Chapter 2: The Baudhayana Pythagoras Theorem

Baudhāyana's elegant answer is to use the diagonal of the given square as the side of the new square.

Suppose the original square has side $$s$$. If we simply double the side, the new square would have side $$2s$$ and area $$(2s)^{2}=4s^{2}$$, which is four times the original area, not twice. So doubling the side does not work.

Instead, draw the diagonal of the given square. If we construct a new square using this diagonal as one of its sides, then, as we shall see in Section 2.1, this new square is made up of $$4$$ small triangles, each congruent to one of the two triangles that make up the original square (which is made of $$2$$ such triangles). So the new square contains twice as many congruent triangles as the original, and therefore its area is exactly $$2s^{2}$$, double the original.

Draw horizontal and vertical (east-west and north-south) lines that extend the sides of the original square. These extensions divide the dotted square into $$4$$ small congruent right-angled triangles, while the same construction divides the original square into $$2$$ such triangles (along its diagonal).

Since all these small triangles are congruent, each has the same area, call it $$T$$. Then

Therefore

That is, the dotted square has double the area of the original square.

The dotted (tilted) square is drawn on the diagonal of the original square, so the sides of the dotted square make an angle of $$45^{\circ}$$ with the horizontal.

Focus on any one vertex of the dotted square (say the left-most vertex). The two sides meeting there are inclined at $$45^{\circ}$$ to the horizontal — one going up-and-right and one going down-and-right. The horizontal line drawn through this vertex therefore lies exactly halfway between these two sides, i.e., it bisects the $$90^{\circ}$$ angle of the dotted square at that vertex.

By the diagonal property of a square, the line that bisects a vertex angle of a square passes through the opposite vertex. Hence this horizontal line passes through the opposite vertex of the dotted square too. In the same way, the vertical line through a top or bottom vertex of the dotted square bisects that vertex-angle and reaches the opposite vertex.

The horizontal line under consideration is precisely the extension of the horizontal side of the original square (they lie along the same line because the original square is inscribed with its vertices on the sides of the dotted square). Similarly for the vertical extension. So the extensions of the horizontal and vertical sides of the original square pass through the vertices of the dotted square.

Every one of the small triangles is a right-isosceles triangle whose two perpendicular legs have the same length — namely, half the side of the original square.

Look at the figure: the horizontal and vertical lines split the dotted square into four triangles by cutting along its two diagonals. Each of these four triangles has

• a right angle (the angle between the horizontal and vertical directions),
• two equal legs (both equal to half the side of the original square, because the horizontal and vertical lines pass through the midpoints of the dotted square's diagonals which are the sides of the original square).

Two right triangles are congruent whenever their two legs are equal in length (by the SAS congruence rule applied to the right angle and the two legs). Therefore all four small triangles inside the dotted square are congruent, and each is also congruent to each of the two triangles formed by the diagonal of the original square.

Each identical square is cut along both diagonals into $$4$$ congruent right-isosceles triangles. Cutting Square 1 gives pieces $$1,2,3,4$$ and cutting Square 2 gives pieces $$5,6,7,8$$. Each little triangle has its hypotenuse equal to a side of the original square (say length $$s$$) and its two equal legs equal to $$\dfrac{s}{\sqrt{2}}$$.

Now take Square 1 (still intact) and place pieces $$5, 6, 7, 8$$ around it, one on each side, laying the hypotenuse of each triangle flush against one side of Square 1 and pointing the right-angle vertex outward.

The outer boundary of the new figure is a square:

• At every corner of Square 1, two adjacent outer triangles meet. The two $$45^{\circ}$$ angles of the triangles at that corner combine with the $$90^{\circ}$$ corner of Square 1 to give $$45^{\circ}+90^{\circ}+45^{\circ}=180^{\circ}$$ — a straight line. So each pair of adjacent triangle legs lines up into a single straight edge of length $$\dfrac{s}{\sqrt{2}} + \dfrac{s}{\sqrt{2}} = s\sqrt{2}$$.
• The four right angles of the outer triangles form the four corners of the new figure.

Thus the outer figure is a square of side $$s\sqrt{2}$$, hence area $$(s\sqrt{2})^{2}=2s^{2}$$ — that is, twice the area of Square 1.

Total area check: Area of Square 1 + Area of Square 2 $$= s^{2} + s^{2} = 2s^{2}$$, which matches.

Reverse the construction used for doubling. Given a square, join the mid-points of its four sides in order to obtain a smaller, tilted square inscribed inside the original.

If the original square has side $$s$$, then the tilted inner square has each side equal to the distance between two adjacent midpoints of the original square, which is the hypotenuse of a right triangle with legs $$\dfrac{s}{2}$$ and $$\dfrac{s}{2}$$:

So its area is

i.e., exactly half the area of the original square.

Draw the two horizontal and two vertical lines (east-west and north-south) that pass through the midpoints of the sides of the larger square. Together with the sides of the tilted inner square, they cut the larger square into $$8$$ small congruent right-isosceles triangles.

