CLASS 8 MATHS · PART 2 · CHAPTER 2 · STUDY NOTES

The Baudhāyana–Pythagoras Theorem

Understand the baudhāyana–pythagoras theorem through short explanations, labelled models and original practice. Read one topic, explain it in your own words, then check.

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Double the area using the diagonal

Doubling a side makes four times the area. A square built on the diagonal doubles the area instead.

One square has area 1; the other has area 2

y (units)

x (units)

Highlight a line or construction

Unit square · Unit square
Square on its diagonal · Square on its diagonal

1Original area 1 · (0.5, 0.5)
2New area 2 · (0, 1.7)

Exact coordinates and equal axis units. The squares overlap to compare their sides; the areas are measured separately. Equal units on both axes.

Read the plotted values

Unit square

x	y
0	0
1	0
1	1
0	1
0	0

Square on its diagonal

x	y
0	0
1	1
0	2
-1	1
0	0

Diagonal: A segment joining opposite vertices of a polygon.

A unit square has area 1. Its diagonal is √2, so a square with that side has area 2.

The exact plotted diamond below is a square: adjacent edges have equal length √2 and are perpendicular.

Take a moment. What do you think?

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Question

What happens to area when every side of a square triples?

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NCERT reference: chapter PDF pages 1, 2, 3.

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Join the side midpoints to halve the area

Four equal corner triangles surround the smaller square.

The midpoint square keeps half

y (units)

x (units)

Highlight a line or construction

Outer square · Side 4, area 16.
Midpoint square · Midpoint square

1Area 8 · (2, 2)
22 · (1, 0)
32 · (0, 1)

The inner square is exact, not a perspective sketch. Corner triangles together take half the outer area. Equal units on both axes.

Read the plotted values

Outer square

x	y
0	0
4	0
4	4
0	4
0	0

Midpoint square

x	y
2	0
4	2
2	4
0	2
2	0

Midpoint: The point halfway between two endpoints.

A square of side 4 has area 16. Joining its midpoints makes a smaller square of area 8.

Each corner triangle has perpendicular sides 2 and 2, so its area is 2. Removing four such triangles removes 8 square units.

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Question

Would halving the side length halve the area?

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NCERT reference: chapter PDF pages 4, 5.

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Equal perpendicular sides give a√2

The hypotenuse is opposite the right angle, not simply the side drawn at the top.

Identify the opposite side

y (units)

x (units)

Highlight a line or construction

Triangle · Legs 4 and 4; diagonal length 4√2.
Right-angle marker · Perpendicular legs meet at the origin.

1Leg 4 · (2, 0)
2Leg 4 · (0, 2)
3Hypotenuse 4√2 · (2, 2)

Equal unit scales make the isosceles right triangle accurate. Equal units on both axes.

Read the plotted values

Triangle

x	y
0	0
4	0
0	4
0	0

Right-angle marker

x	y
0.4	0
0.4	0.4
0	0.4

Hypotenuse: The side opposite the right angle in a right triangle.
Isosceles right triangle: A right triangle with two equal perpendicular sides.

If each leg is a, then c² = a² + a² = 2a². For positive side lengths, c = a√2.

For legs 4 cm and 4 cm, c = 4√2 cm ≈ 5.657 cm. The radical is exact; the decimal is rounded.

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Question

An isosceles right triangle has hypotenuse 10. Find each leg.

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NCERT reference: chapter PDF pages 5, 6, 8, 9.

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Locate √2 between nearby decimals

Compare the squares of positive decimal values with 2.

Zoom in around the positive root

Value (number)

Number-line marks · Only horizontal position represents the number.

11.414 · (1.414, 0)
21.4142 · (1.4142, 0)
31.415 · (1.415, 0)

The middle mark is √2, with displayed digits rounded. The endpoints are exact terminating decimals.

Read the plotted values

Number-line marks

Value	Height
1.414	0
1.4142135623730951	0
1.415	0

Irrational number: A real number that cannot be written as a fraction of integers with nonzero denominator.

1.414² = 1.999396 and 1.415² = 2.002225. Therefore 1.414 < √2 < 1.415.

√2 has a nonterminating, nonrepeating decimal expansion. 1.41421356… is an approximation when only finitely many digits are used.

