# Given a 3 4 5 triangle, how do you know that it is a right triangle?

Without knowing the Pythagorean theorem, and in presenting reasons why the theorem might be true (without giving a full proof), is there any way to give examples of triangles that are intuitively understandable to be right triangles?

For example, one can easily show (proof by elementary picture with elementary geometry) that two area 1 squares and one area 2 square (side $$\sqrt{2}$$ but no need to mention that) form an isosceles right triangle, by dissecting the unit squares into four smaller congruent isosceles right triangles and the area 2 squares into eight of the same triangles.

It is 'obvious that the abc triangle is right, and that the two smaller squares sum to the larger (with units the triangle tiles).

But I don't know of a similar tiling for a 345 triangle.

The square on the hypotenuse is skew and impossible to visually validate. There's nothing to sum. It is simple calculation to note that $$3^2 + 4^2 = 5^2$$, but its picture doesn't 'say' that the angle between the 3 and 4 sided square must be 90 degrees.

So how do you know that a 3/4/5 triangle is right without proving the full force of the Pythagorean theorem?

• @Addem That is exactly the thing that motivated this question. I can accept the conservation of volume as given but I still can't accept that it is obvious that it is a right triangle. Yes, that image demonstrates that the sum of the two small squares equals the larger and that is why everybody oohs and ahhs over the image, but it doesn't say anything about the angle. Commented Oct 27, 2015 at 19:36
• Are you able to use the Law of Cosines? Commented Oct 27, 2015 at 20:18
• The law of cosines has three problems here: first it is about as complex to prove as Pythagoras's theorem, second is more difficult to apply (even less intuitively true than Pythagoras, and three is logically equivalent (both can be proved using Pythagoras and can be used to prove Pythagoras. My question is about the same as 'can you prove that a 3/4/5 triangle is right without going through the trouble of proving Pythagoras?' Commented Oct 27, 2015 at 21:15
• This question is essentially about proving the converse of Pythagoras' theorem without using the theorem. Wikipedia says it can be done but doesn't say how; I suspect it won't be a simple proof. I recommend grabbing three sticks of the appropriate lengths and showing that they make a right triangle. Commented Oct 27, 2015 at 23:41
• Supposedly the Babylonians knew that the 3/4/5 triangle is right. Did they know the pythagorean theorem? If not, how did they know the first fact? Commented Oct 28, 2015 at 1:58

The Chinese came up with the following a long time ago. Probably something better, but this is the gist of it.

1. Let's start with a right triangle with height b=4 and base a=3. We know it has some hypotenuse, c, but we don't know it's length because that's what we're trying to prove.

1. Now we're going to make 3 copies, and rotate them each 90°.

1. Next we're going to shove them all together so their sides touch. Because we defined them as right triangles, and rotated them exactly 90°, we know all the sides that are touching are parallel.

1. Finally, we construct a square around the entire construct, ensuring the sides are parallel/perpendicular to the a/b sides of the right triangles.

1. We can see that the sides of the large square must be a+b = 7 in length. Necessarily, the area of the square (green+orange+yellow) is 49.

2. One of the orange triangles forms a rectangle with a green triangle, with sides 3 by 4, so the area of either an orange triangle or a green triangle is (3*4)/2 = 6.

3. The four green triangles have a total area of 6*4 = 24. The area of the middle square (orange+yellow) must be 49-24 = 25. Alternately, you could just add the obviously 1*1 = 1 area of the central square (yellow) to the area of the orange triangles which is also 6*4 = 24, for a total of 25.

4. Because of symmetry, the sides of the middle square must be length c. We know the area of the square is c^2 and we also know it's 25, so c^2 = 25. Because 5^2 = 25, c=5. We've also managed to show at least one case of Pythagorean's theorem is true in the process.