Each such small triangle has two equal legs of length $$\dfrac{s}{2}$$, where $$s$$ is the side of the larger square. So each has area $$\dfrac{1}{2}\cdot\dfrac{s}{2}\cdot\dfrac{s}{2}=\dfrac{s^{2}}{8}$$.

The larger square contains $$8$$ of these triangles, giving total area $$8\cdot\dfrac{s^{2}}{8}=s^{2}$$ (as expected). The inner tilted square contains exactly $$4$$ of these same triangles, so its area is

which is half the area of the larger square.

Take the square piece of paper and locate the midpoints of its four sides. This is easily done by folding: fold the paper in half horizontally to crease it, then unfold and fold in half vertically to crease it — the two creases meet the sides at their midpoints.

Now fold each of the four corners of the paper inward so that each corner meets the centre of the paper. Each crease line joins the midpoints of two adjacent sides.

The resulting square-shaped fold in the middle (bounded by the four crease lines joining consecutive midpoints) is the required square. As shown in the previous solution, its side is $$\dfrac{s}{\sqrt{2}}$$ (where $$s$$ is the side of the original square), so its area is $$\dfrac{s^{2}}{2}$$ — exactly half the original.

No. The area of a square depends on the square of the sidelength, not the sidelength itself.

If the original square has side $$s$$, its area is $$s^{2}$$. A square with half the sidelength has side $$\dfrac{s}{2}$$ and area

So its area is one quarter of the original, not one half.

Since each small square has area $$\dfrac{s^{2}}{4}$$, the number of such squares needed to fill the original square of area $$s^{2}$$ is

Indeed, we can fit exactly $$4$$ such small squares in a $$2\times 2$$ arrangement.

Let $$ABCD$$ be the original square paper of side $$s$$, and let $$P, Q, R, S$$ be the midpoints of $$AB, BC, CD, DA$$ respectively. We must show that $$PQRS$$ is a square whose area is $$\dfrac{s^{2}}{2}$$.

All four sides are equal. Consider the four corner triangles $$\triangle APS$$, $$\triangle BQP$$, $$\triangle CRQ$$, $$\triangle DSR$$. Each has:

• a right angle at the corner of the original square,
• two legs of length $$\dfrac{s}{2}$$ (half a side).

By SAS congruence, all four corner triangles are congruent. Their hypotenuses $$PS$$, $$PQ$$, $$QR$$, $$RS$$ are therefore equal in length; each equals $$\sqrt{\left(\tfrac{s}{2}\right)^{2}+\left(\tfrac{s}{2}\right)^{2}} = \dfrac{s}{\sqrt{2}}$$.

All four angles are right angles. At each vertex, say $$P$$, the two triangle angles adjacent to $$P$$ (from $$\triangle APS$$ and $$\triangle BQP$$) are each $$45^{\circ}$$ (base angles of a right-isosceles triangle). Since these two $$45^{\circ}$$ angles and the angle $$\angle SPQ$$ of quadrilateral $$PQRS$$ together lie along the straight line $$AB$$, they add up to $$180^{\circ}$$:

The same reasoning at $$Q, R, S$$ gives four right angles.

Since all sides are equal and all angles are $$90^{\circ}$$, $$PQRS$$ is a square.

Its area is half. By drawing the diagonals $$QS$$ and $$PR$$ (which are respectively parallel to $$AB$$ and $$AD$$), we divide the original square into $$8$$ congruent right-isosceles triangles, each with legs $$\dfrac{s}{2}$$. The square $$PQRS$$ contains $$4$$ of them and the four outer corner triangles are the other $$4$$. So

Let the hypotenuse be $$c$$. The square built on this hypotenuse has area $$c\times c = c^{2}$$.

Now, by the doubling-a-square construction, the square built on the diagonal (hypotenuse) of a unit square is exactly twice the area of the unit square. The unit square has area $$1\times 1 = 1$$, so

Therefore

By definition, $$\sqrt{2}$$ is the positive number whose square is $$2$$; i.e., $$(\sqrt{2})^{2} = 2$$.

Its decimal expansion is non-terminating and non-repeating. Continuing the successive-bounds argument gives

To one decimal place, $$\sqrt{2}\approx 1.4$$; to two decimal places, $$\sqrt{2}\approx 1.41$$; and so on. There is no exact finite decimal or fractional form for it (as the later sections show).

Compare the two positive numbers by squaring them: $$1^{2} = 1$$ and $$(\sqrt{2})^{2} = 2$$. Since $$1 < 2$$, and both numbers are positive, the number with the smaller square is smaller:

Geometrically: a square of side $$1$$ has area $$1$$ sq. unit, whereas a square of side $$\sqrt{2}$$ has area $$2$$ sq. units. The bigger square has the bigger side, so $$\sqrt{2} > 1$$.

Again compare via squares: $$(\sqrt{2})^{2}=2$$ and $$2^{2}=4$$. Since $$2 < 4$$ and both numbers are positive,

Geometrically: a square of side $$\sqrt{2}$$ has area $$2$$ sq. units, while a square of side $$2$$ has area $$4$$ sq. units. The smaller area corresponds to the smaller side.

