edges of your right triangle are squares. They can be any figure, since the area of a figure goes with the square of its side. So you can put pentagons, stars, or a camel's shape... In the following picture the area of the large pentagon is the sum of the areas of the two smaller ones:

But you can also put a triangle similar to the original one, and you can put it inside instead of outside. So here comes the proof of Pitagora's Theorem.

Let $ABC$ be your triangle with a right angle in $C$. Let $H$ be the projection of $C$ onto $AB$. You can easily notice that $ABC$, $ACH$ and $CBH$ are similar. And they are the triangles constructed inside the edges of $ABC$. Clearly the area of $ABC$ is the sum of the areas of $ACH$ and $CBH$, so the theorem is proven.

In my opionion this is the proof which gives the essence of the theorem.

Answer 3

Bulberage's diagram is a common way to visualize it. I like this one too. Make four copies of your right triangle and arrange them like this:

The total area of the square is $C^{2} $

The area of the white square is $(A-B)^{2} = A^{2}-2AB+B^{2}$. (Note that it doesn't matter if $A-B$ is negative because we're squaring it.)

The area of the four triangles is $2AB$.

The total area of the outer square is equal to the four triangles plus the inner square:

$C^{2} = A^{2} - 2AB + B^{2} + 2AB $

and we're done.

Answer 4

This is an extended addendum to all the neat proofs listed in the other answers to the post (and prompted by a reference request from @fish.frog):

There is no free lunch: All the proofs which use the notion of area (unlike the one via similarities) depend on the following

Theorem. There exists a function $A$ (usually called "area") on the set of all planar polygonal regions which takes values in ${\mathbb R}_+$, and which satisfies the following properties:

1. The function is invariant under congruences, that is, if $P, Q$ are congruent polygonal regions then $A(P)=A(Q)$.

2. The function $A$ is additive in the sense that if we subdivide a polygonal region $P$ in finitely many non-overlapping polygonal regions $P_1,...,P_n$ then $$ A(P)=\sum_{i=1}^n A(P_i). $$

3. If $P$ is a rectangle with the base $a$ and height $b$, then $A(P)=ab$.

4. Edit: the following is not needed, it follows from 1, 2 and 3 as noted by Hagen von Eitzen: $A(kP)=k^2A(P)$ where $kP$ is the dilation if $P$ by $k$.

Furthermore, if one has a well-developed notion of a limit (which Euclid did not have) then 3 can be replaced by a single normalization condition:

3'. $A(Q)=1$, where $Q$ is a unit square.

Existence of such function is not at all obvious (to a mathematician; anybody else would say that it is completely obvious, but this site is dedicated to mathematics). The shortest proof I know (using only axioms of Euclidean geometry) can be found in the book

E. Moise, "Elementary Geometry from an Advanced Standpoint" (3rd Edition).

The proof is long (about 10 pages) and rather tedious. (Somebody has to pay for a free lunch!)

Edit.: Proofs of existence of the area function for planar polygonal regions that one might see in a typical textbook or a class go as follows:

a. Compute area of triangle from the area of rectangles (Property 3). No attempt is made to show that different way to do so result in the same value of area of triangles.

b. Triangulate a polygonal region $P$ (in the convex case). No attempt is made to do this in the nonconvex case. (That's OK if one reduces attention only to convex polygons.) Proof of existence of a triangulation in the nonconvex case requires few clever but not difficult ideas. Archimedes could have done it (and maybe did!).

c. End of the story: No attempt is made to show that the result is independent of a triangulation, i.e., that the constructed area function is well-defined.

(This is not meant as a criticism of the common treatments of the notion of area, since there is only so much time and space one can devote to this topic, but to show where the problems arise and why a rigorous treatment is needed.)

