NCERT Solutions Class 8 Maths Chapter 9 the Baudh Yana Pythagoras Theorem

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 8 · M AT H S

NCERT Solutions

Chapter 9: The Baudhāyana-
Pythagoras Theorem

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 33 – 54 19 59 English

Solutions, notes, sample papers & more at 47 pages

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

CLASS 8 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 9: The Baudhāyana-
Pythagoras Theorem
Chapter 2 of Ganita Prakash Grade 8 Part-II starts from a single question Baudhāyana asked in his Śulba-Sūtra
around 800 BCE — how do you build a square of double the area? — and follows it all the way to a² + b² = c², to
the first irrational number √2, and to Fermat's Last Theorem.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 8) Part II, 33 – 54

SECTIONS QUESTIONS

19 59

MEDIUM

English

In-text Questions — Page 33
2.1 Doubling a Square

Q1 How can one construct a square having double the area of a given square?

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:

“The diagonal of a square produces a square of double the area of the original square.”

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.

Page 1 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

new square

diagonal

original square

The diagonal of the shaded square is a side of the dashed square, and the dashed square has double
the area.

Q2 A first guess might be to simply double the length of each side of the square. Will
this new square have double the area of the original square?

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

Page 2 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Original square: side = s, area = s²

New square: side = 2s, area = 2s × 2s = 4s²

So the area becomes 4 times the original, not 2 times.

Why it happens: area is a product of two lengths. When each length is multiplied by
2, the product is multiplied by 2 × 2 = 4. You can see it directly: the bigger square is
made of four copies of the original square. To double an area we need the side to be
multiplied by a number whose square is 2 — and that number is √2, not 2.

In-text Questions — Page 34
2.1 Doubling a Square

MATH TALK

Q1 Why does the new dotted square have double the area of the original square?

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.

Original square = 2 small triangles

Dotted square = 4 small triangles

All 4 triangles are congruent, so

Area of dotted square = 2 × area of original square

Page 3 of 47

as e
a
Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem
g l AglaSem · NCERT Solutions

co m
e m.
m l as
m .co a g
l a se
a g

com
e m . ag
g l as
a
3 co m
em.
m 2 l as
m .co a g
l a se
ag
4 s
com a
1 m . agl
ase
agl

co m
m .
m as e
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se m
g l a
a
se m
com g l a
.
The two green lines are the extended sides of the original square. They are the diagonals of the
m a
ase
dotted square and cut it into four congruent triangles: the original square is 1 + 2, the dotted square

agl
is 1 + 2 + 3 + 4.

co m
m .
as e
comdouble the area of the original square? l
Can you draw some horizontal and vertical lines to see why the new square has
g
Q2

m . a
ase
agl c
.

s e m
m a
agl
Yes. Draw the two lines Baudhāyana calls the ‘east-west’ and ‘north-south’ lines — that is, extend
. co
m
the horizontal side and the vertical side of the original square right across the figure.

as e
a g l
Each of them starts at one vertex of the dotted square and ends at the opposite vertex, so
together they are the two diagonals of the dotted square.

co m
m .
m ase
.co


a g l Page 4 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

The diagonals of a square cut it into four congruent triangles. So the dotted square is 4
triangles.
One of those four triangles is exactly half of the original square (the half on the far side of
the diagonal), and the original square is 2 such triangles.

Dotted square = 4 triangles

Original square = 2 triangles

Ratio = 4 : 2 = 2 : 1

Tip: the book adds still more horizontal and vertical lines on page 34 and gets 8 tiny
triangles in the dotted square and 4 in the original — the ratio is still 2 : 1.

Q3 Why should the extension of the vertical and horizontal sides of the original square
pass through the vertices of the dotted square? [Hint: From the diagonal property
of a square, the line that bisects an angle passes through the opposite vertex.
Argue why the vertical and horizontal sides of the original square bisect the two
angles of the dotted square.]

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.
Take the vertex of the dotted square that is also the bottom-right corner of the original square.
Call it C. Two things meet at C:

the side of the dotted square that lies along the diagonal of the original square, and
the other side of the dotted square, which goes off to the right.

The diagonal of the original square makes an angle of 45° with each of its sides. So the vertical
side of the original square at C makes 45° with one side of the dotted square. The full angle of
the dotted square at C is 90°, so the vertical side makes 45° with the other side too — it bisects
the 90° angle at C.

Angle of dotted square at C = 90°

Angle between the vertical side and one side of the dotted square = 45°

So the vertical side is the bisector at C

In a square, the bisector at a vertex is the diagonal through that vertex

A diagonal ends at the opposite vertex ⇒ the extended side passes through it

Page 5 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

The same argument at the top-left corner shows the extended horizontal side passes through
the opposite vertex of the dotted square.

Why it happens: a square's diagonal splits its 90° corner into two 45° halves. The
original square's diagonal is a side of the new square, so anything at 45° to it is at
45° to the new square's side as well — and 45° is exactly half of the new square's
corner.

Q4 Moreover, all these small triangles are congruent to each other. Can you explain
why?

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 lines we drew meet at the centre of the dotted
square, which is the corner of the original square that is not on the diagonal.

The four triangles of the dotted square each have the right angle at that meeting point,
because the two lines are perpendicular (one is ‘east-west’, the other ‘north-south’).
Their two perpendicular sides are half-diagonals of the dotted square. Each half-diagonal has
length s, since it runs from that corner to a corner of the original square or to the far vertex.

Each triangle: right-angled, with the two shorter sides = s and s

By SAS, all four are congruent

The two halves of the original square also have shorter sides s and s with a right angle

between them

So all six triangles in the picture are congruent

Check it yourself: each triangle has area ½ × s × s = s²/2. Four of them give 2s², and
two of them give s² — the doubling again.

In-text Questions — Page 35

Page 6 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Doubling a Square Using Paper

Q1 Cut out two identical squares of paper. Draw, label, and cut as follows: Square 1 into
pieces 1, 2, 3, 4 and Identical Square 2 into pieces 5, 6, 7, 8. Now place the pieces 5, 6,
7, and 8 around Square 1 to get a square with double the area.

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.

Side of each paper square = s

Each of pieces 5, 6, 7, 8: longest edge = s, area = s²/4

Total area = s² + 4 × s²/4 = 2s²
Side of the new square = s√2

Why the result is a square: the right-angle tips of the four triangles become the
four corners of the new figure. At every corner of Square 1, the two edges arriving
from neighbouring triangles are in a straight line, because each triangle's base angle
is 45° and 45° + 90° + 45° = 180°. So the boundary is four straight sides of equal
length meeting at right angles — a square. Square 1 now sits tilted inside it, with its
corners at the midpoints of the new square's sides.

Tip: this is the same picture as the drawn construction, read backwards — the tilted
inner square is half of the outer one, so the outer one is double the inner one.