Combining with the previous result, we have the bounds $$1 < \sqrt{2} < 2$$.

Yes — we can trap $$\sqrt{2}$$ between closer and closer decimals by squaring candidate numbers between $$1$$ and $$2$$.

One decimal place. Compute squares of $$1.1, 1.2, 1.3, 1.4, 1.5$$:

Since $$1.96 < 2 < 2.25$$, we get $$1.4 < \sqrt{2} < 1.5$$.

Two decimal places. Try squares of $$1.41, 1.42$$: $$1.41^{2}=1.9881$$ and $$1.42^{2}=2.0164$$. Since $$1.9881 < 2 < 2.0164$$, we get $$1.41 < \sqrt{2} < 1.42$$.

Three decimal places. $$1.414^{2}=1.999396$$ and $$1.415^{2}=2.002225$$, giving $$1.414 < \sqrt{2} < 1.415$$.

Repeating this idea (a decimal at a time), we can trap $$\sqrt{2}$$ within as narrow an interval as we like.

No. Suppose for contradiction some terminating decimal, say $$d = 1.414\ldots 4$$ (some finite string of digits ending in a non-zero digit), has $$d^{2}=2$$.

When we multiply $$d$$ by itself, the last non-zero digit of $$d^{2}$$ is determined by the last non-zero digit of $$d$$: if $$d$$'s last non-zero digit is $$k$$, then $$d^{2}$$'s last non-zero digit is the last digit of $$k\times k$$.

For example, if $$d$$ ends in $$4$$, then $$d^{2}$$ ends in $$6$$ (since $$4\times 4 = 16$$); if $$d$$ ends in $$1$$, $$d^{2}$$ ends in $$1$$; and so on. In every case, the last non-zero digit of $$d^{2}$$ is non-zero.

But $$2 = 2.000\ldots$$ has $$2$$ as its last non-zero digit followed by zeros — its last non-zero digit is $$2$$. Meanwhile, checking each ending digit $$1$$–$$9$$ for $$d$$, the last non-zero digit of $$d^{2}$$ is $$1,4,9,6,5,6,9,4,1$$ respectively — none of these can produce $$2$$ followed only by zeros.

So no terminating decimal squares to exactly $$2$$; the decimal expansion of $$\sqrt{2}$$ never terminates.

No — $$\sqrt{2}$$ cannot be written as any such fraction. Here is the classical proof (essentially due to Euclid).

Assume for contradiction that $$\sqrt{2} = \dfrac{m}{n}$$ for some counting numbers $$m,n$$. Then squaring both sides,

Count the number of $$2$$'s in the prime factorisation. Any perfect square has each prime appearing an even number of times in its factorisation (because a square is formed by pairing up the prime factors).

• In $$m^{2}$$, the prime $$2$$ occurs an even number of times.
• In $$n^{2}$$, the prime $$2$$ occurs an even number of times. Multiplying by $$2$$ adds one more $$2$$, so $$2n^{2}$$ has an odd number of $$2$$'s.

Equation $$(\ast)$$ says these two counts must be equal — an even count on the left equalling an odd count on the right — which is impossible.

The assumption must be wrong. Therefore $$\sqrt{2}$$ cannot be written as $$\dfrac{m}{n}$$ with $$m, n$$ counting numbers.

Yes. Each of the two identical squares (side $$s$$) is cut along one diagonal, producing $$4$$ congruent right-isosceles triangles labelled $$1, 2, 3, 4$$. Each such triangle has two equal legs of length $$s$$ (the sides of the original square) and hypotenuse of length $$s\sqrt{2}$$ (the diagonal).

Arrange the four triangles so that their hypotenuses form the boundary of a new figure and the four right-angle vertices meet at a single central point:

• Place triangle $$1$$ with its hypotenuse forming the top edge and its right angle pointing down toward the centre.
• Place triangle $$2$$ with its hypotenuse forming the right edge and its right angle pointing left toward the centre.
• Place triangle $$3$$ with its hypotenuse forming the bottom edge and its right angle pointing up.
• Place triangle $$4$$ with its hypotenuse forming the left edge and its right angle pointing right.

At the centre, four right angles meet ($$4\times 90^{\circ}=360^{\circ}$$), so the pieces fit together with no gap. Each pair of adjacent triangles shares a leg of length $$s$$.

The outer figure has four equal sides (each a hypotenuse, of length $$s\sqrt{2}$$) and four right angles (the right angles of the original triangles are now at the outer corners), so it is a square of side $$s\sqrt{2}$$. Its area is

which is exactly double the area of either original square.

Let $$a$$ be the length of each equal leg and $$c=10$$ the hypotenuse. Two such right-isosceles triangles put together (along their hypotenuses) form a square whose side equals $$a$$; and separately, we know the relation $$c^{2}=2a^{2}$$ from the doubling-a-square formula.

