Class 8 › Mathematics › Chapter
The Baudhāyana-Pythagoras Theorem
Doubling and halving a square, the hypotenuse of an isosceles right triangle and √2, Baudhāyana's theorem $a^2 + b^2 = c^2$, integer triples and applications.
1. Doubling and Halving a Square
Baudhāyana (Šulba-Sūtra, c. 800 BCE): the diagonal of a square produces a square of double the area of the original.
- Doubling the side gives $4\times$ the area ($2 \times 2$), not double.
- Why the diagonal square is double: draw east-west and north-south lines through the vertices. The original square is made of 2 congruent small triangles; the square on its diagonal is made of 4 of them. Repeating gives squares of $2, 4, 8$ triangles, each double the previous.
- Halving: join the midpoints of the sides of a square to get a tilted inner square; it is made of $2$ of the $4$ small triangles, so its area is half. It cannot be done by halving the side (that gives $\tfrac14$).
- Paper folding: fold the corners inward so the creases pass through the midpoints of the sides; the resulting square PQRS has half the area (proved by congruent triangles).
2. Hypotenuse of an Isosceles Right Triangle
Isosceles right triangle (equal sides $a$, hypotenuse $c$)
$$c^2 = 2a^2 \quad\Longrightarrow\quad c = a\sqrt{2}$$
- The hypotenuse is the side opposite the right angle (the longest side).
- Reason: a square of side $a$ is made of two such triangles; the square on the diagonal has area $2a^2$, and equals $c \times c$.
- Unit square: $a = 1 \Rightarrow c^2 = 2 \Rightarrow c = \sqrt2$.
- Find $c$ from $a$: $a = 12$: $c = \sqrt{2 \times 144} = \sqrt{288}$. Since $16^2 = 256$ and $17^2 = 289$, $c$ lies between $16$ and $17$.
- Find $a$ from $c$: $c = \sqrt{72}$: $72 = 2a^2 \Rightarrow a^2 = 36 \Rightarrow a = 6$.
- For hypotenuse $10$: $a^2 = 50$, so each side is $\sqrt{50}$.
3. The Number $\sqrt{2}$
Bounds
Squeezing $\sqrt2$
$1^2 = 1 \lt 2 \lt 4 = 2^2$, so $1 \lt \sqrt2 \lt 2$. Refining: $1.4^2 = 1.96$, $1.5^2 = 2.25$; $1.41^2 = 1.9881$, $1.42^2 = 2.0164$; $1.414^2 = 1.999396$, $1.415^2 = 2.002225$. So $1.414 \lt \sqrt2 \lt 1.415$.
Value
Non-terminating
$\sqrt2 = 1.41421356\ldots$ goes on forever. It has no terminating decimal form.
Why $\sqrt2$ is not a terminating decimal
If a terminating decimal had square $2$, its last non-zero digit would give a non-zero last digit in the square too (e.g. a number ending in $4$ squares to a number ending in $6$), so the square could not be exactly $2.000\ldots$
Why $\sqrt2$ is not a fraction $\tfrac mn$ (Euclid)
If $\sqrt2 = \tfrac mn$ then $2n^2 = m^2$.
In the prime factorisation of a square, each prime occurs an even number of times. On the left, $2$ occurs an odd number of times; on the right, an even number. Contradiction, so $\sqrt2$ cannot be written as $\tfrac mn$.
4. Combining Two Squares: Baudhāyana's Theorem
Baudhāyana-Pythagoras Theorem
$$a^2 + b^2 = c^2$$
$a, b$ are the shorter sides of a right triangle and $c$ is the hypotenuse.
The 3-4-5 triangle: $3^2 + 4^2 = 5^2$.
Baudhāyana's proof (Verse 2.1)
1. Join squares of sides $a$ (smaller) and $b$ (larger). Mark a rectangle in the larger square using side $a$; its diagonal makes a right triangle with legs $a$ and $b$, and hypotenuse $c$.