• Isn't this just a proof of the Pythagorean theorem, which the question specifically asks not to use? Commented Oct 28, 2015 at 11:25
• @javier No, this doesn't use PT, but it does look almost as complex as PT. With little extra effort it is one of the classic proofs of PT. It may be that what I'm asking for ends up being equivalent to proving PT. I'm hoping for something simpler (like a one-off diagram of a dissection). Commented Oct 28, 2015 at 13:31
• This is the only thing posted so far that doesn't rely on an "eyeballing" argument. The argumentation is similar to a proof of PT. But that's because it has to be. Commented Oct 28, 2015 at 15:01
• After step 3, instead of drawing a big square around the whole picture (step 4), you could use the fact that you've already got a $c$-by-$c$ square. All four sides are $c$, and all the angles are right because you rotated the original triangle in $90^\circ$ increments. Then you can finish the proof by saying that the area of this square is, on the one hand, $c^2$ and, on the other hand, $(4\times6)+1=25$ (from the 4 triangles of area 6 each, plus the 1-by-1 square in the middle). So $c=5$. Commented Oct 28, 2015 at 20:06
• Related: Zhou Bi Suan Jing. Commented Oct 28, 2015 at 20:49

The fact that there is a 3-4-5 triangle that is a right triangle is unique to the Euclidean plane. There is no such triangle in the spherical or hyperbolic planes. Since the Pythagorean theorem is equivalent to the parallel postulate, any proof that a 3-4-5 triangle is a right triangle will somehow depend on the Pythagorean theorem/parallel postulate.

• This is a good observation, because all of the "proofs with pictures" are unknowingly assuming the pythagorean theorem.
– Ryan
Commented Oct 29, 2015 at 21:12
• This post says, "Szmielew proved (assuming continuity) that every statement which is false in hyperbolic geometry and correct in Euclidean geometry is equivalent to the fifth parallel postulate." That means we can take the statement that every 3-4-5 triangle is right as a replacement for the parallel postulate. In this system the statement about the 3-4-5 triangle has a (far) shorter proof than does the Pythagorean theorem. Also, in general, a proof of equivalence of two statements may be shorter/more intuitive in one direction than the other. Commented Jan 18, 2022 at 14:40

Here is a simple construction. Adjust to taste.

1. Draw a line $l$ passing through a point O.

2. Construct circles of radius 3 and 4 with centre O. Call them $C_3$ and $C_4$. Let the intersection of $C_4$ and $l$ be A. (Note that OA is 4 units.)

3. Construct an interval of length 5 from A to $C_3$. Call it B. (Note that OB is 3 units, and AB is 5 units.)

4. Construct an interval of length 5 from B to $C_4$. Call it C. Do not assume C lies on $l$. (Note that OC is 4 units, and BC is 5 units.)

OAB and OBC are both 3-4-5 triangles. By symmetry, $\angle\text{AOB}=\angle\text{BOC}$. So if C lies on $l$, then $\angle\text{AOB}$ must be a right angle.

If you draw everything accurately, you will of course find that C lies on $l$.

If 3-4-5 were not a right triangle, you would get a result were C did not lie on $l$, like this (a 4-4-5 triangle):

5 (optional). For further assurance, go all the way round the circle and see if you get back to A.

The intuitive argument here is that anything but a right triangle would either go "too far" or "not far enough" if you put 2 or 4 of them together. Of course, there will be a lot of "not quites" due to inaccuracies (those confounded cheap school compasses always seem to open up slightly during use!), which is a good opportunity to discuss the limitations of proof-by-picture. The missing square puzzle is a fantastic example.