Substituting $$c=10$$:

So $$a = \sqrt{50}$$. We can simplify: $$\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$$.

Estimate: since $$7^{2}=49<50<64=8^{2}$$, $$a$$ is between $$7$$ and $$8$$; refining, $$7.0^{2}=49$$ and $$7.1^{2}=50.41$$, so $$7.0

Let $$a=12$$ and $$c$$ be the hypotenuse. Using $$c^{2}=2a^{2}$$:

To bound $$\sqrt{288}$$: $$16^{2}=256$$ and $$17^{2}=289$$, so $$16 < \sqrt{288} < 17$$. Indeed $$\sqrt{288}$$ is very close to $$17$$ (since $$288$$ is just $$1$$ less than $$289$$).

Let each equal leg have length $$a$$ and the hypotenuse be $$c=\sqrt{72}$$. Using $$c^{2}=2a^{2}$$:

Taking the positive square root gives

Therefore each of the two equal sides has length $$6$$ units.

Baudhāyana's beautiful answer, stated in Śulba-Sūtra Verse 1.12, is:

"The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides."

Concretely: given two squares of sides $$a$$ and $$b$$, form a right-angled triangle whose two perpendicular legs have lengths $$a$$ and $$b$$. The square built on the hypotenuse of this right triangle has area equal to the sum of the areas of the two given squares, i.e., $$a^{2}+b^{2}$$.

Practically, place the two given squares side by side so that they share an edge along the bottom (with the smaller square, side $$a$$, on the left and the larger, side $$b$$, on the right). Mark a rectangular portion of the larger square of width $$a$$ (adjacent to the smaller square) and draw its diagonal — this diagonal has length equal to the hypotenuse of a right triangle with legs $$a$$ and $$b$$. Constructing a square on this diagonal (or, equivalently, tiling three appropriate right-triangular pieces around it) produces the desired combined square.

Place the two squares side by side with the smaller (side $$a$$) on the left and the larger (side $$b$$) on the right, sharing part of a common base. Inside the larger square, mark a rectangle of width $$a$$ and height $$b$$ against the smaller square, and draw its diagonal. This diagonal is the hypotenuse of a right-triangle with perpendicular legs $$a$$ and $$b$$ (call it the base triangle).

Now cut and re-attach three more congruent copies of this base triangle so that all four hypotenuses form a new $$4$$-sided figure over the original hypotenuse:

• Triangles $$T, U, V, W$$ are all congruent copies of the base triangle (each has one leg $$a$$ and one leg $$b$$ and hypotenuse $$c$$).
• The four hypotenuses become the four sides of the new figure — so all four sides have the same length $$c$$.
• By tracking angles (see next question), each corner of the figure is a right angle, so the figure is a square.

Now count areas. The rearrangement uses exactly the pieces $$T, U, V, W$$ plus the remaining central piece: together they cover the same total area as the two original squares. So the area of the new square (on the hypotenuse) equals

That is why Baudhāyana's method works: the four congruent right-triangles fit together so their common hypotenuse becomes the side of a square whose area is exactly $$a^{2}+b^{2}$$.

Yes. When $$a=b=s$$, the two given squares are identical (side $$s$$). The right-angled triangle used in Baudhāyana's method has legs $$s$$ and $$s$$ — i.e., it is an isosceles right triangle. Its hypotenuse has length

which is exactly the diagonal of a square of side $$s$$.

The new square built on this hypotenuse has area

the sum of the two identical squares' areas — as required.

This is precisely the earlier construction for doubling a square: use the diagonal of the given square (equivalently, the hypotenuse of the right-isosceles triangle) as the side of the new square. So Baudhāyana's general method reduces to the doubling construction in the special case of equal squares.

The $$4$$-sided figure is bounded by the four hypotenuses of four congruent right-triangles $$T, U, V, W$$ (each with legs $$a$$ and $$b$$). Since all four hypotenuses have the same length $$c=\sqrt{a^{2}+b^{2}}$$, the figure has four equal sides.

The four corners of the figure are the right angles of the four triangles, so each corner is $$90^{\circ}$$. (This is proved in the next question by an angle count.)

A quadrilateral with four equal sides and four right angles is a square. So the figure is a square of side $$c$$, with area $$c^{2}=a^{2}+b^{2}$$.

Why is this area equal to the sum of the two smaller squares? The construction rearranges pieces from the two smaller squares (joined side-by-side) into the new figure without adding or removing area. So:

Focus on any one corner of the $$4$$-sided figure. Two adjacent congruent right-triangles meet there. Each right-triangle has a right angle ($$90^{\circ}$$) and two acute angles which sum to $$90^{\circ}$$; call these acute angles $$x$$ and $$90^{\circ}-x$$.

At the corner in question, the following three angles together sit along a straight line (because the outer boundary of the original two side-by-side squares was straight):

where $$\alpha$$ is the angle of the new figure at that corner. Simplifying,

So the angle at that corner is $$90^{\circ}$$. The same reasoning applies at each of the four corners, so all four angles of the figure are right angles.