Moise does a very thorough job here. His proof of c is very modern, it depends in the set of ideas used to prove "Schoenflies theorem" for simple closed polygonal curves in the plane: that every such curve bounds a planar polygon (that's Jordan's separation theorem in this context) which is homeomorphic to a convex polygon via a piecewise-linear homeomorphism. (Moise was a topologist.) I do not believe anybody could have done a proof that area is well-defined rigorously before, say, middle of the 19th century (Riemann integration or, alternatively, appearance of topology as a field).

By the way, Moise also gives several proofs of Pythagorean formula: Two using areas and one using similarities.

There are alternative ways to prove the theorem on existence of the area function, but, from what I know, they all require introduction of Cartesian coordinates on the Euclidean plane, which, in turn, depend on the Pythagorean formula. (One such approach is via Riemann or Lebesgue integration, developing which takes much more than 10 pages; another proof uses determinants.)

Answer 5

Surely no set of answers to this question is complete without mentioning Euclid's proof?

In this figure, the right angle is at A, and AL is perpendicular to BC.

Euclid shows a little more than just the Pythagorean theorem; namely, that the two pink regions have equal areas, as do the two blue regions.

First, observe that triangle FBC can be rotated about B to become triangle ABD. This follows from side-angle-side (FB=AB, angle FBC = angle ABD, BC = BD). So these two triangles are congruent.

(Aside: That rotation is a perfect 90 degrees, so we have just shown in passing that segments CF and DA are (a) equal and (b) perpendicular. Strictly speaking, this diversion is unnecessary... But getting sidetracked is one of the joys of Euclidean geometry, in my opinion. I only wish had I learned this in school. End aside.)

Next, observe that both the pink square (FBAG) and the triangle FBC have base FB and height BA. So the pink square has twice the area of the triangle.

Next, observe that both the pink rectangle (BDLK) and the triangle ABD have base BD and height DL. So the pink rectangle also has twice the area of the triangle.

Since the triangles are equal, the pink square has the same area as the pink rectangle. A symmetric argument holds for the blue square and blue rectangle, which completes the proof.

Of course, as @studiosus points out, this does assume that the blue and pink rectangles add up to the square of BC...

Answer 6

I like Garfield's proof the best. By the way, do you know that he became the president of the US five years after giving this proof! I like because it uses simple geometry and no rearrangement stuff, that others here have resorted to. Here it goes:

What I have done here is constructed a trapezium $ADCE$, such that $AD=a+b$, $AC = a$ and $DE = b$. Now we place a point $B$ on $AD$, such that $AB= b$ and $BD = a$. So, we can see by $SAS$ criteria, that triangles $ABC$ and $BDE$ are congruent.

This gives us $BC = BE = c$. Similarly we can show that, $\angle CBE = 180^{\circ} -(\angle ABC + \angle DBE) = 90^{\circ}$. Now we come to the crux of the proof, equating the area of the trapezoid and the individual triangles.

We know the area of the trapezoid, using the area formula of the trapezoid, is $\frac{(a+b)^2}{2}$. You can visualize it by extending the sides [as shown in the diagram], and observing that it becomes a square of side $a+b$, and seeing that the two trapezoids formed are congruent. Now:

$$\begin{align} (ACDE) & = (ABC) + (BCE) + (BDE) \\ \frac{(a+b)^2}{2} & = \frac{1}{2}ab + \frac{1}{2}ab + \frac{1}{2}c^2 \\ (a+b)^2 & = ab + ab +c^2 \\ a^2+b^2+2ab & = 2ab + c^2 \\ a^2+b^2 & = c^2 \end{align}$$

Got it! And Bhaskara's proof, as suggested by Free, is worth as look as well. You might be surprised to learn that this theorem has the most number of proofs for any theorem in mathematics! You can see 100 of them here: http://www.cut-the-knot.org/pythagoras/index.shtml .

Answer 7

Similar Triangle Approach

Here is the simplest proof I have seen that doesn't use areas. It just uses similar triangles. However, what is most intuitive is a subjective decision.

