The Baudhayana-Pythagoras Theorem

1. Doubling a square — the diagonal trick

Suppose you are given a square and asked to draw a new square with double the area. A first guess is to double the side — but doubling the side multiplies the area by $2\times2=4$, not $2$. So that is wrong.

Baudhāyana's elegant answer (Śulba-Sūtra, Verse 1.9):

Why it works: draw the square on the diagonal (a tilted square). The original square is made of 2 equal small triangles, while the new tilted square is made of 4 of the same small triangles. So the new square is exactly $\tfrac{4}{2}=2$ times the area. Repeating this gives a sequence of squares of areas made from $2,4,8,\dots$ small triangles — each double the last.

2. Halving a square

To make a square with half the area, just reverse the idea: draw a tilted square inside the given one, joining the midpoints of the four sides. That inner square $PQRS$ has half the area.

Note again: a square with half the side does not have half the area — it has one-quarter the area, so four such squares fill the original. Halving the area needs the midpoint (tilted) square, not the half-side square.

3. Hypotenuse of an isosceles right triangle

In any right triangle, the side opposite the right angle is the hypotenuse — the longest side. Take an isosceles right triangle with the two equal (perpendicular) sides each $=1$. What is the hypotenuse $c$?

A unit square is made of two such triangles. The square built on the hypotenuse (the diagonal) has double the area, so its area $=2\times1=2$. But the area of a square of side $c$ is $c^{2}$, so:

The hypotenuse is $\sqrt2$ units long.

4. What number is $\sqrt2$?

We can trap $\sqrt2$ between bounds by squaring decimals:

• $1^{2}=1$ and $2^{2}=4$, while $(\sqrt2)^{2}=2$, so $1<\sqrt2<2$.
• $1.4^{2}=1.96$ and $1.5^{2}=2.25$, so $1.4<\sqrt2<1.5$.
• $1.41^{2}=1.9881$ and $1.42^{2}=2.0164$, so $1.41<\sqrt2<1.42$.
• $1.414^{2}=1.999396$ and $1.415^{2}=2.002225$, so $1.414<\sqrt2<1.415$.

The squares creep towards $2$ but never hit it exactly. In fact $\sqrt2$ has a non-terminating, non-repeating decimal: $\sqrt2=1.41421356\dots$ It also cannot be written as a fraction $\dfrac{m}{n}$ — such numbers are called irrational.

5. The hypotenuse formula for isosceles triangles — worked NCERT examples

For equal sides $a$, use $c^{2}=2a^{2}$. This finds $c$ from $a$, or $a$ from $c$.

6. Combining two different squares — Baudhāyana's general rule

Two equal squares can be merged into one bigger square (the diagonal trick). Baudhāyana (Śulba-Sūtra, Verse 1.12) solved the harder case of two different squares too:

In other words: make a right-angled triangle whose two perpendicular sides equal the sides of the two squares. The square on its hypotenuse has area equal to the sum of the two original squares. Cutting and rearranging the pieces shows the new four-sided figure is a perfect square whose side is the hypotenuse — this is the heart of the theorem.

7. The Baudhāyana-Pythagoras Theorem

The result of §6, stated cleanly:

Baudhāyana was the first in history to state this in full generality. It is also named after the Greek mathematician Pythagoras (c. 500 BCE), who studied it a couple of centuries later — hence the combined name Baudhāyana-Pythagoras Theorem.

8. Finding a missing side

Rearrange the theorem depending on what is unknown:

• Hypotenuse missing: $c=\sqrt{a^{2}+b^{2}}$.
• A shorter side missing: $a=\sqrt{c^{2}-b^{2}}$.

9. Baudhāyana (Pythagorean) triples

A Baudhāyana triple is a set of three whole numbers $(a,b,c)$ with $a^{2}+b^{2}=c^{2}$ — i.e. the integer side-lengths of a right triangle. Baudhāyana himself listed several (Śulba-Sūtra, Verse 1.13):

Scaling makes more: if $(a,b,c)$ is a triple, so is $(ka,kb,kc)$ for any positive integer $k$, because $(ka)^{2}+(kb)^{2}=k^{2}(a^{2}+b^{2})=k^{2}c^{2}=(kc)^{2}$. So from $(3,4,5)$ we get $(6,8,10),(9,12,15),(12,16,20),\dots$ — proving there are infinitely many triples.

A triple with no common factor >1 is called primitive (e.g. $(3,4,5)$, $(5,12,13)$ are primitive; $(9,12,15)$ is not — it is $(3,4,5)$ scaled by $3$). Every triple is either primitive or a scaled-up primitive.

10. Applications of the theorem

The theorem turns many real situations into a quick calculation. The trick is to spot the right triangle.

11. A glimpse beyond — Fermat's Last Theorem

Since infinitely many squares are the sum of two squares ($a^{2}+b^{2}=c^{2}$), the mathematician Fermat asked: is there a cube that is the sum of two cubes, a fourth power that is the sum of two fourth powers, and so on? I.e. does $x^{n}+y^{n}=z^{n}$ have whole-number solutions for $n>2$?

Fermat claimed the answer is no for every $n>2$, scribbling that he had "a marvellous proof" too big for the margin. No one found his proof. After 300 years of failed attempts, Andrew Wiles finally proved it in 1994. This is Fermat's Last Theorem. (Not for the exam — but a great story.)

12. NCERT Figure-it-Out — find the missing side (fully solved)

$c$ is the hypotenuse. Use $c=\sqrt{a^{2}+b^{2}}$ or a shorter side $=\sqrt{c^{2}-(\text{other side})^{2}}$.

• (i) $a=5,b=7$: $c=\sqrt{25+49}=\sqrt{74}\approx8.6$.
• (ii) $a=8,b=12$: $c=\sqrt{64+144}=\sqrt{208}\approx14.4$.
• (iii) $a=9,c=15$: $b=\sqrt{225-81}=\sqrt{144}=12$.
• (iv) $a=7,b=12$: $c=\sqrt{49+144}=\sqrt{193}\approx13.9$.
• (v) $a=1.5,b=3.5$: $c=\sqrt{2.25+12.25}=\sqrt{14.5}\approx3.8$.

13. Common mistakes to avoid

• Doubling the side to double a square's area — that gives $4\times$ the area. Use the diagonal.
• Treating any side as the hypotenuse — $c$ is always the longest side, opposite the right angle.
• Writing $a+b=c$ instead of $a^{2}+b^{2}=c^{2}$. The squares add, not the lengths ($3+4\neq5$, but $9+16=25$).
• For a missing shorter side, wrongly adding: it must be $c^{2}-b^{2}$ (subtract), not $c^{2}+b^{2}$.
• Saying $\sqrt2$ "ends" somewhere — it is non-terminating and non-repeating (irrational).
• Calling $(6,8,10)$ primitive — it shares the factor $2$; only $(3,4,5)$ is its primitive form.

14. Quick revision checklist

• Right triangle: $a^{2}+b^{2}=c^{2}$ ($c$ = hypotenuse, longest side).
• Converse: if $a^{2}+b^{2}=c^{2}$, the triangle is right-angled.
• Square of side $s$: diagonal $=s\sqrt2$; isosceles right triangle of equal sides $a$: hypotenuse $=a\sqrt2$.
• $\sqrt2=1.41421356\dots$ — irrational, between $1.414$ and $1.415$.
• Triples: $(3,4,5),(5,12,13),(8,15,17),(7,24,25)$; scale by $k$ to get more; infinitely many.
• Word problems: draw the right triangle, label hypotenuse, then apply the formula.