2. Add three more congruent copies of this triangle around the hypotenuse to make a 4-sided figure.
3. All four sides equal $c$. At each corner, the two acute angles of the triangles ($x$ and $90^\circ - x$) and the corner angle lie on a straight line, so $x + (90^\circ - x) + \text{corner} = 180^\circ$ and the corner angle is $90^\circ$. So it is a square of side $c$.
4. Area of the square on $c$ = T + U + V = V + W + X (as T, U, W, X are congruent) = area of the two given squares. Hence $c^2 = a^2 + b^2$.
- Same-size squares: $a = b$ gives $c^2 = 2a^2$, matching the doubling result.
- Paper activity: two cuts on the joined squares give three pieces that rearrange into the larger square.
- Also called the Pythagorean theorem (Pythagoras, c. 500 BCE, a couple of hundred years after Baudhāyana); the transitional name is 'Baudhāyana-Pythagoras Theorem'.
- Verse 1.9, 1.12: the diagonal's square is the sum of the squares of the two sides.
5. Using the Theorem
| Find | Formula | Example |
|---|---|---|
| Hypotenuse | $c = \sqrt{a^2 + b^2}$ | $a = 3, b = 4$: $c = \sqrt{25} = 5$. $a = 5, b = 12$: $c = 13$ |
| A shorter side | $a = \sqrt{c^2 - b^2}$ | $b = 8, c = 17$: $a = \sqrt{289 - 64} = 15$ |
| Square of area $3\times$ / $5\times$ | Right triangle with legs $\sqrt{\ }$ of the smaller areas | Triple: legs $1$ and $\sqrt2$ (units of the side) give $c^2 = 3$. Five times: legs $1$ and $2$ give $c^2 = 5$ |
| Square of area $a^2 - b^2$ | Use $a$ as hypotenuse, $b$ as a leg | Squares of side $7$ and $5$: other leg $= \sqrt{49 - 25} = \sqrt{24}$ |
- Hypotenuse is always the longest side of a right triangle ($c^2 = a^2 + b^2 \gt a^2, b^2$).
- The theorem applies only to right-angled triangles; the largest side must be $c$.
- When you cannot take an exact root, bound it: $\sqrt{288}$ lies between $16$ and $17$.
6. Baudhāyana Triples
Baudhāyana triple: positive integers $(a, b, c)$ with $a^2 + b^2 = c^2$. Baudhāyana listed $(3,4,5)$, $(5,12,13)$, $(8,15,17)$, $(7,24,25)$, $(12,35,37)$, $(15,36,39)$.
- Scaling: if $(a, b, c)$ is a triple, so is $(ka, kb, kc)$ for any positive integer $k$, since $(ka)^2 + (kb)^2 = k^2(a^2 + b^2) = (kc)^2$. So there are infinitely many triples (e.g. $(6,8,10)$, $(9,12,15)$, $(12,16,20)$).
- Primitive triple: no common factor greater than $1$. $(3,4,5)$ and $(5,12,13)$ are primitive; $(9,12,15)$ is not. Every triple is primitive or a scaled copy of a primitive one.
- Primitive triples with numbers up to $20$: $(3,4,5)$, $(5,12,13)$, $(8,15,17)$. Scaled versions are not primitive.
- If $f \gt 1$ divides all three numbers of a triple, $(\tfrac af, \tfrac bf, \tfrac cf)$ is also a triple (divide the equation by $f^2$).
Generating triples with odd numbers
The sum of the first $n$ odd numbers is $n^2$, and the $n$th odd number is $2n - 1$. So $(n-1)^2 + (2n - 1) = n^2$.
If $2n - 1$ is itself a square, we get a triple. $9$ is the $5$th odd number: $4^2 + 3^2 = 5^2$. $25$ is the $13$th: $12^2 + 5^2 = 13^2$.
More: $49$ ($n = 25$): $(7, 24, 25)$; $81$ ($n = 41$): $(9, 40, 41)$; $121$ ($n = 61$): $(11, 60, 61)$.
