Baudhayana-Pythagoras theorem

Turn a right-angled triangle inside out and a strange fact appears: draw a square on each of its three sides, and the two smaller squares always add up, in area, to exactly the largest one. This is the Baudhayana-Pythagoras theorem — one of the oldest and most useful results in mathematics, recorded by the Indian mathematician Baudhāyana in his Śulba-Sūtra around 800 BCE, centuries before the Greek mathematician Pythagoras, to whom it is often credited alone in Western texts. Builders use it to square a corner, sailors and surveyors used it to lay out right angles without any protractor, and it still sits at the heart of geometry, trigonometry, and coordinate distance formulas taught in later grades. This section builds the theorem from the ground up: what legs and hypotenuse mean, a hands-on cut-and-rearrange proof, a fully worked hypotenuse calculation, and the converse test that tells you whether any triangle is right-angled just by checking three numbers.

The core insight

In a right-angled triangle with legs $a,b$ and hypotenuse $c$ , the areas satisfy $a^2+b^2=c^2$ , a relation known since Baudhāyana's Śulba-Sūtra roughly 800 BCE and provable by cutting and rearranging squares built on each side.

The core idea

Legs and hypotenuse. In a right-angled triangle, the side opposite the $90°$ angle is the hypotenuse — always the longest side. The two sides that form the right angle are the legs. If the legs have lengths $a$ and $b$ , and the hypotenuse has length $c$ , the theorem states:

a^2+b^2=c^2

where: $a$ and $b$ are the lengths of the two legs (the sides meeting at the right angle), and $c$ is the length of the hypotenuse (the side opposite the right angle).

The area picture. The same relation can be read as a statement about areas. If you draw a square on each of the three sides of the right triangle, the areas of the two smaller squares — built on the legs — add up exactly to the area of the largest square, built on the hypotenuse:

\text{Area on } a + \text{Area on } b = \text{Area on } c

This area reading is what makes the theorem provable by pure cutting and rearranging, with no algebra at all — see the figure below and the derivation that follows.

Baudhāyana before Pythagoras. The Śulba-Sūtras were practical manuals for constructing fire altars of precise shape and area. Baudhāyana's text states the $a^2+b^2=c^2$ relationship, and shows how to construct a square equal in area to a given rectangle using this same idea, roughly three centuries before Pythagoras is traditionally said to have lived. Babylonian, Chinese, and Egyptian sources independently show evidence of using the same relationship in surveying and construction long before it had one named discoverer. Calling the result the Baudhāyana–Pythagoras theorem credits this older, independently-arrived-at insight rather than crediting a single culture.

The converse — a right-angle test. The theorem also runs backwards: if three side lengths $p,q,r$ (with $r$ the largest) satisfy $p^2+q^2=r^2$ , then a triangle with those three sides is guaranteed right-angled, with the right angle opposite $r$ . This converse is what makes the theorem so practically useful — it lets you check for a right angle using only a measuring tape, with no protractor or angle tool at all.

Figure

A right triangle with legs a and b and hypotenuse c, with a square drawn on each side, showing that the two smaller square areas together equal the largest square area

Derivation AThe dissection proof of a squared plus b squared equals c squared

Show, using only cutting and rearranging of shapes with no algebraic manipulation of unknowns, that the two squares built on the legs of a right triangle together have exactly the same total area as the single square built on its hypotenuse, so that $a^2+b^2=c^2$ for every right-angled triangle regardless of its size or proportions.

Step 1. Start with a right-angled triangle whose legs have lengths $a$ and $b$ and whose hypotenuse has length $c$ . Construct a square of side $a$ on one leg, a square of side $b$ on the other leg, and a square of side $c$ on the hypotenuse, each square drawn outward from the triangle so none overlaps it.

Step 2. Take four identical copies of the original right triangle and arrange them inside a large square of side $(a+b)$ , placed so their hypotenuses form a tilted square of side $c$ in the middle, with the right angle of each triangle tucked into a corner of the big square. The big square's total area is $(a+b)^2$ , and it is made up of the four triangles plus the tilted inner square of area $c^2$ .

Step 3. Rearrange the same four triangles inside a second, identically sized square of side $(a+b)$ , but this time push them into two of the corners so they leave behind two separate uncovered squares — one of side $a$ and one of side $b$ — instead of one tilted square in the middle. Because both arrangements use the same four triangles inside squares of the same total size $(a+b)^2$ , the leftover uncovered area must be the same in both cases.

