What is the Baudhayana-Pythagoras Theorem?

For a right-angled triangle with legs a and b and hypotenuse c, a squared plus b squared equals c squared. Baudhayana stated this in full generality in his Sulba-Sutra centuries before the Greek mathematician Pythagoras.

What is Fermat's Last Theorem?

It states that the equation with exponent n has no solution in positive integers when n is greater than 2, unlike n equals 2 where infinitely many Pythagorean triples exist. It was proven in 1994 by Andrew Wiles after more than 300 years of failed attempts.

Class 8 Maths NCERT Solutions Part 2 Chapter 2: The Baudhāyana-Pythagoras Theorem (Ganita Prakash, Part 2)

Key Concepts & Formulae at a Glance

Doubling a square: the square constructed on the diagonal of a given square has exactly double its area (Baudhāyana, Śulba-Sūtra 1.9) — simply doubling the side length instead gives 4× the area, not 2×.
Halving a square: reversing this idea, the square formed by joining the midpoints of a square's sides has exactly half its area.
Isosceles right triangle: if the two equal (leg) sides have length \(a\), the hypotenuse \(c\) satisfies \(c^2=2a^2\), i.e. \(c=a\sqrt2\).
\(\sqrt2\) lies between 1.414 and 1.415. It cannot be written as a terminating decimal, nor as a fraction \(\frac{m}{n}\) with \(m,n\) positive integers (it is irrational).
Baudhāyana's Theorem (the Pythagoras theorem): for a right triangle with legs \(a,b\) and hypotenuse \(c\): \(a^2+b^2=c^2\). Baudhāyana stated this in full generality centuries before Pythagoras.
Baudhāyana (Pythagorean) triple: a triple of positive integers \((a,b,c)\) with \(a^2+b^2=c^2\) — e.g. (3,4,5), (5,12,13), (8,15,17). Scaling any triple by a positive integer \(k\) gives another triple \((ka,kb,kc)\). A triple with no common factor greater than 1 is primitive.
Generating triples from odd squares: if the \(n\)th odd number \((2n-1)\) is itself a perfect square, then \((n-1,\ \sqrt{2n-1},\ n)\) is a Baudhāyana triple, since \((n-1)^2+(2n-1)=n^2\) always.
Fermat's Last Theorem: \(a^n+b^n=c^n\) has no positive integer solutions for \(n\gt2\) — proven by Andrew Wiles in 1994, over 300 years after Fermat's claim.

\[a^2+b^2=c^2 \qquad \text{(isosceles case)}\ c=a\sqrt2 \qquad \text{(scaling)}\ (a,b,c)\to(ka,kb,kc)\]

2.1 Doubling a Square

In Baudhāyana's Śulba-Sūtra (c. 800 BCE), Baudhāyana asks: how can one construct a square having double the area of a given square?

MTMath Talk — A first guess might be to double the side length. Will this new square have double the area?

No — doubling the side length gives a new square of side \(2s\), with area \((2s)^2=4s^2\), which is 4 times the original area, not double. The dashed grid lines show the new square is made of 4 copies of the original square.

Doubling the side length quadruples the area (4×), it does not double it.

So how can one make a square with exactly double the area? Baudhāyana's Śulba-Sūtra (Verse 1.9) gives an elegant answer:

"The diagonal of a square produces a square of double the area of the original square."

The key is to construct a new square on the diagonal of the original square. One elegant way to see this: place the original square as one quarter of a larger square (side \(2s\)), and draw the tilted square connecting the midpoints of the larger square's four sides.

The original square (solid, teal fill) sits in one quarter of the outer dashed square. The tilted dotted square joins the midpoints of the outer square's sides — its side is exactly the diagonal of the original square.

MTMath Talk — Why does the new dotted square have double the area of the original square? Draw some horizontal and vertical ("east-west" and "north-south") lines to see why.

Adding the horizontal and vertical midlines splits the big square (side \(2s\), area \(4s^2\)) into 4 equal small squares (each side \(s\), area \(s^2\)) — the bottom-left one is our original square. Each of the 4 small squares is cut in half by a diagonal of the tilted dotted square, contributing exactly one triangle (half its area, \(\frac12 s^2\)) to the dotted square. So the dotted square's total area \(=4\times\frac12s^2=2s^2\) — exactly double the original square's area \(s^2\).

The dotted square is made of 4 half-small-squares (one triangular half from each of the 4 quadrants), totalling 2s² — exactly double the original s².

MTMath Talk — Why should the extension of the vertical and horizontal sides of the original square pass through the vertices of the dotted square? [Hint: the line that bisects an angle of a square passes through the opposite vertex.]

The centre point (where the horizontal and vertical midlines cross) is the centre of the big square. The dotted square's diagonals are exactly these same horizontal and vertical midlines (by construction, since the dotted square's vertices are the midpoints of the big square's sides, directly above/below/left/right of the centre).

A square's diagonal always bisects the angle at each vertex it passes through and lands exactly on the opposite vertex. Since the horizontal and vertical lines here are the dotted square's diagonals, they must pass through its vertices — which is exactly where the original square's sides (extended) meet the dotted square's boundary.

The horizontal/vertical midlines are precisely the diagonals of the dotted square, and a square's diagonal always passes through its opposite vertex — so the extended sides of the original square necessarily hit the dotted square's corners.

So the new square has double the area of the original because the original square is made up of two small triangles, while the new (dotted) square is made up of four small triangles.

MTMath Talk — Moreover, all these small triangles are congruent to each other. Can you explain why?

Each of the 4 small (side-\(s\)) squares is cut by the same diagonal-type line of the dotted square, so every triangle is a right triangle with legs \(s,\ s\) (half of a small square, cut corner-to-corner) — identical dimensions in every quadrant. Two triangles with the same two leg lengths and the same right angle between them are always congruent (SAS), so all 8 triangles across the whole figure are congruent to one another.

Every triangle is a right triangle with both legs equal to s (half of a small s×s square cut along its diagonal) — identical shape and size everywhere, hence all congruent by SAS.

