Ganita Prakash II · Chapter 2 — The Baudhāyana-Pythagoras Theorem

About 800 BCE, the Indian mathematician Baudhāyana wrote the Śulba-Sūtra, a set of rules for building fire altars of exact shapes and sizes. One of his questions sounds simple: how do you build a square with exactly twice the area of a given square?

The tempting guess. Double each side. But a square of side 2 units has area 2 × 2 = 4 square units, while a square of side 1 has area 1. Doubling the side makes the area four times as large, not twice. Area grows with side × side, so the side must grow by less than double.

Baudhāyana’s answer. Draw a square on the diagonal of the given square. In his words (in our own paraphrase): the square drawn on the diagonal has double the area of the original square.

Why it works. Take a square and draw one diagonal. It splits the square into 2 identical right triangles. Now build the big square on the diagonal. If you extend the sides of the small square and draw the other diagonal of the big square, you can see the big square is made of 4 of those same triangles. So:

• Small square = 2 small triangles.
• Square on the diagonal = 4 small triangles.
• So the square on the diagonal has twice the area.

All the small triangles are congruent: each is half of a copy of the original square, cut along a diagonal. Each is an isosceles right triangle, with two equal sides and angles of 90°, 45° and 45°.

A chain of doubles. Repeating the trick gives squares made of 2, 4, 8, 16, ... small triangles, each one double the last.

A paper activity. Cut out two identical paper squares. Cut the second one along both its diagonals into 4 small triangles. Lay each small triangle against one side of the first square, with the triangle’s long side along the square’s side and its corner pointing outwards. You get a bigger, tilted square. Its side is the diagonal of the original square, and its area is exactly two of the original squares.

Points to remember. Doubling the side gives 4 times the area. Doubling the area needs the side to become the diagonal, which is about 1.41 times the side, as we will see soon.

Now reverse the question: given a square, how can we make a square with half its area?

Not by halving the side. A square with half the side has one quarter of the area: 4 such squares fill the original one. So halving the side is too much.

Run Baudhāyana’s idea backwards. In the last topic, a small square was half of the square on its diagonal. So start with the big square and find the small square tilted inside it. Mark the midpoints of the four sides of the given square and join them in order. The result is a tilted square that stands on its corner inside the original.

Why is it half? Draw the two lines joining the midpoints of opposite sides (an east-west line and a north-south line). They cut the big square into 4 small squares. Each small square is cut by a side of the tilted square into 2 equal triangles: one inside the tilted square and one outside. So the inside and the outside have equal area, and the tilted square is exactly half of the big square.

With paper. Take a square sheet. Fold each corner in so that it touches the centre of the square. The four creases pass through the midpoints of the sides, and the folded sheet is a smaller square PQRS. The four flaps exactly cover PQRS, which shows that PQRS has half the area of the sheet.

Why is PQRS a square? Join the midpoints of opposite sides, which cross at the centre O. The four corner triangles cut off are all congruent right triangles with two equal legs (each half the side of the sheet), by SAS. So the four sides of PQRS are equal. Each corner triangle has 45° angles at its long side, and at each corner of PQRS two 45° angles meet, making 90°. So PQRS has equal sides and right angles: it is a square.

Linking the two topics.

• Side of the big square = s. Side of the tilted square = the diagonal of a small square of side s/2.
• Area of big square = s². Area of tilted square = s²/2.
• So halving and doubling are the same construction seen from opposite ends.

Example. A square sheet has side 10 cm, area 100 sq cm. Folding the corners to the centre gives a square of area 50 sq cm. Its side is not 5 cm (that would give 25 sq cm); it is about 7.07 cm, because 7.07 × 7.07 ≈ 50.

In a right triangle, the side opposite the right angle is the hypotenuse. It is the longest side. Let us find the hypotenuse of an isosceles right triangle whose equal sides are 1 unit long.