By similar triangles in the diagram below, we have $$ \large\frac{c_a}{a}=\frac ac\qquad\normalsize\text{and}\large\qquad\frac{c_b}{b}=\frac bc $$

$\hspace{3.5cm}$

Thus, $$ \large c=c_a+c_b=a\frac ac+b\frac bc $$ Therefore, $$ \large c^2=a^2+b^2 $$

---

Ratio of Areas Approach

Consider a right triangle whose hypotenuse is coincident with the side of a square. Let the legs of the triangle be $a$ and $b$ and the hypotenuse, $c$. The ratio of the area of the triangle to that of the square is $$ \frac{ab}{2c^2}=\frac12\frac ac\frac bc $$ Since $\frac ac$ and $\frac bc$ are the same for all similar triangles, and the red, green, and purple triangles below are similar, the ratios of the areas of each triangle to the areas of each corresponding square are equal.

$\hspace{3.5cm}$

since the area of the purple triangle is the sum of the areas of the exterior red and green triangles, the area of the purple square is the sum of the areas of the red and green squares.

Answer 8

I would have posted this as a comment, but I can't yet; so please forgive me!

I strongly suggest following and watching the user Vi Hart on YouTube, she has many fascinating videos on math, and a couple about Geometry with specific relevance to squaring and why the Greeks invented a lot a math based on the shapes of objects; they used the areas of shapes to work out the answers.

In a very literal sense, whilst we know Pythagoras's theory as $A^2+B^2=C^2$; he wouldn't have squared a number on a page; he would have drawn a triangle and extend each of the faces of the triangle into a square and add up their respective areas.

I hope this helps!

Do check out Vi Hart though, she's fascinating and your teacher might learn something too. :)

Answer 9

You can profitably use Eli Maor, The Pythagorean theorem A 4,000-year history (2007) about the

well over four hundred proofs of it [today] known.

It is interesting to read the Ch.2 : Pythagoras [page 17-on]: it is not very clear what is the real contribution of Pythagoas itself to the question, due to the paucity of information rlated to his historical personality, but we can surely assert that the Pythagorean theorem is a milestone of ancient Greek mathematics and geometry.

One of greek "reflection" about it has the following aspect :

Of particular interest to the Pythagoreans were figurative numbers [... they are obtained as patterns of physical objects, like pebbles, that show special properties; among them square numbers, like $4=2 \times 2$ and $9 = 3 \times 3$]. Such explorations led the Pythagoreans to develop a kind of primitive algebra, based on the interrelation of various figures. For example, the familiar formula $(a + b)^2 = a^2 + 2ab + b^2$ can be proved geometrically by considering a square of side $(a + b)$, as shown in figure [see Bulberage's figure]. We can dissect this square into two smaller squares of areas $a^2$ and $b^2$ and two rectangles of areas $a \times b$ and $b \times a$. But the two rectangles are congruent, so their areas are equal. Thus the total area of the dissected parts is $a^2 + 2ab + b^2$, and this is equal to the area of the original square, $(a + b)^2$.

We may inspect the figures [see Bulberage's answer] and notes that the LH is a big square (call it $SQ$) with inside two squares (of sides $a$ and $b$ respectively) and four congruent triangles (of sides $a$ and $b$).

With the algebraic expression of the area we have that :

$SQ = a^2 + b^2 +4 \times ab/2$.

The RH figure has the same square $SQ$ withe the four triangles "rearranged": what remains is an inside square (call it $sq$).

Again we have .

$SQ = sq + 4 \times ab/2$.

With algebraic manipulations, we get :

$a^2 + b^2 = sq$

where the square $sq$ is precisely the square constructed on the hypotenuse of the triangle of sides $a$ and $b$, i.e. a formulation of the Pythagorean theorem.