Repeating this construction gives a sequence of ever-doubling squares:

Each square has double the area of the previous one, made up of 2, 4, and 8 small congruent triangles respectively.

Doubling a Square Using Paper

MTMath Talk — Cut two identical squares of paper. Cut Square 1 along both diagonals into 4 triangles (1,2,3,4); cut Square 2 into an X-shape leaving 4 corner triangles (5,6,7,8). Place pieces 5,6,7,8 around Square 1 to get a square of double the area. Why does this work?

Each corner triangle (5, 6, 7, or 8) of Square 2 is congruent to each quarter-triangle (1, 2, 3, or 4) of Square 1 — both are right triangles with legs equal to half the square's diagonal-halves. Placing pieces 5–8 around the outside edges of Square 1 exactly recreates the "4 quadrants, cut diagonally" pattern seen above: the result is the same tilted square of double area, just assembled from physical paper pieces instead of drawn lines.

This is a hands-on version of the same proof: 4 triangles from Square 1 plus 4 congruent triangles from Square 2 combine into one square of double the area, since 4+4=8 triangles fit together exactly as in the "8 small triangles" diagram above.

2.2 Halving a Square

Now suppose we're given a square and want to construct a square with half its area. One way is to reverse the doubling construction: draw a tilted smaller square inside the larger one, connecting the midpoints of the larger square's sides.

A tilted square (dotted, teal) joining the midpoints of the larger square's sides — visibly half the area of the outer square.

MTMath Talk — Why is the smaller inside square exactly half the area of the larger square?

The horizontal and vertical midlines cut the big square (side \(2s\)) into 4 small squares (each side \(s\)). Each small square is split in half by a side of the tilted inner square, and exactly 2 of these 4 half-small-squares (i.e. triangles) lie inside the tilted square, while the other 2 lie outside it (in the four outer corners). So the tilted square's area \(=2\times\frac12s^2=s^2\), and the big square's total area is \(4\times\frac12s^2 \times 2\)... more directly: the tilted square is made of 4 triangles (one per quadrant) each of area \(\frac12s^2\), totalling \(2s^2\) — while the outer square has area \((2s)^2=4s^2\). So the tilted square is exactly \(\frac{2s^2}{4s^2}=\frac12\) of the outer square's area.

The tilted inner square is built from 4 half-small-square triangles (area 2s² total), while the outer square has area 4s² — exactly half.

Halving a Square Using Paper

MTMath Talk — Cut a square from paper. Will a square with half the sidelength have half the area? Why not? Fold the square inward so the creases pass through the midpoints of the sides, forming square PQRS. Why is PQRS a square, and why is its area half the original?

No, a square with half the sidelength does NOT have half the area — it has \(\left(\frac12\right)^2=\frac14\) the area, since halving the side scales the area by the square of the factor. It would take exactly 4 such quarter-sidelength squares to fill the original.

Why PQRS is a square, with half the area: connect \(QS\) and \(PR\) (the two fold-diagonals). Each of the 4 corner flaps is a right triangle (right angle at the original square's corner) with both legs equal to half the original side — so all 4 flap-triangles are congruent by SAS. Since they're congruent, \(PQ=QR=RS=SP\) (all equal, being the hypotenuses of congruent triangles), so PQRS has 4 equal sides. The angles at P, Q, R, S are each formed by two of these congruent right triangles meeting at a midpoint of the original square's side, and by symmetry (the two angles on either side of each midpoint must add to 180° along the straight original edge, and are equal to each other by the congruent triangles) each works out to exactly 90°. A four-sided figure with all sides equal and all angles 90° is a square.

Since the 4 outer flap-triangles (folded inward) exactly cover the 4 "cut off corner" regions and PQRS is what remains uncovered by folding... more directly, PQRS is built from 4 of the same congruent right-triangle halves used in the doubling construction, so by the identical reasoning as before, its area is exactly half the original square.

A half-sidelength square has only 1/4 the area (needs 4 copies to fill the original), not 1/2. PQRS is a square (by triangle congruence — all 4 sides equal, all angles 90°) with exactly half the area of the original paper.

2.3 Hypotenuse of an Isosceles Right Triangle

Recall that in a right triangle, the side opposite the right angle is called the hypotenuse. Let's find the hypotenuse of an isosceles right triangle with both legs of length 1 unit.

MTMath Talk — Find the hypotenuse of an isosceles right triangle with both legs 1 unit, using the square PEAR (side 1) and the square REST built on its diagonal.

Square PEAR has side 1, so its area is 1 sq. unit — and it's made of two isosceles right triangles like the one we want. By the doubling construction, the square REST built on PEAR's diagonal has double the area of PEAR:

\(\text{Area of REST}=2\times\text{Area of PEAR}=2\times1=2\) sq. units.

If \(c\) is the hypotenuse (the diagonal ER), then \(\text{Area of REST}=c\times c=c^2\). So \(c^2=2\), which means \(c=\sqrt2\).

The hypotenuse of an isosceles right triangle with legs 1 unit is √2 units.

In the rest of this chapter, we assume all lengths are in a fixed unit unless stated otherwise.

Decimal Representation of √2

MTMath Talk — Is √2 less than or greater than 1? Than 2? Can we find closer bounds?

A square of side 1 has area 1; a square of side \(\sqrt2\) has area 2. Since \(1^2=1\) and \((\sqrt2)^2=2\), and \(1\lt2\), we get \(1\lt\sqrt2\).

A square of side 2 has area 4; a square of side \(\sqrt2\) has area 2. Since \(2^2=4\gt2=(\sqrt2)^2\), we get \(\sqrt2\lt2\). So \(1\lt\sqrt2\lt2\) — we call 1 a lower bound and 2 an upper bound.

Refining the bounds by trial squaring:

1.1²=1.21	1.41²=1.9881	1.411²=1.990921
1.2²=1.44	1.42²=2.0164	1.412²=1.993744
1.3²=1.69	1.41 < √2 < 1.42	1.413²=1.996569
1.4²=1.96		1.414²=1.999396
1.5²=2.25		1.415²=2.002225
1.4 < √2 < 1.5		1.414 < √2 < 1.415

1 < √2 < 2, refined to 1.4 < √2 < 1.5, then 1.41 < √2 < 1.42, then 1.414 < √2 < 1.415 — each round of squaring narrows the bounds further.