In-text Questions — Page 36

Page 7 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

2.2 Halving a Square

Q1 Now suppose we are given a square, and we want to construct a square whose area
is half that of the original square. How would you do it?

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.

Original square: side s, area s²

Side of the tilted square = half-diagonal of a corner triangle = s/√2
Area of tilted square = (s/√2)² = s²/2

Page 8 of 47

as e
a
Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem
g l AglaSem · NCERT Solutions

co m
e m.
m l as
m .co a g
l a se
a g

com
e m . ag
g l as
a

co m
em.
m l as
m .co a g
l a se
a g
m a s
m.co agl
l a se
a g

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
.
The blue square joins the four midpoints (red dots). The two green lines cut the picture into 8
m a
ase
congruent triangles — 4 inside the blue square, 4 outside it.

agl

co m
Why is the smaller inside square half the area of the larger square?
m .
e
Q2

m l as
.co a g
a s em ANSWER

agl Draw the east-west and north-south lines through the centre. They cut the whole picture into 8
.c
congruent triangles, and the inner square contains exactly 4 of them.
s e m
. com a
The two lines join opposite midpoints, so they are the diagonals of the inner square and theygla
a s
also cut the outer square intoemfour equal small squares.
agl by a side of the inner square into two congruent right triangles:
Each small square is split
one inside the inner square, one outside it.

co m
m .
m ase
.co


a g l Page 9 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Outer square = 8 congruent triangles

Inner square = 4 of those triangles

Area of inner square = 4/8 = ½ of the outer square

Why it happens: each corner triangle that is cut off is congruent to the triangle that
stays inside next to it — same legs, same right angle. So exactly half the paper is
thrown away and half is kept.

Q3 Cut out a square from a piece of paper. Now make a square whose area is half the
area of the first square.

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.

Side of the paper = s, area = s²

Side of PQRS = s/√2

Area of PQRS = s²/2 = half the area of the paper

Check it yourself: the four folded-in corner flaps exactly cover PQRS with no gap
and no overlap. That is a physical proof that the flaps together are half the paper
and PQRS is the other half.

Q4 Will the square having half the sidelength have half the area? Why not? How many
such squares will fill the original square?

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

Page 10 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Original: side s, area s²

Half-side square: side s/2, area = (s/2) × (s/2) = s²/4

Number needed = s² ÷ s²/4 = 4

Why it happens: area involves the side twice. Multiplying the side by ½ multiplies
the area by ½ × ½ = ¼. This is the same reason doubling the side gave 4 times the
area, seen from the other end. To halve the area you must multiply the side by 1/√2,
not by ½.

In-text Questions — Page 37
2.2 Halving a Square / 2.3 Hypotenuse of an Isosceles Right Triangle

Q1 Why is PQRS a square? Why is its area half that of the original paper? Explain by
connecting QS and PR, finding the different angles formed, and then using tringle
congruence.

P, Q, R, S are the midpoints of the four sides of the paper square. Join QS and PR — these are the
east-west and north-south lines through the centre O.
PQRS is a square. Let the paper square be ABCD with side s. Each of the four corner triangles
(for example the one at corner B, with legs BQ and BR) has legs of length s/2 and a right angle
between them. By SAS all four corner triangles are congruent, so their hypotenuses PQ, QR, RS,
SP are all equal — PQRS is at least a rhombus.
Now look at the angles at Q. The corner triangles are isosceles right triangles, so each of their
base angles is 45°. At Q, two such base angles sit on either side of angle PQR along the straight
side of the paper:

45° + ∠PQR + 45° = 180°
∠PQR = 90°

The same holds at P, R and S. A rhombus with right angles is a square.
Its area is half. QS and PR are perpendicular, equal, and bisect each other at O, so they cut
PQRS into 4 congruent triangles and cut the whole paper into 8 congruent triangles of the same
shape and size. Hence

Page 11 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Area of PQRS = 4 triangles

Area of paper = 8 triangles

Area of PQRS = ½ × area of the paper

Q2 Find the hypotenuse of this isosceles right triangle. (The two equal sides are 1 unit
each.)

The hypotenuse is √2 units.
We cannot measure it directly, but we can get at it through area. A square PEAR of side 1 unit is
made of two such triangles, and the square REST built on its diagonal ER has twice the area of
PEAR.

Area of PEAR = 1 × 1 = 1 sq. unit

Area of REST = 2 × Area of PEAR = 2 × 1 = 2 sq. units

If c is the hypotenuse ER, then REST has side c, so

Area of REST = c × c = c²
So c² = 2

Therefore c = √2

So the hypotenuse of an isosceles right triangle with equal sides 1 unit is √2 units long.

Why this argument is needed: the length itself is not a whole number or a fraction,
so no counting or measuring will produce it exactly. The area, however, is a whole
number — 2 — and the length is recovered from it as the number whose square is 2.

In-text Questions — Page 38

Page 12 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

2.3 Decimal Representation of √2

Q1 What is the value of √2?

√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 = 1.41421356…

The chapter finds it by trapping it between bounds that get closer and closer — first between
1 and 2, then between 1.4 and 1.5, then 1.41 and 1.42, then 1.414 and 1.415, and so on for ever.

Did you know? The Śulba-Sūtra itself gives a very sharp approximation, 1 + 1/3 +
1/(3×4) – 1/(3×4×34) = 1.4142156…, which is correct to five decimal places.

Q2 Is √2 less than or greater than 1?

√2 is greater than 1.

A square of side 1 has area 1 sq. unit
A square of side √2 has area 2 sq. units

2 > 1, so the second square is bigger, so its side is longer

In other words, 1² = 1 and (√2)² = 2

Therefore 1 < √2

Why comparing squares is enough: for positive lengths, a bigger square must
come from a longer side. So comparing the two areas settles which side is longer,
and areas here are ordinary whole numbers we can compare at once.

Page 13 of 47

as e
a
Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem
g l AglaSem · NCERT Solutions

co m
m.
Is √2 less than or greater than 2?
e
Q3

m l as
.co a g
a s em
a islless than 2.
√2 g

co m
. ag
A square of side 2 has area 2 × 2 = 4 sq. units
em
A square of side √2 has area 2 sq. units
g l as
a
2 < 4, so the side √2 is shorter than the side 2

co m
m.
In other words, (√2)² = 2 and 2² = 4

m as e
.co
Therefore √2 < 2
a g l
a s em
a l the two results together:
gPutting
m a s
.co agl
1 < √2 < 2

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

co m
m .
m as e
.co l
Can we find closer bounds for √2?
g
Q4

em a
a s
agl ANSWER

Yes. Keep squaring decimals one place at a time and see where 2 falls.
se m
com g l a
m . a
ase
TRY SQUARES BOUNDS OBTAINED

One decimal place agl
1.4² = 1.96 and 1.5² = 2.25 1.4 < √2 < 1.5

co m
.
Two decimal places 1.41² = 1.9881 and 1.42² = 2.0164 1.41 < √2 < 1.42

m
e1.414 < √2 < 1.415
m l as
.co g
Three decimal places 1.414² = 1.999396 and 1.415² = 2.002225

a
emEach round pins √2 down ten times more tightly. There is no stage at which the process ends, so
a s
agl
.c
the bounds can be made as close as we like — but they never meet at a decimal that stops.

s e m
m a
. co
Check it yourself: 1.4142² = 1.99996164 and 1.4143² = 2.00024449, so 1.4142 < √2 <
e m agl
1.4143.
g l as
a

co m
m .
m ase
.co


a g l Page 14 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Q5 Will we ever get a number with a terminating decimal representation whose square
is 2?