Of course, ancient Greeks did not know about algebraic manipulations; but a "proof by inspection" on the two figures shows that, starting from two equal squares, and subtracting from them the same "quantities" (the four triangles), by an easy application of Euclid's Elements Common notion 3 : "If equals are subtracted from equals, then the remainders are equal", what remains must be equal; i.e.

the square constructed on the hypotenuse of the triangle of sides $a$ and $b$ is equal to the sum of the squares constructed on the sides $a$ and $b$ respectively.

In conclusion, starting from squares (numbers), we ended with squares (geometrical shapes) and we have historical support to the statement that the Pythagorean theorem :

was also about squares, and not only about triangles.

Answer 10

You may not realize it but you have two very different questions here. One is about proof, a lock-tight convincing argument removing all doubt. And then there is one about intuition, what gives you a quick good idea why.

The first, about proof, can be as dull and uninformative as following all the details of long division (e.g. prove that there is a remainder of 1 when dividing 1111 by 37). The second, about intuition, is more informative, but not necessarily lock-tight, say comparing lengths in two pictures.

For the proof of pythagorean's theorem itself, there are many examples of that around:online, Euclid's I.47... in fact at least 101 proofs. So I won't try to repeat them.

One might be able to get 'intuition' by looking at one proof over and over, following each step, understanding not just the truth of each step but also the direction it is going in (one might call this 'why each step is made'). Or you might think 'intuition' is a really easy proof (many of which have been given here).

Also, 'intuition' is just a good guess (looking at pictures and guessing that because two lengths are the same in the picture they're the same for any change in the picture), and it only becomes a reliable guess when backed by understanding well the mechanical details (that's a encouragement not to rely solely on intuition because proofs are too 'hard'). Intuition is not always geometric or visual (e.g. in algebra, your intuition may tell you that certain things will cancel out easily without having to expand a multiplication fully).

Now to actual content about the intuition of Pythagorean's theorem (where it will become more clear what that means). Really, the point of the theorem is that something is special about right triangles. One knows (intuitively) that the legs can be any lengths, but whatever they are the hypotenuse is then defined (that might seem obvious) and there is a simple formula for it.

The weird part is that the formula involves squaring and square roots, right? And it's also weird that it involves (relatedly) going from distances to areas (actual squares) which are very different things (despite the fact that we use numbers for both).

The closest to a 'why' proof is one that is actually more general; that is, that similar (in the technical sense) figures on the feet of a right triangle add to the similar figure on the hypotenuse.

And finally, after a lot of talk, the closest to an actual not-in-quotes intuition for the traditional Pythagorean theorem is to look at 3 examples. One example is the isosceles right triangle. If you construct some examples like the one here, you simple add tiles together and see that the the squares on the feet obviously add up to that on the hypotenuse. The second example is a bit degenerate, suppose one foot is extremely small, then the square on the other foot is almost identical to the square on the hypotenuse (and so as the small foot goes to zero, the sum must approach equality (if they weren't already equal).

Now you take -any- example in between (not equal feet, not one foot zero). If the sum of squares equals the square of the hypotenuse it means the as you move from isosceles to a sliver, at three points the equation holds. Then intuitively it would just be strange for it not to hold at every point in between. (it wouldn't have been strange for it to hold at the two extremes but in between to be off). And since there do exist Pythagorean triples ($3^2 + 4^2 = 5^2$) most likely the Pythagorean theorem holds for all right triangles. Note that this isn't an air-tight proof because maybe it could hold on just those Pythagorean triple but maybe nowhere else).

So that's the intuition: the idea holds for two extremes and then at least one in between, so it probably holds for everything in between.

Answer 11

I think this excellent sequence of answers to the OP's interesting question needs one more observation: the Pythagorean Theorem is logically equivalent to Euclid's Parallel Postulate.

All the proofs above depend on that postulate, either explicitly (e.g. drawing parallel lines with useful properties) or implicitly ( e.g. relying on arguments about similar figures).

The theorem is false in spherical or hyperbolic geometry.

@TomCollinge said as much in a comment above.