MTMath Talk — Will we ever get a number with a terminating decimal representation whose square is exactly 2?

If a terminating decimal starting 1.414... had a square of exactly 2, it would need a non-zero last digit. But a decimal with a non-zero last digit, when squared, also produces a non-zero last digit in its result (e.g. if the number ends "...4", the square's last digit comes from \(4\times4=16\), ending in 6, not 0). So the square of any such terminating decimal can never come out to be exactly \(2.000\ldots\) (which has an all-zero tail after some point).

So a terminating decimal cannot have 2 as its square. Thus the decimal expansion of \(\sqrt2\) must go on forever — it is non-terminating.

No — any terminating decimal's square also terminates with a non-zero digit (if the original had one), so it can never equal exactly 2. √2's decimal expansion never terminates.

TTTry This — Can √2 be expressed as a fraction \(\frac{m}{n}\), where \(m,n\) are counting numbers?

Suppose \(\sqrt2=\frac{m}{n}\). Then \(2=\frac{m^2}{n^2}\), so \(2n^2=m^2\).

In the prime factorisation of any perfect square, every prime occurs an even number of times. In the equation \(2n^2=m^2\): on the right side (\(m^2\)), the prime 2 occurs an even number of times. But on the left side (\(2n^2\)), the prime 2 occurs an odd number of times (the extra "2" out front, plus however many even times it appears in \(n^2\)). Since the same number can't have the prime 2 appearing both an odd and an even number of times, this is a contradiction.

So \(\sqrt2\) cannot be expressed as a fraction \(\frac{m}{n}\) — this classic proof (essentially Euclid's, Elements, c. 300 BCE) shows \(\sqrt2\) is irrational. We can only express it as a non-terminating decimal: \(\sqrt2=1.41421356\ldots\)

No — assuming √2=m/n leads to the prime 2 appearing an odd number of times on one side of 2n²=m² and an even number of times on the other, which is impossible. √2 is irrational.

Figure it Out — Set 1

Three questions from pages 39–40 of the textbook.

1Math Talk — Two identical square papers are each cut along one diagonal into 2 triangles (labelled 1,2 and 3,4). Can you arrange these 4 pieces to create a square with double the area of either original square?

Yes. Each of the 4 triangular pieces (two from each square) is a right triangle with both legs equal to the original square's side \(s\) — all four are congruent. Arranging all 4 around a common centre point, with each triangle's hypotenuse forming one side of a new figure, produces exactly the tilted "diamond" square seen in the doubling construction: a square whose side is the diagonal of the original (\(s\sqrt2\)) and whose area is \(2s^2\) — double either original square's area.

Yes — arranging the 4 congruent right-triangle pieces around a centre point (hypotenuses facing outward) recreates the same tilted square of double area used throughout this section.

2Find the hypotenuse of an isosceles right triangle given the length of its equal legs, with bounds having at least one digit after the decimal point: (i) 3 (ii) 4 (iii) 6 (iv) 8 (v) 9

Using \(c^2=2a^2\) for legs of length \(a\):

(i) \(a=3\): \(c^2=18\). Since \(4.2^2=17.64\) and \(4.3^2=18.49\), we get \(c=\sqrt{18}\), with \(4.2\lt c\lt4.3\).

(ii) \(a=4\): \(c^2=32\). Since \(5.6^2=31.36\) and \(5.7^2=32.49\), we get \(c=\sqrt{32}\), with \(5.6\lt c\lt5.7\).

(iii) \(a=6\): \(c^2=72\). Since \(8.4^2=70.56\) and \(8.5^2=72.25\), we get \(c=\sqrt{72}\), with \(8.4\lt c\lt8.5\).

(iv) \(a=8\): \(c^2=128\). Since \(11.3^2=127.69\) and \(11.4^2=129.96\), we get \(c=\sqrt{128}\), with \(11.3\lt c\lt11.4\).

(v) \(a=9\): \(c^2=162\). Since \(12.7^2=161.29\) and \(12.8^2=163.84\), we get \(c=\sqrt{162}\), with \(12.7\lt c\lt12.8\).

(i) √18, between 4.2 and 4.3 (ii) √32, between 5.6 and 5.7 (iii) √72, between 8.4 and 8.5 (iv) √128, between 11.3 and 11.4 (v) √162, between 12.7 and 12.8

3The hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths? [Hint: find the area of the square made of two such right triangles.]

The square built on the hypotenuse (side 10) has area \(10^2=100\) sq. units. This square is made of 2 copies of our isosceles right triangle, so each leg-square (built on a leg of length \(a\)) has area \(\frac{100}{2}=50\) sq. units — since \(c^2=2a^2 \Rightarrow a^2=\frac{c^2}{2}=\frac{100}{2}=50\).

So \(a=\sqrt{50}=5\sqrt2\approx7.07\).

Each of the other two sides has length √50 = 5√2 ≈ 7.07 units.

General Solution

The relation between a square's area and the area of the square on its diagonal gives a general formula: for an isosceles right triangle with legs \(a\) and hypotenuse \(c\), \(c^2=2a^2\) — usable to find \(c\) from \(a\), or \(a\) from \(c\).

Example 1. Find the hypotenuse when the equal legs have length 12. \(c=\sqrt{2\times12^2}=\sqrt{288}\). Since \(16^2=256\) and \(17^2=289\), \(\sqrt{288}\) lies between 16 and 17.

Example 2. If the hypotenuse is \(\sqrt{72}\), find the other two sides. \(c^2=2a^2 \Rightarrow 72=2a^2 \Rightarrow a^2=36 \Rightarrow a=\sqrt{36}=6\). So each of the other two sides has length 6.

2.4 Combining Two Different Squares

We've seen how to combine two copies of the same square into a larger square (sidelength = the diagonal of either smaller square). What if the two squares are different sizes?