Combined with the fact that all four sides are equal (each is a hypotenuse of a congruent triangle, length $$\sqrt{a^{2}+b^{2}}$$), the figure has four equal sides and four right angles — so it is a square.

Draw a horizontal segment of length $$12$$ cm; at one end, draw a perpendicular vertical segment of length $$5$$ cm. Join the free ends to complete the right-triangle. Measuring the joining segment with a ruler gives approximately $$13$$ cm.

To confirm using Baudhāyana's Theorem with legs $$a=5$$ cm and $$b=12$$ cm and hypotenuse $$c$$:

Taking positive square roots,

So the hypotenuse is exactly $$13$$ cm — matching the ruler measurement.

Let the two legs be $$a=8$$ cm and $$b$$ (unknown), and let $$c=17$$ cm be the hypotenuse. By Baudhāyana's Theorem,

Compute:

So the third side is $$15$$ cm. Drawing a right triangle with legs $$8$$ cm and $$15$$ cm gives a hypotenuse of $$17$$ cm (matching the given data), confirming the answer.

Use Baudhāyana's method for combining two different-sized squares: the square built on the hypotenuse of a right triangle with legs $$a$$ and $$b$$ has area $$a^{2}+b^{2}$$.

Triple area. Let the given square have side $$s$$ and area $$s^{2}$$. First double the given square (using the doubling construction) to produce a square of side $$s\sqrt{2}$$ and area $$2s^{2}$$. Now combine this doubled square with the original given square using Baudhāyana's method: form a right triangle with legs $$s$$ (side of original) and $$s\sqrt{2}$$ (side of doubled). The square on its hypotenuse has area

So the hypotenuse has length $$s\sqrt{3}$$, and the square on it is the required tripled square.

Five times area. First double the given square twice: doubling once gives area $$2s^{2}$$ (side $$s\sqrt{2}$$); doubling again gives area $$4s^{2}$$ (side $$2s$$). Now combine this quadrupled square with the original given square using Baudhāyana's method — legs of the right triangle are $$s$$ and $$2s$$, and

The hypotenuse has length $$s\sqrt{5}$$, and the square built on it has area $$5s^{2}$$.

(Alternatively: combine the doubled square with itself using Baudhāyana to get area $$2s^{2}+2s^{2}=4s^{2}$$, or combine the tripled square with the doubled square to get $$3s^{2}+2s^{2}=5s^{2}$$, etc. Many routes work.)

A Baudhāyana triple $$(a,b,c)$$ consists of positive integers with $$a^{2}+b^{2}=c^{2}$$ (the sidelengths of a right-triangle). We list all such triples whose largest number $$c$$ is at most $$20$$.

Start with the smallest primitive triples and their multiples:

• $$(3,4,5)$$: $$9+16=25$$. Its multiples with $$c\le 20$$: $$(3,4,5)$$, $$(6,8,10)$$, $$(9,12,15)$$, $$(12,16,20)$$.
• $$(5,12,13)$$: $$25+144=169$$. Only $$(5,12,13)$$ fits (next multiple has $$c=26$$).
• $$(8,15,17)$$: $$64+225=289$$. Only $$(8,15,17)$$ fits.

(No other primitives have hypotenuse $$\le 20$$: the next primitive is $$(7,24,25)$$, whose hypotenuse is already $$25$$.)

Therefore the complete list is:

Yes — there are infinitely many Baudhāyana triples. One simple way to see this is to notice that whenever $$(a,b,c)$$ is a triple, so is $$(ka,kb,kc)$$ for every positive integer $$k$$.

Starting from the smallest triple $$(3,4,5)$$, we get the unending sequence

Since there are infinitely many positive integers $$k$$, this list is unending — proving that there are infinitely many Baudhāyana triples.

Check $$(30,40,50)$$:

Yes — the equality holds, so $$(30,40,50)$$ is a Baudhāyana triple. (Note that $$(30,40,50)=10\times(3,4,5)$$.)

Check $$(300,400,500)$$:

Yes — again the equality holds, so $$(300,400,500)$$ is also a Baudhāyana triple. (Note that $$(300,400,500)=100\times(3,4,5)$$.)

Yes. All four listed triples arise from a single starting triple, namely $$(3, 4, 5)$$, by multiplying each entry by the same positive integer:

So each of these triples is of the form $$(3k,4k,5k)$$ for some positive integer $$k$$.

Based on the pattern that $$(3,4,5),(6,8,10),(9,12,15),(12,16,20)$$ are all of the form $$(3k,4k,5k)$$ and each is a Baudhāyana triple, we may conjecture:

Conjecture: For every positive integer $$k$$, the triple $$(3k,4k,5k)$$ is a Baudhāyana triple.

(In the next question we verify this conjecture algebraically.)

Yes. To verify, we check whether $$(3k)^{2}+(4k)^{2}=(5k)^{2}$$ for every positive integer $$k$$.