See http://www.cut-the-knot.org/triangle/pythpar/PTimpliesPP.shtml

Answer 12

A great question, and lots of interesting answers! Mauro ALLEGRANZA's answer that the ancient Greeks played with pebbles (we get our word "calculation" from the Greek word for a pebble) is a good one - that gave them a feeling for the way squares could add up. Like for instance the way that 12² plus 5² would equal 13²: Emanuele Paolini's thoughts about how any shape will do, not just a square, and that way of dividing a triangle he shows, and which robjon elaborates - those seem the likely routes to me. More thoughts here.

Answer 13

Many people seem to confuse proof with explanation. Proof proves; it does not explain. If you think about it you can see that regardless of the shape of a triangle, the lengths of any two sides together with the angle between them determine the length of the third side, i.e., the side opposite the angle. Draw a triangle and you'll see for yourself that this is true. Now it happens that the relationship between these four quantities is known. When you study trigonometry you'll see it. It's called the law of cosines. It is a peculiarity of that relationship that when the angle is ninety degrees (pi over 2 radians), the term in that relationship that involves the angle becomes zero, i.e., the opposite side is ONLY related to the lengths of the other two sides. That is the relationship called the Pythagorean theorem. (Pythagorus didn't know about the law of cosines. He (or whoever) came by the discovery a different way.) This is what explains it to me. Does this help? Of course this leaves the law of cosines to be explained but that's a different question. Mathematics is loaded to the gills with questions. Good for you for asking. Keep it up. We need much more explanation and much less proof. Proof is important to specialists (professional mathematicians) and most of us are not specialists.

Answer 14

The accepted answer provides no explanation for the diagrams:

Each diagram shows an outer square, $\color{red}{(a+b)}$ units on a side, divided into other figures. The diagrams suggest different expressions for the area of the outer square. Equating these expressions leads to the equation $a^2+b^2=c^2$.

Total area of $\color{red}{outer}$ square = Area not covered by triangles + Area covered by triangles

$\color{red}{(a+b)^2} = c^2 + 4(\dfrac1{2}ab)$ ________$\color{red}{(a+b)^2} = a^2 +b^2 + 4(\dfrac1{2}ab)$

$c^2 + 4(\dfrac1{2}ab)= a^2 +b^2 + 4(\dfrac1{2}ab)$

$\therefore c^2 = a^2+b^2$

source: Algebra Structure and Method Book $1$ by Brown, Dolciani et al.

Here is a visual aid to help the new learner move from one diagram to the other:

or

Answer 15

Consider a right triangle ABC with hypotenuse AB (which I'll call c) and legs CB (which I'll call a) and AC (which I'll call b). If we accept that the right angle at C means that vectors $\vec{CA}$ and $\vec{CB}$ are perpendicular and that $\vec{CA}$ and $\vec{CB}$ are perpendicular implies that $\vec{CA}\cdot\vec{CB}=0$ (so $\vec{CB}\cdot\vec{CA}=0$ by symmetry of the inner product on $\mathbb{R}^n$ with the standard metric) and that for any vector $\vec{v}$, the norm squared of $\vec{v}$ is given by $|\vec{v}|^2=\vec{v}\cdot\vec{v}$, then since vectors add tail to heads, $\vec{CA}=\vec{CB}+\vec{BA}$ implies

$\vec{CA}\cdot\vec{CA}=(\vec{CB}+\vec{BA})\cdot(\vec{CB}+\vec{BA})$ $=\vec{CB}\cdot\vec{CB}+\vec{CB}\cdot\vec{BA}+\vec{BA}\cdot\vec{CB}+\vec{BA}\cdot\vec{BA}$ by linearity of the inner product

gives $c^2 = |\vec{CA}|^2 = |\vec{CB}|^2 + 0 + 0 + |\vec{BA}|^2 = a^2 + b^2$.

I say all of this because to me, the intuition behind Pythagoras' theorem is "forces which act in perpendicular directions can be considered separately in physics," or "normal forces don't interact."