In his Śulba-Sūtra (Verse 1.12), Baudhāyana gives a remarkable answer:

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

That is: make a right-angled triangle whose two perpendicular (leg) sides equal the sidelengths of the two squares. The square built on the hypotenuse of this triangle has an area equal to the sum of the two original squares' areas.

Right triangle with legs a and b (hypotenuse c); the tilted square on the hypotenuse has area a²+b² — the sum of the two original squares.

MTMath Talk — Why does Baudhāyana's method work? Does it agree with the earlier method for combining two same-sized squares?

When the two squares are the same size (\(a=b\)), the right triangle has both legs equal to \(a\) — exactly the isosceles right triangle from Section 2.3! Its hypotenuse is \(a\sqrt2\), the diagonal of either square, and the square built on it has area \(2a^2=a^2+a^2\) — matching perfectly with our earlier "double the area via the diagonal" construction. So Baudhāyana's general method for two different squares is a direct generalisation of the same-square doubling method.

Yes — setting a=b recovers exactly the earlier doubling construction, confirming Baudhāyana's general method is a true generalisation of it.

Subsequently (Verse 2.1), Baudhāyana gives another verse explaining why the method works in general:

"To combine different squares, mark a rectangular portion of the larger square using a side of the smaller square. The diagonal of this rectangle is the side of a square that has area equal to the sum of the areas of the smaller squares."

MTMath Talk — Follow Baudhāyana's instructions: mark off a rectangle in a bigger square using the smaller square's side, draw its diagonal to get a right triangle (legs a, b), then repeat around all four sides. Why is the resulting 4-sided figure a square, and why does its area equal a²+b²?

Why it's a square: the outer square has side \(a+b\), and each side is marked at the same distance \(a\) from a consistent corner going around (rotationally symmetric). This creates 4 congruent right triangles (each with legs \(a,b\)) sitting in the 4 corners, by SAS — same two legs, same right angle. Their hypotenuses (all length \(c=\sqrt{a^2+b^2}\), since all 4 triangles are congruent) form the sides of the inner 4-sided figure, so all its sides are equal.

Why the angles are all 90°: at each vertex of the inner figure, the angle is what's "left over" from the straight 180° angle along the big square's side, after removing the two acute angles of the adjacent triangles. If one triangle's acute angle (at that corner) is \(x\)°, the corresponding angle in the next triangle (rotated into position) is \(90-x\)° (the two acute angles of any right triangle add to 90°). So the inner figure's angle there is \(180-x-(90-x)=90\)°, regardless of \(x\). Since this holds at every vertex, all four angles of the inner figure are 90°.

A 4-sided figure with all sides equal and all angles 90° is a square — with sidelength \(c\), the hypotenuse of the right triangle with legs \(a,b\).

Why its area is \(a^2+b^2\): the outer square (side \(a+b\)) has area \((a+b)^2\). It's made of the inner square (area \(c^2\)) plus 4 corner triangles (each area \(\frac12ab\)):

\[(a+b)^2 = c^2 + 4\times\frac12ab = c^2+2ab\]

Expanding the left side: \((a+b)^2=a^2+2ab+b^2\). Setting the two expressions equal:

\[a^2+2ab+b^2=c^2+2ab \;\;\Rightarrow\;\; a^2+b^2=c^2\]

The inner 4-sided figure is a square (equal sides from congruent triangles, 90° angles from the "x + (90−x) + 90" argument), and comparing (a+b)² two ways gives exactly a²+b²=c² — Baudhāyana's assertion.

Baudhāyana's Theorem

Baudhāyana's Theorem on Right-angled Triangles: If a right-angled triangle has sidelengths \(a\), \(b\), and \(c\), where \(c\) is the hypotenuse, then \(a^2+b^2=c^2\).

Baudhāyana was the first person in history to state this theorem in this generality and essentially modern form — centuries before the Greek philosopher-mathematician Pythagoras (c. 500 BCE), who also studied and admired it. The theorem is therefore also called the Pythagorean Theorem, and often the transitional name Baudhāyana-Pythagoras Theorem.

Using the theorem: a right triangle with shorter sides 3 cm and 4 cm has hypotenuse:

\[a^2+b^2=c^2 \;\Rightarrow\; 3^2+4^2=c^2 \;\Rightarrow\; 9+16=c^2 \;\Rightarrow\; 25=c^2 \;\Rightarrow\; c=5\text{ cm}\]

MTMath Talk — Combining Two Squares Using Paper: join two different-sized squares side by side, make two cuts to form 3 pieces, and rearrange them into one larger square. Explain why this works.

Joining the two squares (sides \(a,b\)) edge to edge along their common height creates a step-shaped figure. Marking off a rectangle of width \(a\) inside the bigger square and drawing its diagonal creates exactly the right triangle (legs \(a,b\), hypotenuse \(c\)) from Baudhāyana's method. Cutting along this diagonal produces 3 pieces: the small square untouched, and the big square split into two triangular pieces by the diagonal.

Rearranging these 3 pieces — sliding the two triangular pieces from the big square around the small square — exactly recreates the "4 congruent right triangles + inner tilted square" pattern from the proof above, now assembled as one single square of side \(c\) with area \(a^2+b^2\). This is a hands-on, physical confirmation of Baudhāyana's Theorem.

The 3 cut pieces (the small square plus two triangles from the diagonal cut) rearrange exactly into a single square of side c = √(a²+b²) — a physical demonstration of a²+b²=c².

Figure it Out — Set 2

Four questions from pages 47–48 of the textbook.

1If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm, what is the length of its hypotenuse? (Draw the triangle and measure, then check using the theorem.)

\(5^2+12^2=c^2 \Rightarrow 25+144=c^2 \Rightarrow 169=c^2 \Rightarrow c=13\) cm.

Hypotenuse = 13 cm — measuring your drawn triangle should confirm this.

2If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side?

\(8^2+b^2=17^2 \Rightarrow 64+b^2=289 \Rightarrow b^2=225 \Rightarrow b=15\) cm.

The third side is 15 cm.