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).
When you square such a number, the last digit of the answer is the last digit of (last digit)²,
and that is never 0. For example, if the number ended in 4, the square ends in 6; if it ended in
3, the square ends in 9.
So the square also ends in a non-zero digit after the decimal point.

But 2 = 2.000… has no non-zero digit after the decimal point

So no terminating decimal can have square 2

Hence the decimal expansion of √2 goes on for ever

Why it happens: if the decimal has k digits after the point, its square has exactly 2k
digits after the point, and the very last of them is fixed by the last digit of the original
— it cannot be 0. A number ending in 4 gives …6, in 1 gives …1, in 5 gives …5, and so
on. None of these is 0.

In-text Questions — Page 39
2.3 Decimal Representation of √2

TRY THIS

Q1 Can √2 be expressed as a fraction m/n, where m and n are counting numbers?

No. Suppose it could. Then

√2 = m/n

Squaring both sides: 2 = m²/n²

So 2n² = m²

Page 15 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Now count how many times the prime 2 appears on each side. In the prime factorisation of any
square number, every prime occurs an even number of times.

On the right, m² is a square, so 2 occurs an even number of times.
On the left, n² is a square, so 2 occurs an even number of times in it; the extra factor 2 in
front makes the total odd.

Left side: 2 occurs an odd number of times

Right side: 2 occurs an even number of times

The same number cannot have two different factorisations ⇒ impossible

So no fraction m/n of counting numbers can equal √2. Together with the previous result, √2 is
neither a terminating decimal nor a fraction — yet it is a perfectly definite length, the diagonal
of a unit square.

Did you know? This proof is due to Euclid, in his book Elements (c. 300 BCE).

Figure it Out — Pages 39–40
2.3 Hypotenuse of an Isosceles Right Triangle

MATH TALK

Q1 Earlier, we saw a method to create a square with double the area of a given square
paper. There is another method to do this in which two identical square papers are
cut in the following way (each square cut along a diagonal into pieces 1, 2 and 3, 4).
Can you arrange these pieces to create a square with double the area of either
square?

Yes. Each square is cut along one diagonal, so the four pieces are four congruent isosceles right
triangles with equal sides s and longest side s√2.
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.

Page 16 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Area of each piece = ½ × s × s = s²/2

Total area of the four pieces = 4 × s²/2 = 2s²

Each boundary edge = s√2, so the figure is a square of side s√2

Its area = (s√2)² = 2s² = double the area of either square

Why the boundary really is a square: at the centre the four right angles fill 4 × 90°
= 360°, so the pieces close up with no gap. The four longest edges are equal, and at
each outer vertex two 45° base angles meet, giving 45° + 45° = 90°. Four equal sides
and four right angles — a square. The two lines joining opposite outer corners are
its diagonals.

1 2
2 4
→
1 3
3 4

Two squares, one diagonal cut each
Double the area

The four right-angle corners meet at the centre; the four hypotenuses become the sides of the new
square.

Page 17 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Q2 The length of the two equal sides of an isosceles right triangle is given. Find the
length of the hypotenuse. Find bounds on the length of the hypotenuse such that
they have at least one digit after the decimal point. (i) 3 (ii) 4 (iii) 6 (iv) 8 (v) 9

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

c² = 2a², so c = √(2a²)

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

EQUAL SIDES C² = HYPOTENUSE SQUARES OF THE BOUNDS
A 2A² C BOUNDS

(i) 3 18 √18 ≈ 4.243 4.2² = 17.64, 4.3² = 18.49 4.2 < √18 < 4.3

(ii) 4 32 √32 ≈ 5.657 5.6² = 31.36, 5.7² = 32.49 5.6 < √32 < 5.7

(iii) 6 72 √72 ≈ 8.485 8.4² = 70.56, 8.5² = 72.25 8.4 < √72 < 8.5

(iv) 8 128 √128 ≈ 11.314 11.3² = 127.69, 11.4² = 129.96 11.3 < √128 <
11.4

(v) 9 162 √162 ≈ 12.728 12.7² = 161.29, 12.8² = 163.84 12.7 < √162 <
12.8

Tip: every one of these hypotenuses is a√2 — for example √18 = 3√2 and √162 = 9√2.
Since √2 is not a fraction, none of these five lengths is a whole number or a fraction;
each is irrational.

Q3 The hypotenuse of an isosceles right triangle is 10. What are its other two
sidelengths? [Hint: Find the area of the square composed of two such right
triangles.]

Each of the other two sides is √50 = 5√2 ≈ 7.07 units.
Two such triangles make a square whose diagonal is the hypotenuse 10, so use c² = 2a²
backwards:

Page 18 of 47

as e
a
Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem
g l AglaSem · NCERT Solutions

co m
e m.
c² = 2a²
m l as
10² = 2a²
m .co a g
l a se
g
100 = 2a²
aa² = 50

com
. ag
a = √50 = 5√2

se m
l a
ag49 and 7.1² = 50.41, so 7.0 < √50 < 7.1.
Bounds to one decimal place: 7.0² =

. com
Why the answer is not a whole number: if a were a whole number, a² = 50 would
m and since
make 50 a perfect square, which it is not (7² = 49, 8² = 64). In fact a =e5√2,
s
co m gl a
. a
em
√2 is irrational so is 5√2.

a s
a gl
m a s
.co agl
In-text Questions — Page 41
2.4 Combining Two Different Squares
se m
g l a
a
Q1 What if we wish to combine two squares of ‘different’ sizes to make a large square

co m
.
whose area is the sum of the areas of the two smaller squares?

e m
m l as
.co a g
em

l a s
Baudhāyana answers this in Verse 1.12 of the Śulba-Sūtra:
ag
se m
com l a
“The area of the square produced by the diagonal is the sum of the areas of the squares
. a g
m
ase
produced by the two sides.”

agl
The construction: make a right-angled triangle whose two perpendicular sides are the sides of

co m
the two given squares. The square built on its hypotenuse has area equal to the sum of the two

m .
as e
given areas.
m l
.co
m Squares of sides a and b, areas a² and b² a g
l a se
ag Right triangle with perpendicular sides a and b, hypotenuse c
.c
s e m
m a
co agl
Then a² + b² = c²
m .
as e
a g l

co m
m .
m ase
.co


a g l Page 19 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

When the two squares happen to be the same size, the right triangle is isosceles and c² = 2a² —
the doubling rule of Section 2.1. So this new rule contains the old one as a special case.