Expand each square:

Therefore

The equality holds for every positive integer $$k$$, so $$(3k,4k,5k)$$ is indeed a Baudhāyana triple for every $$k$$. This confirms the conjecture — and, since there are infinitely many positive integers $$k$$, it proves there are infinitely many Baudhāyana triples.

Yes, the statement is true. We must verify that $$(ka)^{2}+(kb)^{2}=(kc)^{2}$$.

Compute each side:

Since $$(a,b,c)$$ is a Baudhāyana triple, $$a^{2}+b^{2}=c^{2}$$. Substituting,

Hence $$(ka,kb,kc)$$ is also a Baudhāyana triple. It is called a scaled version of $$(a,b,c)$$.

A Baudhāyana triple is primitive if its three entries share no common factor greater than $$1$$.

Check $$(5,12,13)$$: First verify it is a Baudhāyana triple: $$5^{2}+12^{2}=25+144=169=13^{2}$$. Yes. Now check the gcd. $$5$$ is prime, so any common factor of $$(5,12,13)$$ must be $$1$$ or $$5$$. But $$5$$ does not divide $$12$$ (since $$12=5\times 2+2$$). So the only common factor is $$1$$ — meaning $$(5,12,13)$$ is primitive.

Other primitives with $$c\le 20$$: From the full list $$\{(3,4,5),(6,8,10),(9,12,15),(12,16,20),(5,12,13),(8,15,17)\}$$, remove those with an obvious common factor greater than $$1$$:

• $$(6,8,10)$$: common factor $$2$$ — not primitive.
• $$(9,12,15)$$: common factor $$3$$ — not primitive.
• $$(12,16,20)$$: common factor $$4$$ — not primitive.
• $$(3,4,5)$$: gcd$$=1$$ — primitive.
• $$(5,12,13)$$: gcd$$=1$$ — primitive.
• $$(8,15,17)$$: gcd$$=1$$ — primitive.

So the primitive Baudhāyana triples with all numbers $$\le 20$$ are $$(3,4,5),\ (5,12,13),\ (8,15,17)$$.

Multiply each entry of a primitive triple by $$k=2,3,4,5,6$$ in turn.

From $$(3,4,5)$$:

From $$(5,12,13)$$:

From $$(8,15,17)$$:

None of these scaled versions is primitive: every entry of $$(ka,kb,kc)$$ is divisible by $$k$$, so the common factor of the three numbers is at least $$k>1$$. So scaled versions of primitives are always non-primitive.

Yes, $$\left(\dfrac{a}{f},\dfrac{b}{f},\dfrac{c}{f}\right)$$ is again a Baudhāyana triple.

Check for $$(9,12,15)$$ with $$f=3$$:

and $$3^{2}+4^{2}=9+16=25=5^{2}$$, so $$(3,4,5)$$ is indeed a Baudhāyana triple.

Justification (general). Since $$f$$ divides $$a,b$$ and $$c$$, the numbers $$\dfrac{a}{f}, \dfrac{b}{f}, \dfrac{c}{f}$$ are positive integers. Now, starting from $$a^{2}+b^{2}=c^{2}$$, divide both sides by $$f^{2}$$:

So $$\left(\dfrac{a}{f},\dfrac{b}{f},\dfrac{c}{f}\right)$$ satisfies the Baudhāyana relation and thus is a Baudhāyana triple.

We use the well-known identity for sums of consecutive odd numbers:

Splitting off the last term $$(2n-1)$$, this can be rewritten as

If the $$n$$th odd number $$2n-1$$ happens to itself be a perfect square — say $$2n-1 = m^{2}$$ for some positive odd integer $$m$$ — then equation $$(\star)$$ becomes

which shows that $$(m,\ n-1,\ n)$$ is a Baudhāyana triple. Moreover, in this triple the two larger entries $$n-1$$ and $$n$$ are consecutive integers, hence coprime, so the whole triple is primitive.

To generate new primitive triples, pick an odd $$m>1$$, compute $$n=\dfrac{m^{2}+1}{2}$$ and $$n-1=\dfrac{m^{2}-1}{2}$$, and read off the primitive triple $$(m,\ n-1,\ n)$$.

The odd numbers in order are $$1, 3, 5, 7, 9, \ldots$$. Each is $$1$$ less than the corresponding even number $$2n$$. So the $$n$$th odd number is

Check: $$n=1$$ gives $$2(1)-1 = 1$$; $$n=2$$ gives $$2(2)-1 = 3$$; $$n=5$$ gives $$2(5)-1 = 9$$. All correct.

We know that the sum of the first $$n$$ odd numbers is a perfect square:

Replacing $$n$$ with $$n-1$$ in this identity, the sum of the first $$(n-1)$$ odd numbers is

i.e., $$1+3+5+\cdots+(2n-3) = (n-1)^{2}$$.

Check: for $$n=5$$, the first $$4$$ odd numbers are $$1,3,5,7$$, and $$1+3+5+7=16=4^{2}=(5-1)^{2}$$. Correct.