Use the doubling square. Two such triangles make a unit square of area 1. The hypotenuse is the diagonal of this square. The square drawn on the diagonal has double the area, so its area is 2. If the hypotenuse is c, then the square on it has area c × c = c². So

c² = 2, and c = √2 units.

We may not yet know the decimal value of √2, but we know exactly what it is: the side of a square of area 2.

The general rule. If the equal sides are a and the hypotenuse is c, two triangles make a square of side a and area a². The square on the hypotenuse has double this area, so

c² = 2a².

This works both ways: it finds c from a, and a from c.

Example 1. The equal sides are 12. Then c² = 2 × 144 = 288, and c = √288. Since 16² = 256 and 17² = 289, √288 lies between 16 and 17 (very close to 17).

Example 2. The hypotenuse is √72. Then 72 = 2a², so a² = 36 and a = 6.

Example 3. The hypotenuse is 10. Then 100 = 2a², a² = 50 and a = √50, which lies between 7 and 8 because 7² = 49 and 8² = 64.

Finding bounds. When the answer is a square root that is not a whole number, find the two squares it lies between.

For instance, 4.2² = 17.64 and 4.3² = 18.49, so √18 lies between 4.2 and 4.3.

Shortcut. Since c² = 2a², c = a × √2. The diagonal of any square is √2 times its side. A square of side 5 cm has a diagonal of 5√2 ≈ 7.07 cm.

What is √2 as a decimal? We can trap it between closer and closer numbers.

Whole numbers. 1² = 1 is less than 2 and 2² = 4 is more than 2. So 1 < √2 < 2. We call 1 a lower bound and 2 an upper bound.

Tenths. 1.4² = 1.96 and 1.5² = 2.25, so 1.4 < √2 < 1.5.

Hundredths. 1.41² = 1.9881 and 1.42² = 2.0164, so 1.41 < √2 < 1.42.

Thousandths. 1.414² = 1.999396 and 1.415² = 2.002225, so 1.414 < √2 < 1.415.

Each step gives one more correct digit. Continuing, √2 = 1.41421356...

Does it ever stop? Suppose √2 were a terminating decimal like 1.414...d, ending in a non-zero digit d. When you square it, the last digit after the decimal point comes from d × d, and the square of a non-zero digit never ends in 0 (1, 4, 9, 6, 5, 6, 9, 4, 1 are the last digits of 1², 2², ..., 9²). So the square would also end in a non-zero digit after the point, and could not be exactly 2.000... Hence the decimal for √2 goes on forever.

Is √2 a fraction? Suppose √2 = m/n, with m and n counting numbers. Squaring, 2 = m²/n², so 2n² = m².

Now think about prime factorisation. In any square number, every prime appears an even number of times (because each factor is repeated twice). So 2 appears an even number of times in m² and in n². But then on the left side, 2n² has one extra 2, so 2 appears an odd number of times. An odd count cannot equal an even count. This is impossible, so √2 cannot be written as a fraction m/n.

This proof is very old: it appears in Euclid’s Elements (about 300 BCE). You will see it in more detail in a later class.

What this means. √2 is a real length: you can draw it as the diagonal of a unit square. But no fraction and no terminating decimal is exactly equal to it. We can only write it as a never-ending decimal, 1.41421356..., or simply as √2.

Using √2 in practice. For measuring, 1.414 is usually accurate enough. A square tile of side 30 cm has a diagonal of about 30 × 1.414 = 42.42 cm.

Two equal squares combine into one square on the diagonal. But what if the two squares are of different sizes, with sides a and b? Can we build one square whose area is a² + b²?

Baudhāyana’s rule (Śulba-Sūtra 1.12). In our words: make a right triangle whose two shorter sides are a and b. The square on its hypotenuse has area equal to the sum of the areas of the squares on the two shorter sides. So if the hypotenuse is c,

a² + b² = c².