In-text Questions — Page 42
2.4 Combining Two Different Squares

Q1 Why does Baudhāyana’s method work? Can you see why the method works in the
case where the two squares are the same size? Does it agree with the method we
used earlier to combine two same sized squares into a bigger square?

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.

By the earlier rule: square on the diagonal = 2 × a²

By the new rule: c² = a² + a² = 2a²

The two agree: same construction, same answer

Why the general method works: Baudhāyana explains it in Verse 2.1 — mark off a
rectangular strip of the larger square using a side of the smaller one, and draw the
strip's diagonal. That diagonal is the hypotenuse of a right triangle with
perpendicular sides a and b. Four copies of this triangle, arranged around a small
square of side b – a, tile the square built on that diagonal — and the very same four
triangles plus the small square also tile the two original squares put together. Same
pieces, two arrangements, so the same total area.

In-text Questions — Page 44

Page 20 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

2.4 Combining Two Different Squares

Q1 The 4-sided figure obtained (T + U + V) is in fact a square with an area equal to the
sum of the areas of the two smaller squares! Why?

Because the four right triangles T, U, X and W drawn in the figure are all congruent — each has
perpendicular sides a and b — so the four sides of the new figure are all the hypotenuse of that
triangle, and therefore equal. Once we also check the angles (next question), the figure is a
square.
For the area, look at the same region counted two ways:

Area of the new square = area of T + area of U + area of V

T is congruent to X and U is congruent to W, so
= area of X + area of W + area of V

But X + W + V is exactly the two given squares joined together

So area of the new square = a² + b²

Why it happens: nothing is added or removed — the pieces T and U are just slid into
the positions X and W. Cutting a shape up and rearranging the pieces never changes
the total area, so the square on the hypotenuse must have the same area as the two
original squares combined.

Q2 Explain why all the angles of this new 4-sided figure are right angles and so it is a
square.

Let the two acute angles of the right triangle with sides a and b be x and 90 – x. In a triangle the
three angles add to 180°, so

x + (90 – x) + 90 = 180 ✓

At each vertex of the new figure, one corner of a triangle meets one corner of the next triangle,
and the two of them stand on a straight line together with the angle of the figure. The two
triangle angles that meet there are x and 90 – x — one from each triangle, because consecutive
triangles are turned by a quarter turn.

Page 21 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

x + ∠(of the figure) + (90 – x) = 180°

∠(of the figure) = 180° – 90° = 90°

This happens at all four vertices. Since we already know the four sides are equal (all are
hypotenuses of congruent triangles), the figure has four equal sides and four right angles —
it is a square, and its side is the hypotenuse of the right triangle with perpendicular sides a and
b.

Tip: notice how the labels x, x, x, x and 90 – x, 90 – x, 90 – x, 90 – x are placed in the
book's figure — every straight edge carries one of each, which is exactly the pairing
used above.

In-text Questions — Pages 45–46
Combining Two Squares Using Paper

Q1 Cut out and join two different sized squares (of sides a and b). Now make two cuts
to make three pieces. Rearrange the three pieces into a larger square. Now make a
right triangle using the two smaller squares. Draw a square on the hypotenuse.
Cover the square on the hypotenuse using your pieces.

Place the two squares side by side so that they share part of an edge, making an L-shaped
hexagon of total area a² + b².

1. Cut 1 and Cut 2: from the top-left corner of the joined figure, cut to the point on the bottom
edge that is a from the right end; from the bottom-right corner, cut to the point where the
two squares meet. Each cut has length c, and the two cuts are equal and perpendicular.
2. You now have three pieces: two congruent right triangles with perpendicular sides a and b,
and one remaining quadrilateral.
3. Rearrange: rotate each triangle by a quarter turn about the vertex it shares with the
quadrilateral. The three pieces close up into a square of side c.

Area before rearranging = a² + b²

Area after rearranging = c²
The pieces are the same, so a² + b² = c²

Page 22 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Finally, build the right triangle with perpendicular sides a and b and draw the square on its
hypotenuse. The three pieces cover that square exactly — no gap, no overlap. That is
Baudhāyana's theorem, demonstrated with paper:

a² + b² = c²

Did you know? Baudhāyana was the first person in history to state this result in this
general, essentially modern form, around 800 BCE. Pythagoras (c. 500 BCE) studied it
a couple of hundred years later, which is why it is also called the Baudhāyana-
Pythagoras Theorem.

Figure it Out — Page 47
Using Baudhāyana’s Theorem

Q1 If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm, then what is
the length of its hypotenuse? First draw the right-angled triangle with these
sidelengths and measure the hypotenuse, then check your answer using
Baudhāyana’s Theorem.

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

a² + b² = c²

5² + 12² = c²

25 + 144 = c²

169 = c²

c = √169 = 13 cm

Tip: (5, 12, 13) is one of the triples Baudhāyana lists in Verse 1.13 — it is a
Baudhāyana triple, so the answer comes out a whole number.

Page 23 of 47

as e
a
Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem
g l AglaSem · NCERT Solutions

co m
m.
If a right-angled triangle has a short side of length 8 cm and hypotenuse of length
se
Q2

o m l a
17 cm, what is the length of the third side? Again, try drawing the triangle and
g
m .c and then check your answer using Baudhāyana’saTheorem.
measuring,

l a se
ag

m
The third side is 15 cm.
. co ag
em
Here the 17 cm side is the hypotenuse, so it goes on the right of the equation, not the left.

g l as
a² + b² = c²
a

co m
m.
8² + b² = 17²

m as e
.co
64 + b² = 289
a g l
se m
a
b² = 289 – 64 = 225

ag l b = √225 = 15 cm

om a s
e m . c agl
s
Why 17 goes on the right: the hypotenuse is the side opposite the right angle, and
la 17 with the 8 on the left would give a triangle
it is always the longest side.gPutting
a
with hypotenuse √353 ≈ 18.8 — a different triangle altogether.

co m
m .
o m l a se
Q3 .cUsing the constructions you have now seen, how woulda gyou construct a square
e m
aglas whose area is triple the area of a given square? Five times the area of a given
square? (Baudhāyana’s Śulba-Sūtra, Verse 1.10)

se m
com g l a
m . a
ase
agl
Keep combining squares with Baudhāyana's rule, one at a time. Let the given square have side 1
and area 1.

m
Triple the area.

. co
em
1. First double it: build the square on the diagonal. Its side is √2 and its area is 2.
m l as
.c+o1 = 3. g
2. Now make a right triangle with perpendicular sides √2 and 1. Its hypotenuse c satisfies c² = 2

em a
a s
agl 3. Build the square on that hypotenuse. Its area is 3.