• C is constructed by taking a pair of compasses opened to 5 units with centre at B, and drawing an arc intersecting with $C_4$. This act ensures BC has length 5 (no Pythagoras needed). OC = 4 because C lies on $C_4$ and O is the centre of $C_4$ (no Pythagoras needed). A great exercise is to try it with different side lengths (e.g. 4-4-5) and observe how C does not end up on $l$, because the angle is not $90^\circ$. Commented Oct 28, 2015 at 7:17
• @DavidEbert The crucial and questionable part is "If you draw everything accurately, you will of course find that C lies on l." This is what is dubious about the entire enterprise here (to someone who does not know PT). Commented Oct 28, 2015 at 13:28
• @MitchHarris So your question assumes that we do not, and cannot, have access to reliable distance measuring tools? Please update your question so we understand what tools we have access to for resolving your question, and what tools are off limits. Commented Oct 28, 2015 at 15:23
• @AdamDavis I am looking for a convincing argument that isn't bound by good drawing (Artelius's Fibonacci triangles are the classic counterexample). I added a picture of the tiling which I believe is convincing (for that single instance). Commented Oct 28, 2015 at 18:01
• It seems to me that this is just a polished version of David Ebert's answer — it can demonstrate that the angle is close to 90°, but doesn't prove that it is exactly 90°. Commented Oct 29, 2015 at 3:57

Graph paper (or square floor tiling) to the rescue!

Proof by picture for a 3 4 5 triangle:

Because the drawing is on the grid and not the skew tiling of the square on the hypotenuse, determining the area is not inscrutable. The blue square (it is a square by adding angles of the triangle) is area 25 by adding the four blue triangles (obviously 6 each) and the single unit square in the middle.

And of course this can be immediately generalized to any right triangle.

Although I asked for the determination of the largest angle of the 3 4 5 triangle (and this visual proof shows the other direction that the hypotenuse is a square on 5), I think the visual intuition is enough to go both directions, that showing a 3 4 rt triangle has hypotenuse 5 is enough (intuitively) to show the 3 4 angle of a 3 4 5 triangle is right.

• ...and this gives a nice picture for the algebraic identity $a^2 + b^2 = (b - a)^2 + 4 \frac{a b}/2$ Commented Oct 29, 2015 at 2:52
• There is much more like this in Henri Picciotto's algebra and geometry books, which are available free at mathed.page. Commented Dec 29, 2021 at 1:53

A complete different approach reasoning with the area $A$ of the triangle 3-4-5.

• Use Heron's Formula to show that $A =6$.
• Conclude that the height to the side with length 4 must be 3 since $A = ah/2$.
• Thus, the length of the side (3) equals the length of the height, which finishes the proof, since the height is perpendicular by definition.

The crucial step of this proof is the use if Heron's formula, which can be shown without using Pythagorian Theorem, see here.

• +1, although the proof of Heron's formula without assuming the Pythagorean Theorem is quite long.
– JRN
Commented Oct 29, 2015 at 6:16

Diagram shows that there exists a 345 triangle that is right-angled.

It is clear by inspection that an angle greater than 90 between 3 and 4 leads to a hypotenuse longer than 5. Similarly an angle less than 90 leads to a hypotenuse of less than 5

• ..and this picture also leads to a general argument for PT. Also, you've shown graphically that $(a+b)^2 - 2 a b = a^2 + b^2$ (or rather used the algebraic identity to substantiate the graphic). Commented Oct 29, 2015 at 13:16

I don't think I can do better than Giles answer, but here is an answer which gets the converse of PT without proving PT first:

Let $AB=3$, $BC=4$ and $AC=5$. Draw a point $D$ on the line segment $AC$ with $AD=9/5$ and $DC=16/5$. Then triangles $ABC$ and $ADB$ are similar, since they have the same angle at $A$ and proportional sides. So $\angle ABC = \angle ADB$. Similarly, $\triangle ABC \sim \triangle BDC$ so $\angle ABC = \angle BDC$.

We have $\angle ADB+\angle BDC = 180^{\circ}$ (a straight line), so $\angle ABC = 90^{\circ}$.

Of course, this can be adapted to any triangle with $a^2+b^2=c^2$. (Mark a point dividing the hypotenuse into $a^2/c$ and $b^2/c$.)