This is the Baudhāyana-Pythagoras Theorem. Baudhāyana stated it in general form around 800 BCE. The Greek thinker Pythagoras (about 500 BCE), after whom it is often named, lived a few centuries later. Using both names makes clear which theorem we mean.

Why is it true? A cut-and-rearrange proof. Baudhāyana also showed how to see it (Śulba-Sūtra 2.1).

• Place the two squares side by side on a line, the small one (side a) on the left and the big one (side b) on the right. Their total area is a² + b².
• Mark a point on the bottom edge at distance b from the left corner. Join it to the top-left corner of the small square and to the top-right corner of the big square. Each cut is the hypotenuse c of a right triangle with legs a and b.
• These two cuts slice off two right triangles. Lift each triangle and turn it so that it fills the notch at the top of the joined squares.
• The pieces now form a four-sided figure with all sides equal to c.

Why is the new figure a square? Its four sides are hypotenuses of congruent right triangles, so they are equal. Look at the point on the bottom edge where the two cuts meet. Three angles sit side by side on a straight line there: an acute angle x of one triangle, the corner of the new figure, and the other acute angle 90° − x of the second triangle. So x + corner + (90° − x) = 180°, which gives corner = 90°. The same reasoning works at every corner, so the figure is a square of side c.

Nothing was added or removed, only moved. So the area of the square on c equals a² + b².

Check with equal squares. If a = b, the rule says c² = a² + a² = 2a², which is exactly the doubling result from earlier.

The statement. If a right triangle has shorter sides a and b and hypotenuse c, then a² + b² = c².

Which side is c? c is always the side opposite the right angle. Because c² equals a² plus something positive, c is longer than a and longer than b: the hypotenuse is the longest side.

Draw a right triangle with legs 3 cm and 4 cm in your notebook and measure the hypotenuse. You should get about 5 cm. The theorem tells us it is exactly 5 cm: 3² + 4² = 9 + 16 = 25, and √25 = 5.

Two kinds of problem.

• Find the hypotenuse: c² = a² + b², then take the square root.
• Find a shorter side: a² = c² − b², then take the square root.

Example 1. Legs 5 cm and 12 cm. c² = 25 + 144 = 169, so c = 13 cm.

Example 2. One leg 8 cm, hypotenuse 17 cm. The other leg b satisfies b² = 289 − 64 = 225, so b = 15 cm.

Example 3. Legs 5 and 7. c² = 25 + 49 = 74. Since 8² = 64 and 9² = 81, c lies between 8 and 9 (about 8.6).

Example 4. Legs 1.5 and 3.5. c² = 2.25 + 12.25 = 14.5, so c = √14.5 ≈ 3.81.

Diagonal of a rectangle. The diagonal of a rectangle is the hypotenuse of a right triangle formed by two sides. A rectangle 9 cm by 12 cm has diagonal √(81 + 144) = √225 = 15 cm.

Side of a rhombus. The diagonals of a rhombus bisect each other at right angles. If the diagonals are 24 and 70, the halves are 12 and 35, and the side is √(144 + 1225) = √1369 = 37.

Height of an equilateral triangle. In an equilateral triangle of side 6, the altitude meets the base at its midpoint (the two halves are congruent by RHS). So the height h satisfies h² + 3² = 6², h² = 27, h = √27 ≈ 5.20. The area is (1/2) × 6 × √27 ≈ 15.6 square units.

Is the angle a right angle? The theorem also works the other way round: if a² + b² = c² for the three sides of a triangle, the angle opposite c is a right angle. Builders use a rope with 12 equal knots, pegged as a 3, 4, 5 triangle, to mark a square corner.

Tips. Always identify the hypotenuse first; square the sides before adding; and take the square root at the end. Never add the sides and then square.

Some right triangles have all three sides whole numbers, like 3, 4, 5. A set of three counting numbers (a, b, c) with a² + b² = c² is called a Baudhāyana triple (also a Pythagorean triple). Baudhāyana himself listed several, including (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (12, 35, 37) and (15, 36, 39).