.c
Five times the area. Carry on in the same spiral.
s e m
m a
e m . co agl
l as
√3 with 1 → c² = 3 + 1 = 4, side 2, area 4
g
a
2 with 1 → c² = 4 + 1 = 5, side √5, area 5

com
m .
m ase
.co


a g l Page 24 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

A shortcut for five: a right triangle with perpendicular sides 2 and 1 gives 4 + 1 = 5 straight
away, and the side 2 is just twice the given side.

1

√4 = 2 √3
√2
1

O 1

Each step adds a perpendicular side of length 1, so the square on the new hypotenuse from O has
one more unit of area than the last: 2, 3, 4, 5, … in turn.

Q4 Let a, b and c denote the length of the sides of a right triangle, with c being the
length of the hypotenuse. Find the missing sidelength in each of the following
cases: (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

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

Page 25 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

CASE WORKING MISSING SIDE BOUNDS

(i) a = 5, b = 7 c² = 25 + 49 = 74 c = √74 ≈ 8.602 8.6² = 73.96, 8.7² = 75.69 ⇒ 8.6 < c < 8.7

(ii) a = 8, b = 12 c² = 64 + 144 = 208 c = √208 = 4√13 ≈ 14.4² = 207.36, 14.5² = 210.25 ⇒ 14.4 < c
14.422 < 14.5

(iii) a = 9, c = 15 b² = 225 – 81 = 144 b = 12 (exact) whole number

(iv) a = 7, b = 12 c² = 49 + 144 = 193 c = √193 ≈ 13.892 13.8² = 190.44, 13.9² = 193.21 ⇒ 13.8 < c
< 13.9

(v) a = 1.5, b = c² = 2.25 + 12.25 = c = √14.5 ≈ 3.808 3.8² = 14.44, 3.9² = 15.21 ⇒ 3.8 < c < 3.9
3.5 14.5

Tip: only case (iii) gives a whole number, because (9, 12, 15) is a Baudhāyana triple —
it is 3 × (3, 4, 5). The others are irrational lengths, perfectly good as lengths even
though no decimal writes them exactly.

In-text Questions — Page 48
2.5 Right-Triangles Having Integer Sidelengths

MATH TALK

Q1 List down all the Baudhāyana triples with numbers less than or equal to 20.

Searching every pair a ≤ b ≤ 20 and keeping those for which a² + b² is a perfect square not bigger
than 20 gives exactly six triples.

TRIPLE CHECK PRIMITIVE?

(3, 4, 5) 9 + 16 = 25 Yes

(5, 12, 13) 25 + 144 = 169 Yes

(6, 8, 10) 36 + 64 = 100 No — 2 × (3, 4, 5)

(8, 15, 17) 64 + 225 = 289 Yes

(9, 12, 15) 81 + 144 = 225 No — 3 × (3, 4, 5)

(12, 16, 20) 144 + 256 = 400 No — 4 × (3, 4, 5)

Page 26 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Tip: the book's own list on this page names four of them — (3, 4, 5), (6, 8, 10), (9, 12,
15), (12, 16, 20) — the ones that make the scaling pattern visible. (5, 12, 13) and (8,
15, 17) belong to the list too, and both are primitive.

Q2 Is there an unending sequence of Baudhāyana triples?

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

(3, 4, 5) is a Baudhāyana triple

So are (6, 8, 10), (9, 12, 15), (12, 16, 20), (15, 20, 25), …
In general (3k, 4k, 5k) for every positive integer k

There is no largest k, so the list never ends

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

Q3 Is (30, 40, 50) a Baudhāyana triple?

Yes.

30² + 40² = 900 + 1600 = 2500

50² = 2500

So 30² + 40² = 50² ✓

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.

Page 27 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Q4 Is (300, 400, 500) a Baudhāyana triple?

Yes — it is (3, 4, 5) scaled by 100.

300² + 400² = 90000 + 160000 = 250000

500² = 250000 ✓

Why scaling always works: multiplying every side by k multiplies every area by k²,
and the relation a² + b² = c² is a statement about areas. Multiplying both sides of a
true equation by the same number k² keeps it true.

Q5 Do you see any pattern among them? [The list (3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16,
20)]

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

TRIPLE AS A MULTIPLE OF (3, 4, 5)

(3, 4, 5) k=1

(6, 8, 10) k=2

(9, 12, 15) k=3

(12, 16, 20) k=4

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.

Q6 Can we form a conjecture on Baudhāyana triples based on this observation?
[Conjecture: (3k, 4k, 5k) is a Baudhāyana triple, where k is any positive integer.] Is
this true?

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

Page 28 of 47

as e
a
Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem
g l AglaSem · NCERT Solutions

co m
e m.
(3k)² + (4k)² = 9k² + 16k²
m l as
= 25k²
m .co a g
l a se
g
= (5k)² ✓
a

com
So (3k, 4k, 5k) satisfies a² + b² = c² for every positive integer k. Since there are infinitely many

m .
values of k, this alone shows there are infinitely many Baudhāyana triples.
e ag
g l as
a
Why algebra is needed here: checking (6, 8, 10), (9, 12, 15) and (12, 16, 20) one by
one can never cover all values of k — there are infinitely many. Working with the
co m
m.
letter k handles them all in one go.

m as e
.co a g l
a s em
a glQ7 Can we further generalise the conjecture?
m a s
m.co agl

l a se
Yes. Nothing in the proof used the particular numbers 3, 4 and 5. The general statement is:
a g
m
If (a, b, c) is a Baudhāyana triple, then (ka, kb, kc) is also a Baudhāyana triple, for every
. co
e m
as
positive integer k.
m l
.co a g
a s emexample, starting from (5, 12, 13) we get (10, 24, 26), (15, 36, 39), (20, 48, 52), … — and (15,
gl 36, 39) is one of the triples Baudhāyana himself lists in Verse 1.13.
For
a
se m
com g l a
se
In-text Questions — Pagea49
m. a
agl
2.5 Right-Triangles Having Integer Sidelengths

co m
m .
e
If (a, b, c) is a Baudhāyana triple, then (ka, kb, kc) is also a Baudhāyana triple where

mk is any positive integer. Is this statement true? glas
Q1

. c o a
a em
s ANSWER
agl c
m .
e
True. We are given a² + b² = c², and we must check whether (ka)² + (kb)² = (kc)².

m a s
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

(ka)² = ka × ka = k²a², and (kb)² = kb × kb = k²b²

So (ka)² + (kb)² = k²a² + k²b²

Taking out the common factor: = k²(a² + b²)
Since a² + b² = c², this is = k²c²

= (kc)² ✓

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

Why taking out k² is the key step: it turns the new sum into the old sum multiplied
by a number. The old sum is already known to be c², so the new one is forced to be
k²c² — and that is exactly the square of kc.