According to the Wikipedia entry for Pythagorean theorem, a proof of the converse of the Pythagorean theorem without assuming the Pythagorean theorem can be found in Stephen Casey, "The converse of the theorem of Pythagoras," The Mathematical Gazette, Vol. 92, No. 524 (July 2008), pp. 309-313.

– JRN
Commented Oct 28, 2015 at 1:44
• Assuming PT, it's elementary (Euclid I 48 is a quick corollary of I 47). Commented Oct 28, 2015 at 2:22

Not an answer; rather an observation.

If you can convince that the blue circle of radius $3$, and the black circle of radius $4$, meet the red circle of radius $\frac{5}{2}$ at the same point, then the right angle follows from Thales' Theorem.

But I have to admit, I don't see how to intersect the circles without using the Pythagorean theorem in some hidden form. If the blue circle is centered at $(0,0)$ and the black at $(5,0)$, they intersect at $(\frac{9}{5},\frac{12}{5})$, which is of course distance $\frac{5}{2}$ from the center of the red circle.

• There may very well be no finite dissections of the three squares into congruent pieces Commented Oct 28, 2015 at 2:28
• By this argument, you could also "prove" that a 4, 7, 8 triangle is a right triangle. Commented Oct 28, 2015 at 15:36
• @mweiss: I don't understand. The corresponding three circles, of radius 4 and 7, and diameter 8, do not all pass through the same point. Or do you mean because $4^2 + 7^2=65$ is close to $8^2$, that one could be tricked by a near miss, if relying, say, on a physical compass? Commented Oct 28, 2015 at 15:46
• The latter. Especially if you are using a physical compass and pencil; there won't be enough precision to tell the difference between a near-miss and a direct hit. Commented Oct 28, 2015 at 15:48
• @JosephO'Rourke: Hey using this wonderful method I found a new right-angled triangle with sides $(7,11,13)$! Commented Oct 30, 2015 at 0:25

If you have Pick's formula at your disposal, you can draw your favourite right triangle on grid paper and count. Actually, you can do the counting for any triangle with grid point vertices, but of course (by what we know) we get equality iff the triangle is a right one.

• Nice. It's a little iffy counting the dots on the edges but I think that would work. How dependent is Picks thm on PT? (it is inter provable with V - E + F = 2) Commented Oct 29, 2015 at 15:12
• The proof of Pick's formula from wikipedia is a little long, but each step is very simple. It certainly doesn't use PT. Actually it hardly uses anything other than a definition of Area as being AxB for a rectangle, and a diagonal bisection of a rectangle creating two right-angled triangles. link[/link] Commented Oct 30, 2015 at 8:39

Cut out four identical 345 triangles and put the biggest angle from each triangle together at a point. Then observe that all 4 identical angles fit to make a revolution, which implies each of the angles is a right angle.

A picture is provided below. More work would be required to prove this construction, but hopefully this is the level of proof you were looking for.

Also, it is interesting to note that ancient Egyptians apparently used 345 triangles for laying foundations, possibly even in religious ceremonies. I'm not sure if they realized a circular rope with 12 equally-spaced knots and be stretched into a 345 triangle made a right angle, but I imagine that they had some idea of the above line of reasoning.

• I think this would be convincing to those who are convinced by the water demonstration. It is dubious (too trusting of execution) but very intuitively correct. Commented Oct 28, 2015 at 18:32
• I’m not sure whether this is the same thing that Mitch is saying, but it seems to me that this is a decent way of demonstrating that the angle is > 89.5° and < 90.5°; not so good a proving that it is equal to 90°. Commented Oct 29, 2015 at 3:49

If you want to convince someone that certain triples are the side lengths of a right triangle, you can exhibit pictures like the following, which show the $$(4,3,5)$$, $$(12,5,13)$$, and $$(8,15,17)$$ right triangles.

The idea here is to let the given triple be $$(a,b,c)$$. Then one of the diagonals of the tilted rectangle is vertical and has length $$2c$$, while half of the other diagonal, which is on a slant, coincides with the hypotenuse of the right triangle with legs $$a$$ and $$b$$. Since the diagonals of a rectangle are equal, this demonstrates that the hypotenuse of the right triangle has length $$c$$.

