The Baudhayana-Pythagoras theorem: a² + b² = c²

Combining two different squares

Doubling a square is easy: draw a square on its diagonal. But what if two different squares, of sides a and b, have to be combined into one square with their total area? In his Shulbasutra (verse 1.12), Baudhayana gave the answer: make a right-angled triangle whose shorter sides are a and b. The square drawn on its hypotenuse has exactly the area of the two squares put together. When a and b are equal, this is the same as the doubling rule, since the hypotenuse is then the diagonal of the square.

The theorem

Written in symbols, Baudhayana’s rule is the theorem: for a right-angled triangle whose shorter sides are a and b and whose hypotenuse is c, a² + b² = c². Baudhayana was the first to state it in this general form, around 800 BCE. The Greek philosopher-mathematician Pythagoras, who lived about three centuries later, also studied it, so it is widely known as the Pythagorean theorem. Calling it the Baudhayana-Pythagoras theorem makes clear which theorem is meant.

Why it is true

Baudhayana’s verse 2.1 points to a proof. Put a square of side b and a square of side a side by side, bottoms in line. Mark a point a units from the left end of the base, and cut from it to the top-left corner of the large square and to the top-right corner of the small one. Each cut is the hypotenuse c of a right triangle with sides a and b. Now slide the two cut-off triangles up into the gaps above. The three pieces fit together into a square of side c. All four of its angles are right angles, because at each corner the two acute angles of a right triangle, x and 90° − x, add up to 90°. Since nothing was added or lost, a² + b² = c².

Finding a missing side

The theorem turns into two quick rules. To find the hypotenuse, add the squares and take the square root: c = √(a² + b²). Draw a right triangle with sides 3 cm and 4 cm and the hypotenuse measures about 5 cm; the theorem says it is exactly 5 cm, since 9 + 16 = 25. To find a shorter side, subtract: if one side is 8 cm and the hypotenuse 17 cm, the third side is √(289 − 64) = √225 = 15 cm. Answers need not be whole numbers: sides 5 and 7 give c = √74, about 8.60.

Squares with three or five times the area

Baudhayana also showed how to make a square three times as big as a given one (verse 1.10). Take a right triangle whose shorter sides are the side a of the square and its diagonal a√2. Then c² = a² + 2a² = 3a², so the square on c has triple the area. With shorter sides a and 2a, c² = a² + 4a² = 5a², five times the area.

Where it is used

The theorem gives the diagonal of a square of side 5 cm as 5√2, about 7.07 cm; the side of a rhombus with diagonals 24 and 70 as √(12² + 35²) = 37; and the depth of the lake in Bhaskaracharya’s lotus problem as 4 units. Learn about whole-number answers in Pythagorean triples and about √2 in the square root of 2. The whole chapter is in the Chapter 2 notes, with every exercise in the NCERT solutions.