Q2 Is (5, 12, 13) a primitive Baudhāyana triple? What are the other primitive
Baudhāyana triples with numbers less than or equal to 20?

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

5 = 5, 12 = 2 × 2 × 3, 13 = 13

Common factor = 1 only ⇒ primitive

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:

TRIPLE COMMON FACTOR PRIMITIVE?

(3, 4, 5) 1 Yes

(5, 12, 13) 1 Yes

(8, 15, 17) 1 Yes

(6, 8, 10) 2 No

(9, 12, 15) 3 No

(12, 16, 20) 4 No

Page 30 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

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

Q3 Generate 5 scaled versions of each of these primitive triples. Are these scaled
versions primitive?

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

PRIMITIVE TRIPLE K=2 K=3 K=4 K=5 K=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 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.

Tip: (15, 36, 39) and (12, 35, 37) both appear in Baudhāyana's own list in Verse 1.13.
The first is 3 × (5, 12, 13) and so is not primitive; the second has common factor 1
and is primitive.

Q4 If (a, b, c) is non-primitive, and the integers have f — greater than 1 — as a common
factor, then is (a/f, b/f, c/f) a Baudhāyana triple? Check this statement for (9, 12, 15).
Justify this statement.

Yes, it is. Dividing out a common factor takes you back down to a smaller triple.
Check for (9, 12, 15). Here f = 3.

(9/3, 12/3, 15/3) = (3, 4, 5)

3² + 4² = 9 + 16 = 25 = 5² ✓ a Baudhāyana triple

Justification in general. Write a/f = p, b/f = q, c/f = r. Since f divides all three, p, q, r are whole
numbers, and a = fp, b = fq, c = fr.

Page 31 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

a² + b² = c²

(fp)² + (fq)² = (fr)²

f²p² + f²q² = f²r²
f²(p² + q²) = f²r²

Dividing both sides by f² (which is not 0):

p² + q² = r² ✓

Why this matters: it says every non-primitive triple sits above a smaller one. Keep
dividing by common factors and you must stop, since the numbers shrink each time
— and where you stop is a primitive triple. That is why every Baudhāyana triple is a
scaled version of a primitive one, and why finding all the primitive triples would find
them all.

Q5 How do we generate more primitive triples?

Use the old pattern that the sum of the first n odd numbers is n², and look for an odd number
in that list which is itself a square.

1 = 1²

1 + 3 = 2²
1 + 3 + 5 = 3²

1 + 3 + 5 + … + (2n – 3) + (2n – 1) = 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

(n – 1)² + (2n – 1) = n²

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).

Page 32 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Q6 For this, we need to know the nth odd number. What is it?

The nth odd number is 2n – 1.

N 1 2 3 4 5 6

2N – 1 1 3 5 7 9 11

Why 2n – 1: the nth even number is 2n, and every odd number is one less than the
even number just above it. So counting odd numbers gives 2n – 1.

In-text Questions — Page 50
2.5 Right-Triangles Having Integer Sidelengths

Q1 What is the sum of the first (n – 1) odd numbers?

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

1 + 3 + 5 + … + (2n – 3) + (2n – 1) = n²

(n – 1)² + (2n – 1) = n²

Check it yourself: the same identity drops straight out of algebra — (n – 1)² = n² – 2n
+ 1, and adding 2n – 1 gives n² – 2n + 1 + 2n – 1 = n².

Q2 Could we have obtained this triple using the equation (n – 1)² + (2n – 1) = n²? [for the
odd square 9, which is the 5th odd number]

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

Page 33 of 47

as e
a
Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem
g l AglaSem · NCERT Solutions

co m
e m.
(n – 1)² + (2n – 1) = n²
m l as
.co
(5 – 1)² + 9 = 5²
m a g
l a se
g
4² + 3² = 5² (because 9 = 3²)
a16 + 9 = 25 ✓

. c om ag
s e
The same works for 25, the 13th odd number m(25 = 2 × 13 – 1), with n = 13:
a
agl
(13 – 1)² + 25 = 13²

co m
m.
12² + 5² = 13²

m as e
.co
144 + 25 = 169 ✓
a g l
se m
g l a
a Why the method works: the identity is true for every n. It becomes a triple only

m a s
agl
when the odd number 2n – 1 is itself a square, because then all three quantities in

m.co
se
the equation are squares of whole numbers.

g l a
a
Figure it Out — Page 50
co m
m .
as e
com
2.5 Right-Triangles Having Integer Sidelengths

. a g l
s em
MATH TALK

gl a
a
Q1 Find 5 more Baudhāyana triples using this idea.
se m
com g l a
m . a
ase
agl

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).

. c om
ODD SQUARE M² N = (M² + 1)/2 TRIPLE (M, N – 1, N)
se mCHECK
m l a
m .49co= 7² ag
ase
25 (7, 24, 25) 49 + 576 = 625 = 25²

agl c
.
81 = 9² 41 (9, 40, 41) 81 + 1600 = 1681 = 41²

s e m
m a
.co agl
121 = 11² 61 (11, 60, 61) 121 + 3600 = 3721 = 61²

se m
169 = 13²
g
85
l a (13, 84, 85) 169 + 7056 = 7225 = 85²

225 = 15²
a113 (15, 112, 113) 225 + 12544 = 12769 = 113²

co m
m .
m ase
.co


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Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

The recipe never runs out — 17² = 289 gives (17, 144, 145), 19² = 361 gives (19, 180, 181), and so
on.

Tip: (7, 24, 25) is on Baudhāyana's own list in Verse 1.13, so this method rediscovers
triples he already knew.

Q2 Does this method yield non-primitive Baudhāyana triples? [Hint: Observe that
among the triples generated, one of the smaller sidelengths is one less than the
hypotenuse.]

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.

Suppose some number f > 1 divided all three

Then f divides n and f divides n – 1

So f divides their difference, n – (n – 1) = 1

But no number greater than 1 divides 1 — contradiction

So the only common factor is 1, and the triple is primitive. You can see it in the list: (3, 4, 5), (5,
12, 13), (7, 24, 25), (9, 40, 41), (11, 60, 61) — every one has no common factor.

Q3 Are there primitive triples that cannot be obtained through this method? If yes,
give examples.

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.

Page 35 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

PRIMITIVE TRIPLE CHECK C–B FROM THIS METHOD?

(8, 15, 17) 64 + 225 = 289 17 – 15 = 2 No

(12, 35, 37) 144 + 1225 = 1369 37 – 35 = 2 No

(20, 21, 29) 400 + 441 = 841 29 – 21 = 8 No

(9, 40, 41) 81 + 1600 = 1681 41 – 40 = 1 Yes

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.

Why the method misses them: it builds a triple out of an odd square that is one of
the odd numbers in the sum 1 + 3 + 5 + …. That forces the other two numbers to be n
– 1 and n. A triple like (8, 15, 17) has an even smallest side, so it can never be the odd
square in that sum.

