Baudhāyana

Baudhāyana was an ancient Indian mathematician who discovered Pythagoras theorem. This might come as a surprise to many, but it’s true that Pythagoras theorem was known […]

pythagorean-theoremBaudhāyana was an ancient Indian mathematician who discovered Pythagoras theorem. This might come as a surprise to many, but it’s true that Pythagoras theorem was known much before Pythagoras and it was It was Baudhāyana who actually discovered it at least 1000 years before Pythagoras was born!

Baudhāyana listed Pythagoras theorem in his book called Baudhāyana Śulbasûtra (800 BCE). Incidentally, Baudhāyana Śulbasûtra is also one of the oldest books on advanced Mathematics. The actual shloka (verse) in Baudhāyana Śulbasûtra that describes Pythagoras theorem is given below :

“dīrghasyākaayā rajjuH pārśvamānī, tiryaDaM mānī, cha yatpthagbhUte kurutastadubhayākaroti.”

Interestingly, Baudhāyana used a rope as an example in the above shloka which can be translated as – A rope stretched along the length of the diagonal produces an area which the vertical and horizontal sides make together. As you see, it becomes clear that this is perhaps the most intuitive way of understanding and visualizing Pythagoras theorem (and geometry in general) and Baudhāyana seems to have simplified the process of learning by encapsulating the mathematical result in a simple shloka in a layman’s language.

Another problem tackled by Baudhāyana is that of finding a circle whose area is the same as that of a square (the reverse of squaring the circle). His sūtra i.58 gives this construction:

Draw half its diagonal about the centre towards the East-West line; then describe a circle together with a third part of that which lies outside the square.

Explanation:

  • Draw the half-diagonal of the square, which is larger than the half-side by x = {a \over 2}\sqrt{2}- {a \over 2}.
  • Then draw a circle with radius {a \over 2} + {x \over 3}, or {a \over 2} + {a \over 6}(\sqrt{2}-1), which equals {a \over 6}(2 + \sqrt{2}).
  • Now (2+\sqrt{2})^2 \approx 11.66 \approx {36.6\over \pi}, so the area {\pi}r^2 \approx \pi \times {a^2 \over 6^2} \times {36.6\over \pi} \approx a^2.

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