3How would you construct a square whose area is triple the area of a given square? Five times the area? (Baudhāyana's Śulba-Sūtra, Verse 1.10)

Triple the area: start with the given square (side \(s\), area \(s^2\)). First construct a square of double its area (side \(s\sqrt2\), using the diagonal method from Section 2.1). Now combine the original square (side \(s\)) with this doubled square (side \(s\sqrt2\)) using Baudhāyana's combining method: make a right triangle with legs \(s\) and \(s\sqrt2\). Its hypotenuse is \(\sqrt{s^2+2s^2}=\sqrt{3s^2}=s\sqrt3\), and the square built on this hypotenuse has area \(3s^2\) — triple the original.

Five times the area: combine the original square (side \(s\)) with a square of side \(2s\) (i.e. simply double the side length, giving area \(4s^2\)) using the same method: right triangle with legs \(s\) and \(2s\), hypotenuse \(\sqrt{s^2+4s^2}=\sqrt{5s^2}=s\sqrt5\). The square on this hypotenuse has area \(5s^2\) — five times the original.

Triple: combine the original square with a doubled square (legs s, s√2 → hypotenuse s√3). Quintuple: combine the original square with one of double sidelength 2s (legs s, 2s → hypotenuse s√5).

4Find the missing sidelength for a right triangle with legs \(a,b\) and hypotenuse \(c\): (i) \(a=5,b=7\) (ii) \(a=8,b=12\) (iii) \(a=9,c=15\) (iv) \(a=7,b=12\) (v) \(a=1.5,b=3.5\)

(i) \(c^2=5^2+7^2=25+49=74 \Rightarrow c=\sqrt{74}\approx8.60\)

(ii) \(c^2=8^2+12^2=64+144=208 \Rightarrow c=\sqrt{208}=4\sqrt{13}\approx14.42\)

(iii) \(b^2=15^2-9^2=225-81=144 \Rightarrow b=\sqrt{144}=12\)

(iv) \(c^2=7^2+12^2=49+144=193 \Rightarrow c=\sqrt{193}\approx13.89\)

(v) \(c^2=1.5^2+3.5^2=2.25+12.25=14.5 \Rightarrow c=\sqrt{14.5}=\dfrac{\sqrt{58}}{2}\approx3.81\)

(i) c=√74≈8.60 (ii) c=4√13≈14.42 (iii) b=12 (iv) c=√193≈13.89 (v) c=√58/2≈3.81

2.5 Right Triangles Having Integer Sidelengths

In his Śulba-Sūtra (Verse 1.13), Baudhāyana lists integer triples \((a,b,c)\) that satisfy \(a^2+b^2=c^2\): (3,4,5), (5,12,13), (8,15,17), (7,24,25), (12,35,37), and (15,36,39). Such triples are called Baudhāyana triples (also Baudhāyana-Pythagoras triples, Pythagorean triples, or right-angled triangle triples).

MTMath Talk — List all the Baudhāyana triples with numbers less than or equal to 20.

Checking every combination systematically: \((3,4,5)\), \((6,8,10)\), \((9,12,15)\), \((12,16,20)\), \((5,12,13)\), and \((8,15,17)\).

(3,4,5), (6,8,10), (9,12,15), (12,16,20), (5,12,13), (8,15,17) — six triples in all with every number ≤ 20.

MTMath Talk — Is (30, 40, 50) a Baudhāyana triple? Is (300, 400, 500)?

\(30^2+40^2=900+1600=2500=50^2\) ✓. \(300^2+400^2=90{,}000+160{,}000=250{,}000=500^2\) ✓.

Both are Baudhāyana triples — each is just (3,4,5) with every term multiplied by the same factor (10, then 100).

Yes to both — they're (3,4,5) scaled by 10 and by 100 respectively.

Looking at the list of triples with numbers ≤ 20 — (3,4,5), (6,8,10), (9,12,15), (12,16,20) — notice that four of these six can all be obtained by multiplying every term of (3,4,5) by a positive integer (1, 2, 3, or 4). This suggests a conjecture:

Conjecture: (3k, 4k, 5k) is a Baudhāyana triple, for any positive integer k.

MTMath Talk — Is this conjecture true? Can we generalise it further to any Baudhāyana triple (a,b,c)?

Check: \((3k)^2+(4k)^2=9k^2+16k^2=25k^2=(5k)^2\) ✓ — true for every positive integer \(k\). This alone shows there are infinitely many Baudhāyana triples.

General version: if \((a,b,c)\) is any Baudhāyana triple (so \(a^2+b^2=c^2\)), is \((ka,kb,kc)\) also one? Check: \((ka)^2+(kb)^2=k^2a^2+k^2b^2=k^2(a^2+b^2)=k^2c^2=(kc)^2\) ✓ — true in general, for any Baudhāyana triple and any positive integer \(k\).

Yes, both the specific and general conjectures are true: (ka, kb, kc) is always a Baudhāyana triple whenever (a,b,c) is one — this "scaled version" gives infinitely many triples from just one.

A Baudhāyana triple with no common factor greater than 1 is called a primitive Baudhāyana triple. So (3,4,5) is primitive, while (9,12,15) — a scaled version — is not.

MTMath Talk — Is (5, 12, 13) a primitive Baudhāyana triple? What are the other primitive triples with numbers ≤ 20?

The GCD of 5, 12, and 13 is 1, so (5, 12, 13) is primitive.

From our full list — (3,4,5), (6,8,10), (9,12,15), (12,16,20), (5,12,13), (8,15,17) — checking each for a common factor: (6,8,10) has GCD 2, (9,12,15) has GCD 3, and (12,16,20) has GCD 4 — none of these three are primitive. That leaves (3,4,5), (5,12,13), and (8,15,17) as the primitive triples with numbers ≤ 20.

Yes, (5,12,13) is primitive. The primitive triples ≤ 20 are: (3,4,5), (5,12,13), and (8,15,17).

MTMath Talk — Generate 5 scaled versions of each of these primitive triples. Are these scaled versions primitive?