Some details: the corners of the tilted rectangle are at $$(0,c)$$, $$(a,b)$$, $$(0,-c)$$, and $$(-a,-b)$$. You know this is a rectangle because the slope of the line joining the first and second points is $$-\frac{c-b}{a}$$, while the slope of the line joining the second and third points is $$\frac{c+b}{a}$$. These are negative reciprocals of each other because the equality $$-\left(-\frac{c-b}{a}\right)^{-1}=\frac{a}{c-b}=\frac{c+b}{a}$$ is equivalent to the Pythagorean theorem. Although we are pretending we don't know the Pythagorean theorem, the given triple must satisfy it nonetheless. At this point, the negative reciprocal property is just be a property that happens to be satisfied by the given numbers. The corners of the right triangle are at $$(0,0)$$, $$(0,b)$$, and $$(a,b)$$.

This is actually only a few steps shy of a proof of the Pythagorean theorem, as long as one is willing to countenance use of properties of similar triangles and a bit of algebra.

Given $$\triangle ACB$$ with opposite sides $$a$$, $$c$$, and $$b$$, extend the line through $$\overline{AC}$$ and draw the circle with center $$A$$ and radius $$\overline{AB}$$, intersecting the line at $$D$$ and $$E$$. The rectangle is now easily constructed. Since $$\triangle DCB$$ and $$\triangle BCE$$ are similar, we have the result, $$\frac{c-b}{a}=\frac{a}{c+b}.$$

If one wishes to stick with purely geometric reasoning, construct three squares. Construct $$ACJK$$ on $$\overline{AC}$$ (area is $$b^2$$). Construct $$ADLM$$ on $$\overline{AD}$$ (area is $$c^2$$). Then, drawing the perpendicular to $$\overline{CB}$$ through $$B$$ and the perpendicular to $$\overline{DE}$$ through $$E$$, the two perpendiculars meeting at $$F$$, subdivide rectangle $$BCEF$$ by a line $$\overline{GH}$$ into a square $$EFGH$$ (area is $$a^2$$) and a rectangle $$BCHG$$. The Pythagorean theorem is the statement that the gnomon (L-shaped region) $$CDLMKJ$$ and the square $$EFGH$$ have equal area.

This can be proved by cut-and-paste geometry with the aid of the line perpendicular to $$\overline{GH}$$ through the point $$I$$ where $$\overline{GH}$$ and $$\overline{BE}$$ intersect. Let this line intersect $$\overline{CB}$$ at $$N$$ and $$\overline{EF}$$ at $$P$$. Then rectangle $$CNPE$$ equals the gnomon in area. Since this rectangle's area equals the area of $$\triangle BCE$$ plus the area of $$\triangle EPI$$ minus the area of $$\triangle BNI$$, while the area of square $$EFGH$$ equals the area of $$\triangle EFB$$ plus the area of $$\triangle IHE$$ minus the area of $$\triangle IGB$$, and corresponding triangles in these calculations are congruent in pairs, the result is established.

Building off of the circle construction, but physical like you request:

The only possible triangle that has sides of 3, 4, and 5 must be "right". There are blocks out there (you could replicate blocks with strips of paper if unattainable) and place those down with the corners touching. (https://en.wikipedia.org/wiki/Cuisenaire_rods).

Take a 3 a 4 and a 5, and arrange it into a triangle. We can't "see" that it's right. But, if you continue around the circle, and create a triangle on it's opposite side, then on the opposite of the new triangle, etc... all the way around, you should end up with 4 triangles, completing a "360 degree" turn, and splitting it into 4 pieces... which by definition is 360/4, or a 90 degree triangle. Proof by visuo-physical means.