Limit: in every triple from this method, one leg is $1$ less than the hypotenuse. So primitive triples like $(8, 15, 17)$ are not obtained this way.
7. A Long-Standing Open Problem
- Fermat wondered if $x^n + y^n = z^n$ has solutions in natural numbers for $n \gt 2$ (a cube as a sum of two cubes, etc.).
- Fermat's Last Theorem: there are no solutions for $n \gt 2$. Fermat claimed a proof that 'the margin is too small to contain'.
- It resisted proof for more than $300$ years. Andrew Wiles, who got interested at age $10$ (1963), proved it in 1994.
8. Further Applications
Bhāskarāchārya's lotus problem (Līlāvatī)
A lotus stands $1$ unit above water. Swayed by the breeze, its tip touches the water $3$ units away. Find the depth.
Let depth $= x$, so stem length $= x + 1$. Assume the stem was perpendicular to the water. Right triangle with sides $3, x, x + 1$: $3^2 + x^2 = (x+1)^2 \Rightarrow 9 = 2x + 1 \Rightarrow x = 4$.
- Diagonal of a square of side $a$ is $a\sqrt2$ (side $5$ cm gives $5\sqrt2$ cm).
- Rhombus from diagonals: diagonals bisect each other at right angles, so side $= \sqrt{(d_1/2)^2 + (d_2/2)^2}$. Diagonals $24$ and $70$: $\sqrt{12^2 + 35^2} = \sqrt{1369} = 37$.
- Equilateral triangle, side $6$: the altitude bisects the base, so $h^2 = 6^2 - 3^2 = 27$, $h = \sqrt{27}$ and area $= \tfrac12 \times 6 \times \sqrt{27} = 9\sqrt3$.
- Rectangles with integer sides and diagonal: use triples, e.g. $3 \times 4$ (diagonal $5$), $5 \times 12$ ($13$), $6 \times 8$ ($10$), $8 \times 15$ ($17$), $7 \times 24$ ($25$).
- Squares on a grid: a grid square with vertices on dots has area $a^2 + b^2$ for some integers $a, b$ (e.g. area $2$, $4$, $5$ are possible; $3$ is not).
9. Key Results and Quick Revision
| Topic | Key fact |
|---|---|
| Baudhāyana-Pythagoras theorem | $a^2 + b^2 = c^2$ ($c$ = hypotenuse) |
| Isosceles right triangle | $c = a\sqrt2$ (from $2a^2 = c^2$) |
| Diagonal of a square | $a\sqrt2$; the diagonal square has double the area |
| $\sqrt2$ | $1.414 \lt \sqrt2 \lt 1.415$; non-terminating, not a fraction |
| Triples | $(3,4,5)$, $(5,12,13)$, $(8,15,17)$, $(7,24,25)$; scaling gives infinitely many |
| Odd-number method | $(n-1)^2 + (2n-1) = n^2$ with $2n-1$ a square |
| Fermat | $x^n + y^n = z^n$ has no natural-number solution for $n \gt 2$ (proved 1994) |
Common mistakes:
• Using $a^2 + b^2 = c^2$ in a triangle that is not right-angled.
• Putting a shorter side in place of $c$ (the hypotenuse must be the longest side).
• Writing $\sqrt{a^2 + b^2} = a + b$.
• Doubling the side to double the area (this gives $4\times$ the area).
• Treating $\sqrt{2}$ as exactly $1.414$ or $\tfrac{7}{5}$.
• Assuming all primitive triples have a leg one less than the hypotenuse.
• Using $a^2 + b^2 = c^2$ in a triangle that is not right-angled.
• Putting a shorter side in place of $c$ (the hypotenuse must be the longest side).
• Writing $\sqrt{a^2 + b^2} = a + b$.
• Doubling the side to double the area (this gives $4\times$ the area).
• Treating $\sqrt{2}$ as exactly $1.414$ or $\tfrac{7}{5}$.
• Assuming all primitive triples have a leg one less than the hypotenuse.