Multiplying each primitive triple by \(k=2,3,4,5,6\):

(3,4,5): (6,8,10), (9,12,15), (12,16,20), (15,20,25), (18,24,30)

(5,12,13): (10,24,26), (15,36,39), (20,48,52), (25,60,65), (30,72,78)

(8,15,17): (16,30,34), (24,45,51), (32,60,68), (40,75,85), (48,90,102)

None of these scaled versions are primitive — each has a common factor of at least \(k\) (the very factor used to scale it), by construction.

5 scaled versions listed for each primitive triple above; none of them are primitive, since they all share the scaling factor k as a common factor.

MTMath Talk — If (a,b,c) is non-primitive with common factor f > 1, is \(\left(\frac{a}{f},\frac{b}{f},\frac{c}{f}\right)\) a Baudhāyana triple? Check for (9,12,15).

For \((9,12,15)\): \(f=\gcd(9,12,15)=3\). Dividing: \(\left(\frac93,\frac{12}3,\frac{15}3\right)=(3,4,5)\), which is indeed a Baudhāyana triple (\(3^2+4^2=5^2\)) ✓.

General justification: since \(f\) divides \(a,b,c\), write \(a=fA,\,b=fB,\,c=fC\) for integers \(A,B,C\). From \(a^2+b^2=c^2\): \(f^2A^2+f^2B^2=f^2C^2\). Dividing both sides by \(f^2\) (which is valid since \(f\ne0\)): \(A^2+B^2=C^2\) — so \(\left(\frac af,\frac bf,\frac cf\right)=(A,B,C)\) is indeed a Baudhāyana triple.

Yes — dividing any Baudhāyana triple by a common factor f always gives another Baudhāyana triple (this is simply the reverse of the earlier scaling result). Every non-primitive triple can be reduced down to a primitive one this way.

If we can find all the primitive triples, we can find every Baudhāyana triple (as a scaled version of some primitive one). How do we generate more primitive triples? We use the relation between consecutive odd numbers and square numbers: \(1=1^2\), \(1+3=2^2\), \(1+3+5=3^2\) — the sum of the first \(n\) odd numbers is \(n^2\). Since the \(n\)th odd number is \(2n-1\):

\[\underbrace{1+3+5+\dots+(2n-3)}_{\text{sum of first }(n-1)\text{ odd numbers}}+(2n-1)=n^2 \;\;\Rightarrow\;\; (n-1)^2+(2n-1)=n^2\]

If the \(n\)th odd number \(2n-1\) is itself a perfect square, this equation becomes a sum of two squares equal to a third square — a Baudhāyana triple!

Example. 9 is an odd square — it's the 5th odd number (\(9=2\times5-1\)). Substituting \(n=5\): \((5-1)^2+9=5^2 \Rightarrow 4^2+3^2=5^2\).
Another: 25 is an odd square — the 13th odd number (\(25=2\times13-1\)). Substituting \(n=13\): \((13-1)^2+25=13^2 \Rightarrow 12^2+5^2=13^2\).

Figure it Out — Set 3

Three questions from page 50 of the textbook.

1Find 5 more Baudhāyana triples using the odd-square idea.

Using odd perfect squares \(k^2\) (for odd \(k=7,9,11,13,15\)), with \(2n-1=k^2 \Rightarrow n=\frac{k^2+1}{2}\), giving the triple \((n-1,\,k,\,n)\):

\(k=7\): \(n=25\) → \((24,7,25)\), check: \(24^2+7^2=576+49=625=25^2\) ✓

\(k=9\): \(n=41\) → \((40,9,41)\), check: \(40^2+9^2=1600+81=1681=41^2\) ✓

\(k=11\): \(n=61\) → \((60,11,61)\), check: \(60^2+11^2=3600+121=3721=61^2\) ✓

\(k=13\): \(n=85\) → \((84,13,85)\), check: \(84^2+13^2=7056+169=7225=85^2\) ✓

\(k=15\): \(n=113\) → \((112,15,113)\), check: \(112^2+15^2=12{,}544+225=12{,}769=113^2\) ✓

(24,7,25), (40,9,41), (60,11,61), (84,13,85), (112,15,113) — five more triples generated from the odd squares 49, 81, 121, 169, 225.

2Does this method yield non-primitive Baudhāyana triples? [Hint: one of the smaller sidelengths is always one less than the hypotenuse.]

No — this method always yields primitive triples. Every triple it produces has the form \((n-1,\,k,\,n)\), where one leg (\(n-1\)) is exactly one less than the hypotenuse (\(n\)). Since consecutive integers \(n-1\) and \(n\) can never share a common factor greater than 1 (any common factor of both would have to divide their difference, which is 1), the whole triple \((n-1,k,n)\) can have no common factor greater than 1 either — so it must be primitive.

No, it can never yield a non-primitive triple, precisely because one leg and the hypotenuse always differ by exactly 1, forcing them (and hence the whole triple) to be coprime.

3Are there primitive triples that cannot be obtained through this method? If yes, give examples.

Yes. This method only ever produces triples where one leg is exactly one less than the hypotenuse (\(n-1\) and \(n\)). But there exist perfectly valid primitive triples where neither leg is that close to the hypotenuse.

Example: \((20,21,29)\). Check: \(20^2+21^2=400+441=841=29^2\) ✓, and \(\gcd(20,21,29)=1\), so it's primitive. But \(29-21=8\) and \(29-20=9\) — neither leg is just 1 less than the hypotenuse, so this triple can never arise from the "odd square" method above.

Yes — e.g. (20, 21, 29) is primitive but has no side just 1 less than the hypotenuse, so it cannot be produced by this particular method.

2.6 A Long-Standing Open Problem

The study of Baudhāyana triples inspired the 17th-century French mathematician Fermat to ask a bold generalisation: we've seen infinitely many square numbers that are sums of two square numbers (\(x^2+y^2=z^2\)) — is there a perfect cube that's a sum of two perfect cubes? A fourth power that's a sum of two fourth powers? In other words, does \(x^n+y^n=z^n\) have any solution in natural numbers \(x,y,z\) when \(n\gt2\)?

In the margin of a book, Fermat wrote that no such solution exists for any \(n\gt2\), adding: "I have found a truly marvellous proof of this statement, but the margin is too small to contain it." No one ever found Fermat's proof, and this claim became known as Fermat's Last Theorem. More than 300 years of attempts by the world's greatest mathematicians failed to prove it. In 1963, a 10-year-old boy named Andrew Wiles read about this problem and resolved to solve it — he eventually succeeded in 1994, finally proving Fermat's Last Theorem after over three centuries.

2.7 Further Applications of the Baudhāyana-Pythagoras Theorem

A Problem from Bhāskarāchārya's Līlāvatī

MTMath Talk — "In a lake, a lotus flower peeps 1 unit above the water. Swayed by a breeze, its tip touches the water 3 units from its original position. Find the depth of the lake."

Let \(x\) = depth of the lake (length of stem underwater). Since the stem sticks 1 unit above water, its total length is \(x+1\). Assuming the stem stood perpendicular to the water initially, when swayed, the tip traces to a point 3 units away — forming a right triangle with legs 3 and \(x\), and hypotenuse \(x+1\) (the stem's fixed total length, now lying at an angle).

By the Baudhāyana-Pythagoras theorem: \(3^2+x^2=(x+1)^2 \Rightarrow 9+x^2=x^2+2x+1\). Subtracting \(x^2\) from both sides: \(9=2x+1 \Rightarrow x=4\).

The depth of the lake is 4 units.

Figure it Out — Set 4

Nine questions from pages 52–54 of the textbook.

1Find the diagonal of a square with sidelength 5 cm.

A square's diagonal is the hypotenuse of an isosceles right triangle with legs equal to the side: \(d=5\sqrt2\approx7.07\) cm (using \(c=a\sqrt2\) from Section 2.3).

Diagonal = 5√2 ≈ 7.07 cm.

2Find the missing sidelengths in six right triangles with given values: (a) legs 7, 9 (b) legs 4, 10 (c) leg 40, hypotenuse 41 (d) leg 10, hypotenuse √200 (e) leg 10, hypotenuse √150 (f) leg 27, hypotenuse 45

Exactly which two sides are given (both legs, or a leg and the hypotenuse) depends on where the right-angle mark sits in the printed figure — check your copy. The values below assume the most natural reading of each triangle's shape (e.g. a very "thin," elongated triangle like 40–41 strongly suggests those are a leg and the nearby hypotenuse, since Pythagorean triples with two large close values, like 9-40-41, produce exactly that shape).

(a),(b): both given values are legs, right angle between them. (c),(d),(e),(f): the larger value is the hypotenuse, right angle is between the given leg and the (unmarked) missing leg.

(a) legs 7, 9 → hypotenuse \(=\sqrt{7^2+9^2}=\sqrt{130}\approx11.40\)

(b) legs 4, 10 → hypotenuse \(=\sqrt{4^2+10^2}=\sqrt{116}=2\sqrt{29}\approx10.77\)

(d) leg 10, hypotenuse \(\sqrt{200}\) → other leg \(=\sqrt{200-100}=\sqrt{100}=10\) (an isosceles right triangle, both legs 10, matching \(c=a\sqrt2\))

(e) leg 10, hypotenuse \(\sqrt{150}\) → other leg \(=\sqrt{150-100}=\sqrt{50}=5\sqrt2\approx7.07\)

(f) leg 27, hypotenuse 45 → other leg \(=\sqrt{45^2-27^2}=\sqrt{2025-729}=\sqrt{1296}=36\) (this is 9×(3,4,5): 27=9×3, 36=9×4, 45=9×5)

(a) √130≈11.40 (b) 2√29≈10.77 (c) 9 (d) 10 (e) 5√2≈7.07 (f) 36

3Find the sidelength of a rhombus whose diagonals are 24 units and 70 units.

A rhombus's diagonals bisect each other at right angles, splitting it into 4 congruent right triangles with legs \(\frac{24}{2}=12\) and \(\frac{70}{2}=35\). The rhombus's side is the hypotenuse of one such triangle: \(\sqrt{12^2+35^2}=\sqrt{144+1225}=\sqrt{1369}=37\).

Sidelength = 37 units (using the triple 12-35-37).

4Is the hypotenuse the longest side of a right triangle? Justify your answer.

Yes, always. Since \(c^2=a^2+b^2\) and both \(a^2,b^2\) are positive (assuming \(a,b\gt0\)), we have \(c^2\gt a^2\) and \(c^2\gt b^2\). Since all three lengths are positive, taking square roots preserves the inequality: \(c\gt a\) and \(c\gt b\).

Yes — since c²=a²+b² is strictly greater than both a² and b² individually (as the other term is always positive), c must be strictly greater than both a and b.

5True or False — Every Baudhāyana triple is either a primitive triple or a scaled version of a primitive triple.

True. Take any Baudhāyana triple \((a,b,c)\) and let \(f=\gcd(a,b,c)\). If \(f=1\), the triple is already primitive. If \(f\gt1\), then — as shown earlier — \(\left(\frac af,\frac bf,\frac cf\right)\) is itself a Baudhāyana triple, and by construction it has no common factor left (dividing out the full GCD), so it's primitive. Either way, \((a,b,c)=f\times\left(\frac af,\frac bf,\frac cf\right)\) is exactly a scaled version of a primitive triple (with \(f=1\) covering the "already primitive" case).

True — every Baudhāyana triple, divided by its own GCD, becomes primitive; so every triple is some primitive triple scaled by its GCD.

6Give 5 examples of rectangles whose sidelengths and diagonals are all integers.

A rectangle's diagonal is the hypotenuse of a right triangle formed by its two sides — so any Baudhāyana triple \((a,b,c)\) gives a rectangle of sides \(a,b\) with integer diagonal \(c\):

3×4 rectangle (diagonal 5); 6×8 rectangle (diagonal 10); 5×12 rectangle (diagonal 13); 8×15 rectangle (diagonal 17); 9×12 rectangle (diagonal 15).

3×4 (d=5), 6×8 (d=10), 5×12 (d=13), 8×15 (d=17), 9×12 (d=15) — each pair taken directly from known Baudhāyana triples.