Figure it Out — Pages 52–54
2.7 Further Applications of the Baudhāyana-Pythagoras Theorem

MATH TALK TRY THIS

Q1 Find the diagonal of a square with sidelength 5 cm.

The diagonal is √50 = 5√2 ≈ 7.07 cm.
A diagonal cuts the square into two isosceles right triangles with equal sides 5 cm, so the
diagonal is their hypotenuse.

c² = 5² + 5²

c² = 25 + 25 = 50

c = √50 = 5√2

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

Tip: the diagonal of any square is its side multiplied by √2. Since √2 is irrational, a
square with a whole-number side never has a whole-number diagonal.

Page 36 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Q2 Find the missing sidelengths in the following right triangles: (i) legs 7 and 9; (ii) legs
4 and 10; (iii) leg 40 with hypotenuse 41; (iv) leg 10 with hypotenuse √200; (v) legs 10
and √150; (vi) leg 27 with hypotenuse 45.

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.

TRIANGLE GIVEN WORKING MISSING SIDE

(i) legs 7 and 9 c² = 49 + 81 = 130 √130 ≈ 11.40 (11.4 < c < 11.5)

(ii) legs 4 and 10 c² = 16 + 100 = 116 √116 = 2√29 ≈ 10.77 (10.7 < c < 10.8)

(iii) leg 40, hypotenuse 41 b² = 1681 – 1600 = 81 9 (exact)

(iv) leg 10, hypotenuse √200 b² = 200 – 100 = 100 10 (exact)

(v) legs 10 and √150 c² = 100 + 150 = 250 √250 = 5√10 ≈ 15.81 (15.8 < c < 15.9)

(vi) leg 27, hypotenuse 45 b² = 2025 – 729 = 1296 36 (exact)

Tip: triangle (iv) has legs 10 and 10, so it is an isosceles right triangle — its
hypotenuse √200 = 10√2 fits c² = 2a² exactly. Triangle (iii) is (9, 40, 41) and triangle (vi)
is 9 × (3, 4, 5) = (27, 36, 45); both are Baudhāyana triples, which is why those answers
are whole numbers.

Q3 Find the sidelength of a rhombus whose diagonals are of length 24 units and 70
units.

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.

Page 37 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Half-diagonals: 24 ÷ 2 = 12 and 70 ÷ 2 = 35

s² = 12² + 35²

s² = 144 + 1225 = 1369
s = √1369 = 37

Why the half-diagonals are perpendicular: a rhombus has all four sides equal, so
each diagonal splits it into two congruent isosceles triangles. In an isosceles triangle
the line from the apex to the midpoint of the base is perpendicular to the base —
and that line is the other diagonal.

Did you know? (12, 35, 37) is one of the triples Baudhāyana lists in Verse 1.13, and it
is primitive.

Q4 Is the hypotenuse the longest side of a right triangle? Justify your answer.

Yes, always. The theorem itself proves it.

a² + b² = c²

Since b² > 0, we get c² = a² + b² > a², so c > a

Since a² > 0, we get c² = a² + b² > b², so c > b

Therefore c is longer than both a and b

Why it also makes sense from the angles: the right angle is 90°, and the other two
angles add to 90°, so each of them is less than 90°. The right angle is therefore the
largest angle in the triangle, and the largest angle always faces the longest side. The
side facing the right angle is the hypotenuse.

Page 38 of 47

as e
a
Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem
g l AglaSem · NCERT Solutions

co m
m.
True or False — Every Baudhāyana triple is either a primitive triple or a scaled
e
Q5

m l as
.co
version of a primitive triple.

a g
se m
g l a
a

True.

om ag
Take any Baudhāyana triple (a, b, c). Let f be the largest number that divides all three.
. c
If f = 1, the triple has no common factor
a s em
greater than 1, so it is already primitive.

aglBaudhāyana triple (shown on page 49), and it has no
If f > 1, then (a/f, b/f, c/f) is again a
common factor left, so it is primitive. Multiplying it back by f returns (a, b, c) — a scaled
version of a primitive triple.
co m
em.
m l as
.co
(9, 12, 15): f = 3 ⇒ 3 × (3, 4, 5) — scaled
a g
a s em72, 78): f = 6 ⇒ 6 × (5, 12, 13) — scaled
gl
(30,
a (8, 15, 17): f = 1 — primitive

om a s
e
. c
m so the statement is true. agl
a s
Every triple falls into one of the two cases,

agl

co m
.
Give 5 examples of rectangles whose sidelengths and diagonals are all integers.
m
Q6

o m l a se
.c a g
m
sediagonal of a rectangle splits it into two right triangles with the sides as perpendicular sides.

g l a
a
A
So a rectangle has an integer diagonal exactly when its two sides form a Baudhāyana triple

se m
com l a
with it.
. a g
m
ase
agl
RECTANGLE (LENGTH × BREADTH) DIAGONAL CHECK

4×3 5 16 + 9 = 25

. om+ 25 = 169
c144
em
12 × 5 13

m l as
.co a g
em 24 × 7
15 × 8 17 225 + 64 = 289

a s
agl 25 576 + 49 = 625

.c
s e m
m a
35 × 12 37 1225 + 144 = 1369

e m . co agl
g l as
a

com
m .
m ase
.co


a g l Page 39 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Tip: scaling any of these gives more — 8 × 6 with diagonal 10, 24 × 10 with diagonal
26, and so on. Since there are infinitely many Baudhāyana triples, there are infinitely
many such rectangles.

Q7 Construct a square whose area is equal to the difference of the areas of squares of
sidelengths 5 units and 7 units.

The required area is 7² – 5² = 49 – 25 = 24 sq. units, so the square must have side √24 = 2√6 ≈ 4.9
units.
Baudhāyana's rule is run backwards: instead of building the hypotenuse from the two legs, we
build a leg from the hypotenuse and the other leg.

1. Draw a segment AB of length 5 units.
2. At A, draw a line perpendicular to AB (an ‘east-west’ and ‘north-south’ pair, as in the Śulba-
Sūtra).
3. With B as centre and radius 7 units, draw an arc cutting that perpendicular at C.
4. Then AC is the side of the required square. Build the square on AC.

In right triangle BAC, right-angled at A:

AB² + AC² = BC²

5² + AC² = 7²

AC² = 49 – 25 = 24

Area of the square on AC = 24 sq. units ✓

Page 40 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

C

7
√24
arc of
radius 7

A 5 B

The hypotenuse is the bigger side, 7; one leg is the smaller side, 5; the other leg √24 is the side of the
required square.

Why the bigger length must be the hypotenuse: the hypotenuse is the longest
side, and we want its square to be the sum of the other two. Since 49 = 25 + 24, the 7
belongs on the hypotenuse.