Scaling. Is (30, 40, 50) a triple? 900 + 1600 = 2500 = 50². Yes. In general, if (a, b, c) is a triple, so is (ka, kb, kc) for any counting number k, because

(ka)² + (kb)² = k²a² + k²b² = k²(a² + b²) = k²c² = (kc)².

So there are infinitely many triples. The triples with all numbers up to 20 are (3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16, 20), (5, 12, 13) and (8, 15, 17).

Primitive triples. A triple whose three numbers have no common factor greater than 1 is called primitive. (3, 4, 5) and (5, 12, 13) are primitive; (9, 12, 15) is not, since 3 divides all three. Dividing (9, 12, 15) by 3 gives back (3, 4, 5). Every triple is either primitive or a scaled copy of a primitive one, so finding the primitive triples finds them all.

A recipe from odd numbers. Adding odd numbers gives squares: 1 = 1², 1 + 3 = 2², 1 + 3 + 5 = 3². In general the first n odd numbers add to n², and the n-th odd number is 2n − 1. So

(n − 1)² + (2n − 1) = n².

(You can also check this by expanding: n² − 2n + 1 + 2n − 1 = n².) Now, if the odd number 2n − 1 is itself a square, we get a triple.

In every triple from this recipe the hypotenuse is just 1 more than the longer leg, so the triple is always primitive. But the recipe does not give every primitive triple: (8, 15, 17) has 17 − 15 = 2, so it never appears.

Rectangles with whole-number diagonals. Each triple gives one: 3 × 4 (diagonal 5), 6 × 8 (10), 5 × 12 (13), 8 × 15 (17), 7 × 24 (25).

A puzzle from Līlāvatī. In his book Līlāvatī (12th century), Bhāskarāchārya poses this problem (told here in our own words). A lotus bud stands 1 unit above the still water of a lake. A breeze pushes the stem over, and the bud touches the water 3 units away from where it stood. How deep is the lake?

It looks as if there is not enough information, but there is. Let the depth be x. The stem is x + 1 long. When it tips over, the stem (still x + 1 long) becomes the hypotenuse of a right triangle whose legs are the depth x and the 3 units along the surface. So

x² + 3² = (x + 1)², which gives x² + 9 = x² + 2x + 1, so 9 = 2x + 1 and x = 4.

The lake is 4 units deep, and the stem is 5 units long: a 3, 4, 5 triangle in disguise! We assumed the stem first stood straight up, at right angles to the water.

Squares on a dot grid. On a square dot grid, a tilted square whose side goes p steps across and q steps up has area p² + q². So the possible areas are 1 (1² + 0²), 2 (1² + 1²), 4, 5 (1² + 2²), 8, 9, 10, 13, ... A square of area 3 cannot be drawn, because 3 is not a sum of two squares.

Triple and five times a square. To make a square of triple the area of a square of side a: first make the square on its diagonal (side a√2, area 2a²). Then use a right triangle with legs a and a√2; the square on its hypotenuse has area a² + 2a² = 3a². With legs a and 2a, you get 5a².

The difference of two squares. For a square with area 7² − 5² = 24, draw a right triangle with hypotenuse 7 and one leg 5. The other leg is √24, and the square on it has area 24.

A famous unsolved problem, finally solved. There are infinitely many squares that are sums of two squares. In the 17th century the French mathematician Pierre de Fermat claimed that no cube is a sum of two cubes, no fourth power a sum of two fourth powers, and so on: xⁿ + yⁿ = zⁿ has no solution in counting numbers when n is more than 2. He wrote that he had a marvellous proof, but the margin of his book was too small to hold it. For more than 300 years nobody could prove it. As a 10-year-old, Andrew Wiles read about the problem and decided to solve it, and in 1994 he finally proved Fermat’s Last Theorem.