Let’s look a little closer.

Go deeper: Why a fraction cannot equal √2

Suppose √2 = m/n with positive integers. Then m² = 2n². In a perfect square, every prime factor occurs an even number of times. On the right, the exponent of prime 2 is odd; on the left it is even. This contradiction rules out the fraction.

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Question

Is 1.42 above or below √2?

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NCERT reference: chapter PDF pages 6, 7.

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The three side-squares reveal the theorem

Square each side length, then add the two leg squares.

9 + 16 = 25

y (units)

x (units)

Highlight a line or construction

Right triangle · Legs 3 and 4.
Square on leg 3 · Square on leg 3
Square on leg 4 · Square on leg 4
Square on hypotenuse · Square on hypotenuse

19 · (1.5, -1.5)
216 · (-2, 2)
325 · (3.5, 3.5)

Each square is constructed outward on a triangle side using exact perpendicular coordinates and equal scales. Equal units on both axes.

Read the plotted values

Right triangle

x	y
0	0
3	0
0	4
0	0

Square on leg 3

x	y
0	-3
3	-3
3	0
0	0
0	-3

Square on leg 4

x	y
-4	0
0	0
0	4
-4	4
-4	0

Square on hypotenuse

x	y
3	0
7	3
4	7
0	4
3	0

Baudhāyana–Pythagoras theorem: In a right triangle with legs a and b and hypotenuse c, a² + b² = c².

For a right triangle with legs 3 and 4, the attached squares have areas 9 and 16. Together these equal the hypotenuse-square’s area 25.

This is a relationship of areas, not a rule that 3 + 4 = 5. The right angle is essential.

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Question

Find the hypotenuse when the legs are 6 and 8.

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NCERT reference: chapter PDF pages 9, 10, 11, 12, 13, 14, 15.

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Four triangles leave a square in the middle

Count the large square’s area in two ways to explain a² + b² = c².

Four congruent triangles; one central square

y (units)

x (units)

Highlight a line or construction

Triangle 1 · Triangle 1
Triangle 2 · Triangle 2
Triangle 3 · Triangle 3
Triangle 4 · Triangle 4
Central square · Central square

1c² = 25 · (3.5, 3.5)
2a + b = 7 · (3.5, 7)

For a = 3 and b = 4, the outer area 49 minus four triangle areas 6 each leaves 25. No pieces overlap or leave unintended gaps. Equal units on both axes.

Read the plotted values

Triangle 1

x	y
0	0
3	0
0	4
0	0

Triangle 2

x	y
3	0
7	0
7	3
3	0

Triangle 3

x	y
7	3
7	7
4	7
7	3

Triangle 4

x	y
4	7
0	7
0	4
4	7

Central square

x	y
3	0
7	3
4	7
0	4
3	0

Congruent triangles: Triangles with matching sides and angles, so their areas are equal.

Arrange four right triangles of legs a and b inside a square of side a + b. The remaining central shape is a square of side c.

Outer area is (a + b)². The four triangles total 4 × ab/2 = 2ab. Subtracting gives c² = a² + 2ab + b² − 2ab = a² + b².

1
Find the outer square area: (a + b)².
2
Subtract four triangle areas: 4 × ab/2.
3
Expand and cancel the two ab terms.
4
The remaining square gives c² = a² + b².

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Question

Why is the middle shape a square, rather than just four equal sides?

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NCERT reference: chapter PDF pages 11, 12, 13, 14.

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Choose addition for the hypotenuse; subtraction for a leg

First label which side is the hypotenuse.

Find the side opposite the known leg

y (units)

x (units)

Highlight a line or construction

9–12–15 right triangle · Equal scales and exact coordinates.
Right angle · The two legs are perpendicular.

1Unknown = 12 · (6, 0)
29 · (0, 4.5)
315 · (6, 4.5)

The slanted side is the hypotenuse 15. Subtract its known leg-square to find the other leg-square. Equal units on both axes.

Read the plotted values

9–12–15 right triangle

x	y
0	0
12	0
0	9
0	0

Right angle

x	y
0.7	0
0.7	0.7
0	0.7

Positive root for a length: A geometric length uses the positive square root of its squared value.

For hypotenuse 15 and leg 9, the other leg satisfies b² = 15² − 9² = 144, so b = 12.

For a rectangle 5 by 12, its diagonal is √(25 + 144) = 13. All lengths must use the same unit.

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Question

A right triangle has hypotenuse 13 and one leg 5. Find the other leg.

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NCERT reference: chapter PDF pages 15, 20, 21.

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Integer triples can be scaled without losing the right angle

A triple consists of three positive integers satisfying a² + b² = c².

Scaled triangles keep matching ratios

y (units)

x (units)

Highlight a line or construction

3–4–5 · Small original triangle.
6–8–10 · Every side doubled.

14 · (4, 0)
28 · (8, 0)
33 · (0, 3)
46 · (0, 6)

These are exact similar triangles on equal unit scales. Squared lengths all multiply by four. Equal units on both axes.

Read the plotted values

3–4–5

x	y
0	0
4	0
0	3
0	0

6–8–10

x	y
0	0
8	0
0	6
0	0

Primitive triple: A right-triangle integer triple whose three numbers have no common factor greater than 1.

(3,4,5) is primitive; multiplying all three numbers by 2 gives (6,8,10), which is not primitive.

For an odd integer m ≥ 3, the numbers m, (m² − 1)/2 and (m² + 1)/2 give a triple. This useful family does not generate every primitive triple.

Go deeper: There are infinitely many triples

Any positive integer scale factor k changes a² + b² = c² into (ka)² + (kb)² = (kc)². This already produces infinitely many distinct triples.

The chapter also introduces Fermat’s Last Theorem: for an integer n > 2, xⁿ + yⁿ = zⁿ has no solution in positive integers x, y, z. This does not forbid zero or arbitrary real values, which are outside that statement.

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Question

Generate the triple from m = 7 and check it.

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NCERT reference: chapter PDF pages 16, 17, 18.

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Turn the situation into a right-triangle model

Decide what is fixed and where the right angle comes from.

A fixed-length stem pivots

y (units)

x (units)

Highlight a line or construction

Water surface · Horizontal y = 3.
Initial stem · Length 5, with 2 above water.
Tilted stem · Same length 5.
Horizontal displacement · Distance 4 along water.

1Depth 3 · (0, 1.5)
2Displacement 4 · (2, 3)
3Stem 5 · (2, 1.5)

Exact rigid-stem model: bottom (0,0), water level 3, initial tip (0,5), tilted tip (4,3). The stem neither bends nor changes length. Equal units on both axes.

Read the plotted values

Water surface

x	y
-1	3
5	3

Initial stem

x	y
0	0
0	5

Tilted stem

x	y
0	0
4	3

Horizontal displacement

x	y
0	3
4	3

Mathematical model: A simplified description with assumptions chosen to represent a situation.

Imagine a rigid stem fixed at the lake bottom. Its tip begins 2 units above the water and then touches the water 4 units horizontally away. If the depth is x, the stem length stays x + 2.

The model gives x² + 4² = (x + 2)². Expand: x² + 16 = x² + 4x + 4, so x = 3. A real bending stem may not behave as this rigid straight segment.

1
Name the unknown depth x.
2
Use the fixed stem length x + 2 for the hypotenuse.
3
Write the right-triangle equation.
4
Expand, cancel x² and solve; substitute back to check.

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Question

What is the height of an equilateral triangle of side 8?

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NCERT reference: chapter PDF pages 20, 21, 22.

The Baudhāyana–Pythagoras Theorem Class 8 Notes

Chapter notes

The Baudhāyana–Pythagoras Theorem

Double the area using the diagonal

One square has area 1; the other has area 2

Join the side midpoints to halve the area

The midpoint square keeps half

Equal perpendicular sides give a√2

Identify the opposite side

Locate √2 between nearby decimals

Zoom in around the positive root

The three side-squares reveal the theorem

9 + 16 = 25

Four triangles leave a square in the middle

Four congruent triangles; one central square

Choose addition for the hypotenuse; subtraction for a leg

Find the side opposite the known leg

Integer triples can be scaled without losing the right angle

Scaled triangles keep matching ratios

Turn the situation into a right-triangle model

A fixed-length stem pivots

Now practise The Baudhāyana–Pythagoras Theorem, free