Step 4. In the first arrangement the leftover area was the single tilted square, $c^2$ . In the second arrangement the leftover area is the two separate squares, $a^2+b^2$ . Since cutting and rearranging pieces never changes total area, these leftover areas must be equal: $a^2+b^2=c^2$ . This holds for any right triangle, because no property of $a$ , $b$ , or $c$ beyond the right angle was ever used — only that four copies of the same triangle can be packed both ways inside a square of side $(a+b)$ .

Worked example

Problem: A right-angled triangle has legs of length $6\text{ cm}$ and $8\text{ cm}$ . Find the length of the hypotenuse, and then check whether a nearby triangle with sides $9\text{ m}$ , $12\text{ m}$ , and a measured diagonal of $15\text{ m}$ is also right-angled.

Step 1. Identify the legs of the first triangle: $a=6\text{ cm}$ , $b=8\text{ cm}$ . Apply the theorem, $a^2+b^2=c^2$ .

6^2+8^2=c^2 \implies 36+64=c^2 \implies 100=c^2

Step 2. Take the square root of both sides to isolate $c$ : $c=\sqrt{100}=10$ . So the hypotenuse is $10\text{ cm}$ — this is the well-known $6,8,10$ triple, a multiple of the primitive triple $3,4,5$ .

Step 3. Now use the converse to test the second triangle. A builder wants to check whether a rectangular plot with sides $9\text{ m}$ and $12\text{ m}$ has a genuinely right-angled corner, and measures the diagonal as $15\text{ m}$ . Treat $a=9$ , $b=12$ as the two sides and $c=15$ as the diagonal, and compute $a^2+b^2$ and $c^2$ separately: $9^2+12^2=81+144=225$ , and $15^2=225$ .

Step 4. Compare the two results: $a^2+b^2=225=c^2$ , so the condition holds exactly, confirming the corner is a genuine right angle. This is the classic " $3$ - $4$ - $5$ method," here scaled by $3$ to $9,12,15$ , that builders and carpenters use on-site to check square corners with only a measuring tape — no angle tool required.

Competitive stretch

Quiz and olympiad problems often disguise the theorem inside word problems — a ladder leaning against a wall, the diagonal of a rectangular field, or the shortest path across a rectangular park. In every case, first identify which side is the hypotenuse (the one opposite the right angle, and the longest), then apply $a^2+b^2=c^2$ exactly as above; the disguise never changes the underlying calculation.

Common traps

Applying the theorem to a non-right triangle: $a^2+b^2=c^2$ only holds when the triangle has a genuine $90°$ angle; for any other triangle this relation is false.
Mislabelling the hypotenuse: the hypotenuse must be the side opposite the right angle and is always the longest side; plugging the wrong side into $c$ can give a negative or impossible value under the square root.
Confusing $(a+b)^2$ with $a^2+b^2$ : a common slip is adding the legs first and squaring the sum, instead of squaring each leg separately and then adding — these are not equal in general.
Assuming three numbers satisfying $a^2+b^2=c^2$ can form any triangle shape: they always form one specific right-triangle shape (up to scaling); the angles are not a free choice.

Ask the tutor: Want to see the four-triangle dissection proof animated step by step, or try a few more converse-test word problems?

Does the theorem work for triangles that are not right-angled

No. The relation $a^2+b^2=c^2$ is true only when the triangle contains a genuine $90°$ angle, with $c$ being the side opposite that angle. For an acute or obtuse triangle, $a^2+b^2$ will be greater than or less than $c^2$ respectively, and a different relationship (studied in later grades using the cosine rule) is needed instead.

How do I know which side to call c in a word problem

Always look for the side opposite the right angle, or the longest side if lengths are already given — that is the hypotenuse, and it always goes on its own on one side of the equation as $c^2$ . In practical problems like ladders or diagonals, the hypotenuse is usually the slanted or diagonal measurement, while the legs are the horizontal and vertical measurements.

Is the dissection proof the only way to prove the theorem

No, it is simply the proof suited to this stage of school, since it uses only cutting, rearranging, and area — no algebra of unknown quantities. Later grades cover proofs using similar triangles and formal congruence postulates, and there are in fact hundreds of known proofs of this theorem discovered across history, including one credited to a U.S. president.

Baudhayana-Pythagoras theorem

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads