NCERT Solutions for Class 8 Maths Chapter 9 The Baudh Yana Pythagoras Theorem (PDF) – 2027

Draw the diagonal of the given square and build a new square on that diagonal. This is exactly what Baudhāyana states in Verse 1.9 of his Śulba-Sūtra:

So if the given square has side s, its diagonal becomes the side of the new square, and the new square has exactly twice the area of the original.

No. Doubling the side makes the area four times larger, not twice.

Because the original square is made up of two small triangles, while the new square is made up of four triangles of exactly the same kind.

Extend the vertical and horizontal sides of the original square. They pass through the two remaining vertices of the dotted square and become its two diagonals. These two lines cut the dotted square into four triangles.

Yes. Extend the horizontal and the vertical side of the original square (Baudhāyana’s ‘east-west’ and ‘north-south’ lines) right across the dotted square.

• These two lines are the two diagonals of the dotted square, so they cut it into 4 congruent triangles.
• The original square is cut by its own diagonal into 2 of the same triangles.

Because those sides bisect the angles of the dotted square at the two vertices they start from, and in a square the angle bisector at a vertex is the diagonal through that vertex.

Let C be the bottom-right corner of the original square; it is also a vertex of the dotted square.

In the same way, the extended horizontal side bisects the angle at the top-left corner and passes through the opposite vertex.

Yes. Every one of them is a right triangle whose two perpendicular sides are equal in length, and that common length is the same for all of them.

Let the original square have side s. The two extended sides meet at the top-right corner of the original square, which is the centre of the dotted square.

Each square is cut along both its diagonals, giving four congruent right triangles. Keep Square 1 whole (or reassembled from pieces 1–4) and set the four triangles 5, 6, 7, 8 of Square 2 around it.

How to place them: lay each triangle so that its longest edge lies exactly along one side of Square 1, with the triangle pointing outwards. Do this on all four sides.

Reverse the doubling construction: draw the square that joins the midpoints of the four sides of the given square.

1. Mark the midpoint of each side of the given square.
2. Join the four midpoints in order.
3. The tilted square you get has exactly half the area of the original.

Draw the east-west and north-south lines through the centre. They cut the whole picture into 8 congruent triangles, and the inner square contains exactly 4 of them.

• The two lines join opposite midpoints, so they are the diagonals of the inner square and they also cut the outer square into four equal small squares.
• Each small square is split by a side of the inner square into two congruent right triangles: one inside the inner square, one outside it.

Fold the four corners inwards so that the crease lines pass through the midpoints of the sides.

1. Fold the square in half one way and unfold — this marks two midpoints. Fold the other way and unfold — the other two midpoints are marked.
2. Now fold each corner in so that it lands on the centre. Each crease runs from one midpoint to the next.
3. The square PQRS you are left with is the required square.

No. Halving the side gives a quarter of the area, and 4 such squares fill the original square.

P, Q, R, S are the midpoints of the sides of the paper (side s). Join QS and PR, meeting at the centre O.

The hypotenuse is √2 units.

A square PEAR of side 1 is made of two such triangles, and the square REST on its diagonal ER has double its area.

√2 is the number whose square is 2. It is not a whole number and not a fraction; written as a decimal it never stops:

√2 is greater than 1.

√2 is less than 2.

Putting the two results together:

We call 1 a lower bound on √2 and 2 an upper bound.

Yes. Keep squaring decimals one place at a time and see where 2 falls.

No, never. Suppose some decimal that stops had square exactly 2.

• A decimal that stops has a last digit, and that digit is not 0 (a 0 at the end can just be dropped).
• The last digit of its square is the last digit of (last digit)². For the digits 1 to 9 these are 1, 4, 9, 6, 5, 6, 9, 4, 1 — never 0.
• So the square also ends in a non-zero digit after the decimal point.

No. Suppose it could. Then

In the prime factorisation of a square number, every prime occurs an even number of times.

So √2 cannot be written as a fraction m/n of counting numbers.

Yes. The four pieces are congruent isosceles right triangles with equal sides s (the side of the paper square) and longest side s√2 (its diagonal).

The arrangement: bring the four right-angle corners together at one point, and fit the triangles round it so that the equal sides of neighbouring triangles lie along each other. The four longest edges then form the boundary.

For an isosceles right triangle with equal sides a, the square on the hypotenuse has twice the area of the square on a side:

To get bounds with one decimal place, square the one-place decimals on either side of the answer.

Each of the other two sides is √50 units (between 7.0 and 7.1 units).

Two such triangles make a square whose diagonal is the hypotenuse 10, so use c² = 2a² backwards:

Bounds to one decimal place: 7.0² = 49 and 7.1² = 50.41, so 7.0 < √50 < 7.1.

Baudhāyana answers this in Verse 1.12 of the Śulba-Sūtra:

The construction: make a right-angled triangle whose two perpendicular sides are the sides of the two given squares. The square built on its hypotenuse has area equal to the sum of the two given areas.

Yes, it agrees exactly. Take the two squares to be the same size, each of side a.

• The right triangle whose perpendicular sides are a and a is an isosceles right triangle.
• Its hypotenuse is precisely the diagonal of a square of side a.
• So “build a square on the hypotenuse” becomes “build a square on the diagonal of the given square” — which is the rule of Verse 1.9 from Section 2.1.

T, U, X and W are congruent right triangles (each has perpendicular sides a and b), so the four sides of T + U + V are all equal to their hypotenuse. Its angles are right angles (next question), so it is a square.

For the area, look at the same region counted two ways:

T, U, W and X are congruent right triangles, each with acute angles x and 90° – x (marked in the figure).

Four equal sides and four right angles: T + U + V is a square.

Join the squares side by side, bottoms in line.

1. Two cuts: from the point on the bottom edge that is a from the right end, cut 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 sides a and b.
2. Three pieces: two such triangles and a five-sided piece.
3. Rearrange: turn each triangle a quarter turn about its top corner; the pieces form a square of side c.
4. Cover: the pieces exactly cover the square drawn on the hypotenuse of this right triangle.

The hypotenuse is 13 cm. A careful drawing measures about 13 cm, and the theorem confirms it exactly.

The third side is 15 cm.

Here the 17 cm side is the hypotenuse, so it goes on the right of the equation, not the left.

Triple the area: first double the square by building the square on its diagonal (area 2). Make a right triangle whose perpendicular sides are this diagonal (√2) and a side (1) of the given square. The square on its hypotenuse has area 2 + 1 = 3 times the given square.

Five times the area: make a right triangle with perpendicular sides 2 (twice the given side) and 1. The square on its hypotenuse has area 4 + 1 = 5 times the given square.

Use a² + b² = c² each time — adding when the hypotenuse is missing, subtracting when a shorter side is missing.

Checking every pair a ≤ b ≤ 20 for which a² + b² is the square of a whole number c ≤ 20 gives exactly six triples.

Yes. One triple already gives infinitely many, because every scaled copy of a triple is again a triple.

Later in the section a second, deeper source of triples appears: every odd square number greater than 1 produces a fresh primitive triple, so even the primitive triples go on for ever.

Yes.

It is (3, 4, 5) with every number multiplied by 10 — a scaled version of (3, 4, 5). It is not primitive, since 10 is a common factor.

Yes — it is (3, 4, 5) scaled by 100.

Yes — every one of them is (3, 4, 5) multiplied through by a positive integer.

Geometrically these are all the same triangle drawn at different sizes — enlarge a 3-4-5 triangle by a factor k and the angles do not change, so the right angle survives.

Yes, the conjecture is true, and a single line of algebra proves it for every k at once.

So (3k, 4k, 5k) satisfies a² + b² = c² for every positive integer k. Since there are infinitely many values of k, this alone shows there are infinitely many Baudhāyana triples.

Yes. Nothing in the proof used the particular numbers 3, 4 and 5. The general statement is:

For example, starting from (5, 12, 13) we get (10, 24, 26), (15, 36, 39), (20, 48, 52), … — and (15, 36, 39) is one of the triples Baudhāyana himself lists in Verse 1.13.

True. We are given a² + b² = c², and we must check whether (ka)² + (kb)² = (kc)².

So (ka, kb, kc) is a Baudhāyana triple. We call it a scaled version of (a, b, c).

Yes, (5, 12, 13) is primitive. A triple is primitive when its three numbers have no common factor greater than 1.

The complete list of Baudhāyana triples with all numbers at most 20 is (3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15), (12, 16, 20). Testing each for a common factor:

So besides (5, 12, 13), the primitive ones up to 20 are (3, 4, 5) and (8, 15, 17).

Multiply each primitive triple by k = 2, 3, 4, 5, 6.

None of these scaled versions is primitive. Each was made by multiplying all three numbers by k, so k itself is a common factor, and k > 1 in every case.

Yes, it is. Dividing out a common factor takes you back down to a smaller triple.

Check for (9, 12, 15). Here f = 3.

Justification. Since f divides all three, write a = fp, b = fq, c = fr with p, q, r whole numbers.

Use the fact that the sum of the first n odd numbers is n².

Split off the last term. The part before it is the sum of the first (n – 1) odd numbers, which is (n – 1)². So

If the odd number 2n – 1 happens to be a square number, say m², this reads m² + (n – 1)² = n² — a Baudhāyana triple (m, n – 1, n).

The nth odd number is 2n – 1.

It is (n – 1)².

The rule is that the sum of the first N odd numbers is N². Putting N = n – 1 gives (n – 1)². So splitting the last term off the sum of the first n odd numbers gives

Yes. 9 is an odd square and it is the 5th odd number, since 9 = 2 × 5 – 1. So take n = 5.

Take an odd square m², find which odd number it is by solving m² = 2n – 1, that is n = (m² + 1)/2, and read off the triple (m, n – 1, n).

No — every triple this method produces is primitive.

Look at the shape of the answer. The triple is (m, n – 1, n), so two of its three numbers are the consecutive whole numbers n – 1 and n.

Yes. This method always leaves a gap of exactly 1 between the longer leg and the hypotenuse. Any primitive triple in which that gap is not 1 cannot come from it.

All four are primitive, but only the last has consecutive larger numbers. So (8, 15, 17), (12, 35, 37) and (20, 21, 29) are primitive triples this method can never produce.

The depth of the lake is 4 units.

Let x be the depth (the part of the stem under water). The whole stem is x + 1. Taking the upright stem perpendicular to the water, the swayed stem is the hypotenuse of a right triangle with sides 3 and x.

The diagonal is √50 cm (between 7.0 cm and 7.1 cm).

A diagonal cuts the square into two isosceles right triangles with equal sides 5 cm, so the diagonal is their hypotenuse.

Bounds to one decimal place: 7.0² = 49 and 7.1² = 50.41, so 7.0 cm < diagonal < 7.1 cm.

In each triangle, first identify which side is the hypotenuse — it is the one opposite the right angle. Then add the squares if the hypotenuse is missing, and subtract if a shorter side is missing.

The sidelength is 37 units.

The diagonals of a rhombus bisect each other at right angles. So each side of the rhombus is the hypotenuse of a right triangle whose perpendicular sides are the two half-diagonals.

Yes, always. The theorem itself proves it.

True.

Take any Baudhāyana triple (a, b, c). Let f be the largest number that divides all three.

• If f = 1, the triple has no common factor greater than 1, so it is already primitive.
• If f > 1, then (a/f, b/f, c/f) is again a Baudhāyana triple (shown on page 49), and it has no common factor left, so it is primitive. Multiplying it back by f returns (a, b, c) — a scaled version of a primitive triple.

A diagonal of a rectangle splits it into two right triangles with the sides as perpendicular sides. So a rectangle has an integer diagonal exactly when its two sides form a Baudhāyana triple with it.

The required area is 7² – 5² = 49 – 25 = 24 sq. units, so the square must have side √24 units (about 4.9 units).

1. Draw a segment AB of length 5 units.
2. At A, draw a line perpendicular to AB.
3. With B as centre and radius 7 units, draw an arc cutting that perpendicular at C.
4. AC is the side of the required square. Build the square on AC.

(i) Areas 2, 4 and 5 are possible; area 3 is not.

(ii) Any square drawn on the grid has one side going p steps across and q steps up, for some whole numbers p and q. By Baudhāyana’s theorem its side satisfies side² = p² + q². So:

The area is 3√27 sq. units ≈ 15.6 sq. units.

Step 1. Let AD be the altitude from A to BC. Triangles ABD and ACD have AB = AC, AD common and right angles at D, so they are congruent (RHS): BD = DC = 6 ÷ 2 = 3.

Step 2. Height, by Baudhāyana’s theorem in triangle ABD:

Step 3. Area:

Open any one box and take out a single ball. That one ball settles all three boxes.

Say you open the box labelled RED. Its label is wrong, so it does not hold red balls — it holds blue or green.

• If the ball is blue: that box is the blue box. The box labelled GREEN cannot hold green, and blue is already used up, so it must hold red. The last box, labelled BLUE, therefore holds green.
• If the ball is green: that box is the green box. The box labelled BLUE cannot hold blue, and green is used up, so it holds red. The remaining box, labelled GREEN, holds blue.