Yes. The example under discussion uses the fact that $$9$$ is an odd square. Since $$9 = 2\times 5 - 1$$, we see $$9$$ is the $$5$$th odd number, i.e., $$n=5$$.

Substituting $$n=5$$ into the identity $$(n-1)^{2}+(2n-1)=n^{2}$$:

Since $$9 = 3^{2}$$ is itself a square, the equation becomes

which is exactly the Baudhāyana triple $$(3,4,5)$$.

So the identity $$(n-1)^{2}+(2n-1)=n^{2}$$, applied with $$n=5$$ (i.e., $$2n-1 = 9 = 3^{2}$$), directly yields the triple $$(3,4,5)$$.

Recall the method: pick an odd integer $$m > 1$$, let $$2n-1 = m^{2}$$ (so $$n = \dfrac{m^{2}+1}{2}$$ and $$n-1 = \dfrac{m^{2}-1}{2}$$), and read off the triple $$(m, n-1, n)$$.

Take $$m = 5, 7, 9, 11, 13$$ in turn:

All five triples satisfy $$a^{2}+b^{2}=c^{2}$$, as verified in the last column.

No — every triple generated by this method is primitive.

The method produces triples of the form $$(m,\ n-1,\ n)$$. Observe that the two largest entries, $$n-1$$ and $$n$$, are consecutive integers.

Any common factor $$d$$ of $$n-1$$ and $$n$$ must also divide their difference, $$n - (n-1) = 1$$. So the only common factor of $$n-1$$ and $$n$$ is $$1$$, i.e., $$\gcd(n-1, n) = 1$$.

A common factor of all three numbers $$(m, n-1, n)$$ would in particular be a common factor of $$n-1$$ and $$n$$, so it must equal $$1$$. Therefore every triple $$(m, n-1, n)$$ is primitive.

Yes. The method always gives triples in which the hypotenuse $$c$$ and one of the legs differ by exactly $$1$$ (since the legs are $$m$$ and $$n-1$$ while the hypotenuse is $$n$$, so $$c - (n-1) = 1$$). But there exist primitive Baudhāyana triples where no such pair of entries differs by exactly $$1$$.

Example 1: $$(8, 15, 17)$$. Here $$17-15=2$$ and $$17-8=9$$, neither is $$1$$. Yet $$8^{2}+15^{2}=64+225=289=17^{2}$$, and $$\gcd(8,15,17)=1$$, so it is primitive. This triple cannot be produced by the given method.

Example 2: $$(20, 21, 29)$$. Check: $$400+441=841=29^{2}$$; and $$\gcd(20,21,29)=1$$. But $$29-21=8$$ and $$29-20=9$$, neither is $$1$$.

Example 3: $$(12, 35, 37)$$. $$144+1225=1369=37^{2}$$; $$\gcd=1$$; and $$37-35=2$$, $$37-12=25$$. Also unreachable by this method.

So the given method misses many primitive triples — specifically, every primitive triple in which no leg is exactly $$1$$ less than the hypotenuse.

Let the depth of the lake be $$x$$ units. Since the tip of the lotus is $$1$$ unit above the water, the length of the stem is

Assume the stem stands vertical (perpendicular to the water surface) in its original position.

When the breeze swings the lotus, the stem stays rigidly rooted at the bottom of the lake and its full length is still $$x+1$$. The tip now touches the water surface at a point $$3$$ units horizontally from the point directly above the root. This creates a right-angled triangle:

• Vertical leg: from the root at the lake bottom up to the water surface directly above it — length $$x$$.
• Horizontal leg: from that point along the water surface to where the tip now rests — length $$3$$.
• Hypotenuse: the stem itself — length $$x+1$$.

By Baudhāyana's Theorem,

Expand the right side and simplify:

So the depth of the lake is $$4$$ units.

The diagonal of a square divides it into two right-isosceles triangles whose two equal legs are the sides of the square (length $$5$$ cm each), and whose hypotenuse is the diagonal.

Let $$d$$ be the diagonal. By Baudhāyana's Theorem,

Taking the positive square root,

Numerically, $$5\sqrt{2}\approx 5\times 1.414 = 7.07$$ cm.

Key property of a rhombus: the two diagonals bisect each other at right angles. So they cut the rhombus into four congruent right-angled triangles whose legs are half the two diagonals.

Here the half-diagonals are

Each side of the rhombus is the hypotenuse of one such right triangle. Let the sidelength be $$s$$. By Baudhāyana's Theorem,

Taking the positive square root,

(Indeed, $$(12,35,37)$$ is a Baudhāyana triple.)

Yes — the hypotenuse is always the longest side of a right-angled triangle.

Let the legs be $$a$$ and $$b$$, both positive, and let $$c$$ be the hypotenuse. By Baudhāyana's Theorem,

Since $$b^{2}>0$$, we have

Both $$a$$ and $$c$$ are positive, so we can take positive square roots to get $$c > a$$. By the same reasoning, $$c^{2}>b^{2}$$ gives $$c > b$$.