Page 41 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

Q8 (i) Using the dots of a grid as the vertices, can you create a square that has an area
of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq. units, and (d) 5 sq. units? (ii) Suppose the
grid extends indefinitely. What are the possible integer-valued areas of squares you
can create in this manner?

The square dot grid printed with this question (page 53) — each small square is 1 sq.
unit.

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

Page 42 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

AREA HOW POSSIBLE?

(a) 2 sq. units Tilted square joining (1, 0), (2, 1), (1, 2), (0, 1) — side √2 Yes

(b) 3 sq. units Would need side √3, i.e. p² + q² = 3 — no whole numbers do this No

(c) 4 sq. units Ordinary 2 × 2 square Yes

(d) 5 sq. units Tilted square joining (0, 2), (1, 4), (3, 3), (2, 1) — side √5 Yes

area 2 (side √2) area 4 (side 2) area 5 (side √5)

Three lattice squares. There is no way to draw one of area 3, because 3 is not a sum of two square
numbers.

(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:

Possible areas = all numbers of the form p² + q², with p, q whole numbers, not both 0

1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, …

Not possible: 3, 6, 7, 11, 12, 14, 15, 19, 21, 22, 23, 24, …

So the achievable integer areas are exactly the integers that can be written as a sum of two
square numbers.

Check it yourself: 25 works two ways — the 5 × 5 upright square (25 = 25 + 0) and
the tilted one with p = 3, q = 4 (25 = 9 + 16).

Page 43 of 47

as e
a
Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem
g l AglaSem · NCERT Solutions

co m
m.
Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an
e
Q9

m l as
.co
altitude bisects the opposite side. Use this to find the height.]

a g
se m
g l a
a

The area is 9√3 ≈ 15.59 sq. units.

com
ag
Step 1 — the altitude bisects the base. Let the triangle be ABC with AB = BC = CA = 6, and let

m .
e
AD be the altitude from A to BC. Triangles ABD and ACD have AB = AC (given), AD common, and

g l as
∠ADB = ∠ADC = 90°. So they are congruent (RHS), giving BD = DC.
a
BD = DC = 6 ÷ 2 = 3
co m
em.
m l as
.co
Step 2 — the height, by Baudhāyana's theorem in triangle ABD.
m a g
l a se
a g BD² + AD² = AB²

m a s
.co agl
3² + AD² = 6²

se m
a
9 + AD² = 36

AD² = 27 a g l

m
AD = √27 = 3√3
. co
e m
m l as
.co a g
Step 3 — the area.

a s em
agl Area = ½ × base × height
se m
com l a
= ½ × 6 × 3√3
. a g
m
ase
= 9√3 ≈ 15.59 sq. units

a gl
Bounds: area² = (9√3)² = 243, and 15.5² = 240.25, 15.6² = 243.36, so 15.5 < area < 15.6.

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 44 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

A

6 6
3√3

B 3 D 3 C

The altitude AD splits the equilateral triangle into two congruent right triangles with sides 3, 3√3 and
6.

Tip: the same working with side a gives height a√3/2 and area a²√3/4 — a formula
worth remembering.

Puzzle Time — Page 54
Find the Colours!

Q1 There are 3 closed boxes — one containing only red balls, the second containing
only blue balls and the third containing only green balls. The boxes are labelled RED,
BLUE and GREEN such that ‘no’ box has the correct label. We need to find which
label goes with which box. How can this be done if we are allowed to open only one
box?

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.

Page 45 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

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.

LABEL ON BOX CASE 1: BALL DRAWN IS BLUE CASE 2: BALL DRAWN IS GREEN

RED (opened) blue green

BLUE green red

GREEN red blue

Why one box is enough: the condition that every label is wrong is very strong. With
three boxes there are only two ways to place the contents so that no label is correct,
and the two ways disagree about what is in every single box. So looking inside any
one box tells you which of the two arrangements you are in — and therefore tells
you the other two as well.

Tip: if even one label were allowed to be correct, the trick would fail — there would
then be more than two possible arrangements and one look would not be enough.

Chapter at a glance
Doubling a square is done by building a square on its diagonal, not by doubling the side
(that gives 4 times the area). Drawing 'east-west' and 'north-south' lines cuts the original
square into 2 congruent triangles and the new square into 4 of the same triangles.
Reversing the construction halves a square: the square joining the midpoints of the sides
has half the area.
The hypotenuse of an isosceles right triangle with equal sides a satisfies c² = 2a², so c = a√2.
For a = 1 this gives √2, a number that is neither a terminating decimal nor a fraction m/n.
Baudhāyana's general rule (Verse 1.12): the square on the diagonal of a right triangle has
area equal to the sum of the squares on the two perpendicular sides — a² + b² = c².
Integer triples (a, b, c) with a² + b² = c² are Baudhāyana triples. If (a, b, c) is one, so is (ka,
kb, kc); a triple with no common factor above 1 is primitive.

Page 46 of 47

Class 8 Maths Chapter 9 The Baudhāyana-Pythagoras Theorem AglaSem · NCERT Solutions

The identity (n – 1)² + (2n – 1) = n² turns every odd square into a primitive triple. Fermat
asked the same question for higher powers; aⁿ + bⁿ = cⁿ has no positive-integer solution for
n > 2, proved by Andrew Wiles in 1994.

Quick revision

IDEA STATEMENT FORMULA / EXAMPLE WHERE IT
APPEARS

Doubling a square Build a square on the diagonal of the Area of new square = 2 × Śulba-Sūtra 1.9,
given square area of old page 33

Doubling the side Gives 4 times the area, not 2 2s × 2s = 4s² Page 33

Halving a square Join the midpoints of the four sides Inner square = half the Page 36
area

Isosceles right Square on the hypotenuse = 2 × square c² = 2a², c = a√2 Page 40
triangle on a side

Value of √2 Non-terminating, and not any fraction √2 = 1.41421356…, 1.414 Pages 38–39
m/n < √2 < 1.415

Why √2 ≠ m/n 2n² = m² needs the prime 2 an odd and Impossible Euclid, page 39
an even number of times

Baudhāyana's Square on hypotenuse = sum of squares a² + b² = c² Śulba-Sūtra 1.12,
theorem on the other two sides page 46

Baudhāyana triple Positive integers with a² + b² = c² (3, 4, 5), (5, 12, 13), (8, 15, Śulba-Sūtra 1.13,
17) page 48

Scaling a triple Multiplying a triple by k gives another (ka)² + (kb)² = k²(a² + b²) = Page 49
triple (kc)²

Primitive triple No common factor greater than 1 (3, 4, 5) is; (9, 12, 15) is not Page 49

Odd-square Every odd square 2n – 1 gives a triple (n – 1)² + (2n – 1) = n² Pages 49–50
method

Fermat's Last aⁿ + bⁿ = cⁿ has no solution in positive Proved by Andrew Wiles, Page 51
Theorem integers for n > 2 1994

Page 47 of 47