(this is nothing more than the circle construction proposed by Artelius, or the Triangle solution proposed by Ebert. The only difference is, instead of starting with a triangle, or compass and straightedge, you have the blocks which serve as lengths in a non-triangle format, which I believe is what you want.)

At this level students benefit from exploring physical models. There is an inexpensive product called geoleg that allows the students to build PHYSICAL triangles with specific side and angle measurements. You specify what the students build and they can see if there is more than one possible triangle (eg SSA) or only one possible triangle. Specify a triangle that is right and has two legs that are 3 and 4, and the students will see that the hypotenuse must be 5. Specify a triangle that is 3, 4, 5 and the students will see that it must be a right triangle (angle measures are built in).

• I feel the use of automated calculations is too magical, too untrustworthy. Commented Oct 28, 2015 at 2:24
• @MitchHarris The geoleg is a physical device that allows you to build triangles with specified sides and angles and see what triangles you can produce. It is not automated in any way. Commented Oct 28, 2015 at 12:08
• As I wrote on another answer: By this argument, you could "prove" that a 4, 7, 8 triangle is a right angle. (Unless the geoleg is precise to within 1°.) Commented Oct 28, 2015 at 15:38
• Oh, I agree entirely that it's worthwhile to do this to gain understanding of what a proposition asserts -- but not to claim that it helps you "know" whether or not it is true. Physical experience can only support the claim that something seems to be true. And there are plenty of triangles that seem to have a 90° angle in them, but don't really. I think the OP was asking for a way to know that a 3, 4, 5 triangle has a 90° angle in it. Commented Oct 28, 2015 at 18:43
• This is an interesting pedagogical suggestion, but it doesn't answer the question.
– user507
Commented Oct 29, 2015 at 14:48

To prove the converse of the Pythagorean theorem, you could use the cosine rule. Let's consider the 3-4-5 triangle. We have $\cos C =\frac{3^2+4^2-5^2}{2\times 3\times 4}=0$. Since $C$ is an interior angle of a planar Euclidean triangle, $0\leq C\leq \pi$. Thus $C=\frac{\pi}{2}$ as required.

• At that point, why invoke the law of cosines when PT is mostly bidirectional anyway? Commented Oct 31, 2015 at 14:24
1. Perpendicular bisector of the 3 sides taken 2 at a time circumscribe circle about the triangle.
2. Note that longest side = 2r = d
3. Big angle is inscribed and =1/2 180
• -1 It seems unclear what is being demonstrated here. Perhaps you could edit add more detail? Commented Mar 15, 2016 at 11:56

Can you use congruence of triangles?

Since any triangle with sides of length 3, 4, 5 have equal length side pairs to the right angled triangle with sides of length 3, 4, 5, this triangle must be congruent to the right angled triangle.

• But how do we know that any 3-4-5 triangle is right-angled? Commented Oct 28, 2015 at 7:24
• I take congruence to be barely mentionable here (as far as intuition goes). That is, proving it for a single 3,4,5 intuitively proves it for all. Doing it for that first one is what I am after. Formally of course it needs to be proven. Commented Oct 28, 2015 at 18:36

I would think you'd approach it the reverse direction: Draw a side straight up of length 4, draw another straight to the right of length 3, and then calculate the distance between the ends and show that it's 5--assuming you have the distance formula or permit yourself to measure the line lengths physically.

• The distance formula? Isn't that just Pythag in disguise? Commented Oct 27, 2015 at 19:36
• @JoeTaxpayer It is, although most students don't realize it--so I figured if they were just swallowing the distance formula as a thing to memorize, then maybe you could get away with it. But alternatively, like I said, you could measure the distances physically. If you can't do those two things, it starts to sound like the only way out is a rigorous proof. Commented Oct 27, 2015 at 19:39
• If you have graph paper, it's not difficult to show that the square of the hypotenuse is 25, Commented Oct 27, 2015 at 23:26
• @DagOskarMadsen Duh! And that picture also leads directly to the general (visual) proof. Commented Oct 28, 2015 at 19:07