Thus $$c$$ is greater than both other sides — i.e., the hypotenuse is the longest side of the right triangle.

True. Let $$(a,b,c)$$ be any Baudhāyana triple. Let $$f = \gcd(a,b,c)$$ be the greatest common divisor of the three numbers.

Case 1: $$f=1$$. Then $$(a,b,c)$$ has no common factor greater than $$1$$, so by definition it is primitive.

Case 2: $$f>1$$. Divide each entry by $$f$$ and consider $$\left(\dfrac{a}{f},\dfrac{b}{f},\dfrac{c}{f}\right)$$. From an earlier question, this smaller triple is still a Baudhāyana triple. Furthermore, since $$f$$ was the greatest common divisor, the new triple has $$\gcd = 1$$, so it is primitive. And clearly $$(a,b,c) = f\times \left(\dfrac{a}{f},\dfrac{b}{f},\dfrac{c}{f}\right)$$, i.e., $$(a,b,c)$$ is $$f$$ times this primitive triple — a scaled version.

Either way, $$(a,b,c)$$ is primitive or a scaled version of a primitive triple.

The diagonal of a rectangle with sides $$a$$ and $$b$$ has length $$\sqrt{a^{2}+b^{2}}$$ (by Baudhāyana's Theorem, since the diagonal is the hypotenuse of a right triangle whose legs are the two sides). For all of $$a, b$$ and the diagonal to be integers, $$(a, b, \text{diagonal})$$ must be a Baudhāyana triple.

Five examples, using known Baudhāyana triples:

The required square has area

so its sidelength $$x$$ must satisfy $$x^{2} = 24$$, giving $$x = \sqrt{24}=2\sqrt{6}$$ units.

Construction. Rearrange Baudhāyana's Theorem: if a right triangle has legs $$5$$ and $$x$$, and hypotenuse $$7$$, then $$5^{2}+x^{2}=7^{2}$$, i.e., $$x^{2}=49-25=24$$, which is exactly the sidelength we need.

So construct a right-triangle in which one leg has length $$5$$ units and the hypotenuse has length $$7$$ units:

1. Draw a segment $$AB$$ of length $$7$$ units — this will be the hypotenuse.
2. Draw a circle with $$AB$$ as diameter (its centre is the midpoint of $$AB$$). Any point on this circle sees $$AB$$ at a right angle.
3. Mark a point $$C$$ on this circle such that $$AC = 5$$ units — this makes $$\angle ACB = 90^{\circ}$$, so $$\triangle ACB$$ is right-angled at $$C$$.
4. Then $$BC = x = \sqrt{24}=2\sqrt{6}$$ units.

Construct a square with side $$BC$$. Its area is $$x^{2}=24$$ sq. units, which is exactly $$7^{2}-5^{2}$$.

Let $$\triangle ABC$$ be an equilateral triangle with side $$6$$. Drop the altitude $$AD$$ from $$A$$ perpendicular to $$BC$$.

Step 1: the altitude bisects the base. Consider $$\triangle ABD$$ and $$\triangle ACD$$:

• $$AB = AC = 6$$ (sides of the equilateral triangle),
• $$AD = AD$$ (common),
• $$\angle ADB = \angle ADC = 90^{\circ}$$ ($$AD$$ is perpendicular to $$BC$$).

By the RHS congruence criterion (right-angle, hypotenuse, side), $$\triangle ABD \cong \triangle ACD$$. Hence $$BD = DC$$, i.e., $$D$$ is the midpoint of $$BC$$, and $$BD = DC = \dfrac{6}{2} = 3$$.

Step 2: find the height using Baudhāyana's Theorem. In right triangle $$\triangle ABD$$ (right-angled at $$D$$):

Substitute $$AB=6$$, $$BD=3$$:

So $$AD = \sqrt{27} = 3\sqrt{3}$$ units.

Step 3: compute the area.

Numerically, $$9\sqrt{3}\approx 9\times 1.732 \approx 15.59$$ sq. units.

Strategy: open the box labelled RED. (Any of the three labels would work by symmetric reasoning, but let us pick RED for definiteness.)

Since no label is correct, the box labelled RED cannot actually contain red balls. So the box labelled RED contains either blue or green balls. Two cases:

Case 1: the box labelled RED contains blue balls.

• The box labelled RED = blue box.
• The remaining two boxes (labelled BLUE and GREEN) contain the red and green balls in some order.
• The box labelled BLUE cannot contain blue balls (its label is wrong), so it contains either red or green. But it also cannot be labelled correctly, so — since blue is already taken — the box labelled BLUE contains green.
• The last box, labelled GREEN, must therefore contain red.

Case 2: the box labelled RED contains green balls.

• The box labelled RED = green box.
• The box labelled GREEN cannot contain green (mislabelled), and green is already taken. So the box labelled GREEN contains blue.
• The last box, labelled BLUE, must therefore contain red.

In both cases, opening a single box (the one labelled RED) is enough to correctly identify the contents of all three boxes.