7Construct a square whose area equals the difference of the areas of squares of sidelengths 5 units and 7 units.

The required square's area is \(7^2-5^2=49-25=24\) sq. units — its sidelength is \(\sqrt{24}=2\sqrt6\).

Construction: using the Baudhāyana-Pythagoras theorem in reverse — draw a right triangle with hypotenuse 7 and one leg 5 (draw a leg of length 5, then use a compass centred at its far end with radius 7 to mark the hypotenuse's endpoint, then join to form the right angle). The other leg of this triangle has length \(\sqrt{7^2-5^2}=\sqrt{24}\) — that's the sidelength of the required square, since a right triangle with hypotenuse 7 and leg 5 satisfies \(5^2+(\text{other leg})^2=7^2\), i.e. \((\text{other leg})^2=7^2-5^2=24\).

Required square's side = √24 = 2√6 units, obtainable as the missing leg of a right triangle with hypotenuse 7 and one leg 5.

8Math Talk / Try This — (i) Using dots of a grid as vertices, can you create a square with area (a) 2 (b) 3 (c) 4 (d) 5 sq. units? (ii) If the grid extends indefinitely, what integer areas are possible this way?

A tilted square with "step vector" \((p,q)\) between adjacent grid-dot vertices has area \(p^2+q^2\) (by the Pythagoras theorem, since each side is the hypotenuse of a right triangle with legs \(p,q\) along the grid lines).

(a) Area 2: use step \((1,1)\): \(1^2+1^2=2\) ✓ — a tilted square touching the midpoints of a 2×2 block.

(b) Area 3: impossible. We'd need integers \(p,q\) with \(p^2+q^2=3\) — checking all small cases (\(0+3,1+2\)), no combination of two perfect squares sums to 3.

(d) Area 5: use step \((1,2)\): \(1^2+2^2=1+4=5\) ✓ — a tilted square.

(ii) General rule: an integer \(n\) is achievable as a grid-square area exactly when \(n\) can be written as \(p^2+q^2\) for some integers \(p,q\) (not both zero). By the "sum of two squares" theorem, this happens precisely when every prime factor of \(n\) that leaves remainder 3 when divided by 4 (like 3, 7, 11, 19, 23, …) appears an even number of times in \(n\)'s prime factorisation. For example, 1,2,4,5,8,9,10,13,16,17,18,20,… are all achievable, but 3,6,7,11,12,14,15,19,… are not.

(a) 2 ✓ (step 1,1) (b) 3 ✗ impossible (c) 4 ✓ (step 2,0) (d) 5 ✓ (step 1,2). General rule: n is achievable iff every prime factor ≡3 (mod 4) appears an even number of times in n.

9Try This — Find the area of an equilateral triangle with sidelength 6 units. [Hint: an altitude bisects the opposite side.]

Drawing the altitude from the top vertex splits the equilateral triangle into two congruent right triangles (by symmetry, since all 3 sides are equal — this is provable via SSS congruence of the two halves), each with hypotenuse 6 and one leg \(\frac62=3\) (half the base, since the altitude bisects the opposite side).

Height \(h=\sqrt{6^2-3^2}=\sqrt{36-9}=\sqrt{27}=3\sqrt3\).

Area \(=\frac12\times\text{base}\times\text{height}=\frac12\times6\times3\sqrt3=9\sqrt3\) sq. units.

Area = 9√3 ≈ 15.59 sq. units.

Puzzle Time — Find the Colours!

Three Mislabeled Boxes

Three closed boxes contain only red, only blue, and only green balls respectively. They're labelled RED, BLUE, and GREEN, but no box has the label matching its actual contents. Find which label belongs to which box, opening only one box.

Solution: Open the box labelled GREEN. Since no label is correct, this box cannot actually contain green balls — it must contain either red or blue.

Case 1 — it contains red balls: then the box labelled GREEN actually holds red. The box labelled BLUE can't hold blue (given) and can't hold red (already taken by the GREEN-labelled box) — so it must hold green. That leaves the box labelled RED to hold the only colour remaining: blue.

Case 2 — GREEN-labelled box contains blue balls: then by the same logic, RED-labelled (can't hold red, and blue is taken) must hold green, leaving BLUE-labelled to hold red.

Either way, opening just the one box (GREEN) tells us which of only two possible full mislabeling patterns applies, and from that single piece of information we can deduce the contents of all three boxes without opening the other two.

The key insight: with exactly 3 items and the rule "no label is correct," there are only 2 possible complete mismatches overall — so a single clue (what's really in one box) is always enough to pin down the other two by elimination.

Frequently Asked Questions

For a right-angled triangle with legs a and b and hypotenuse c (the side opposite the right angle), a² + b² = c². Baudhāyana stated this in full generality in his Śulba-Sūtra centuries before the Greek mathematician Pythagoras, which is why it's also called the Baudhāyana-Pythagoras Theorem.

Construct a new square whose side equals the diagonal of the original square. Since the diagonal of a square with side s has length s√2, the new square's area is (s√2)² = 2s² — exactly double the original.

If √2 equaled a fraction m/n, then 2n² = m² would hold. But in any perfect square's prime factorisation, every prime appears an even number of times — so the prime 2 would need to appear an odd number of times on one side of this equation and an even number of times on the other, which is impossible. So no such fraction can exist.

A triple of positive integers (a, b, c) satisfying a² + b² = c² — such as (3,4,5), (5,12,13), or (8,15,17). Scaling any triple by a positive integer gives another triple, so there are infinitely many. A triple with no common factor greater than 1 is called primitive.

It states that the equation a^n + b^n = c^n has no solution in positive integers when n is greater than 2 — unlike n=2, where infinitely many solutions (Pythagorean triples) exist. Fermat claimed to have a proof in the 17th century, but it was only proven in 1994, by Andrew Wiles, after more than 300 years of failed attempts by other mathematicians.

Continue with Part 2, Chapter 3

Move on to the next Part 2 chapter next.

Part 2, Chapter 3

Part 2, Chapter 2The Baudhāyana-Pythagoras Theorem