The Baudhayana Pythagoras Theorem Class 8 Notes Maths Part 2 Chapter 2 - #NCSOLVE 📚

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Class 8 Maths Chapter 2 The Baudhayana Pythagoras Theorem Notes

Class 8 The Baudhayana Pythagoras Theorem Notes

→ The Baudhayana-Pythagoras Theorem is one of the most fundamental theorems in geometry. It expresses the relationship among the three sides of a right-angled triangle.

→ If a, b, and c are the sidelengths of a right-angled triangle, where c is the length of the hypotenuse, then a² + b² = c²

→ In an isosceles triangle with sidelengths a, a, c, we have the relation a² + a² = 2a² = c², i.e., c = a√2.

→ The number √2 lies between 1.414 and 1.415. However, it cannot be expressed as a terminating decimal. It also cannot be expressed as a fraction \(\frac {m}{n}\) with m, n positive integers.

→ A triple (a, b, c) of positive integers satisfying a² + b² = c² is called a Baudhayana-Pythagoras triple.

→ Examples include (3, 4, 5), (6, 8, 10), and (5, 12, 13). An infinite number of such triples can be constructed.

→ The equation aⁿ + bⁿ = cⁿ has no solution in positive integers when n > 2. This is known as ‘Fermat’s Last Theorem’. It was proven by Andrew Wiles in 1994.

Baudhayana
Baudhayana was an ancient Indian scholar from around 800 BCE. He was part of the Yajurveda tradition and wrote about rituals, mathematics, and geometry. He is known as an early thinker in ritual science. His most famous work, the Sulba-Sutra, includes important ideas about shapes and measurements.

Sulba-Sutra
This is an ancient Indian text that outlines the rules for constructing ritual fire altars. It contains early ideas of geometry and measurement, including methods for making squares, rectangles, and circles. It was composed roughly between 800 and 200 BCE. Major authors are Baudhayana, Apastamba, Manava, and Katyayana.

Sulba
‘Sulba’ means a cord or rope. Geometry was done using ropes to measure and construct ritual fire altars. The Sulba-Sutras explain how to construct Vedic fire altars of precise shapes and areas required for rituals—squares, rectangles, circles, falcons, etc.

Doubling A Square
Verse 1.9 of the Sulba-Sutra: The diagonal of a square produces a square of double the area of the original square.

We are given a square ABCD. We wish to construct a square whose area is twice the area of square ABCD.

We do the following steps:
Extend BC to F such that BC = CF.
Extend DC to E such that DC = CE.
Join BE, EF, DB, and FD.
Then, DBEF is a square, such that ar. DBEF = 2 × ar. ABCD.

Sequence of Squares
We again start with square ABCD.

We mark the mid-points of the sides as P, Q, R, and S.

Now we join, PQ, QR, RS, SP, PR, and SQ.

Consider the following figures one by one.

Note that all the small triangles are congruent isosceles right triangles.
Square 1 is made up of 2 triangles.
Square 2 is made up of 4 triangles.
Square 3 is made up of 8 triangles.
Thus area of square 2 = 2 × area of square 1,
and, area of square 3 = 2 × area of square 2
Hence, we get a sequence of squares where each square has double the area of the previous one.

Activity: To double the area of a square using paper.
1. Cut out two identical squares of paper.

2. Draw diagonals and number the triangles as shown.

3. Cut one of the squares along the diagonals.

4. Now place the pieces 5, 6, 7, and 8 around square 1 as shown.

We get a square with double the area.

Halving A Square
Again, we start with square ABCD.

We mark the midpoints of the sides as M, N, O, and P.

We join, MN, NO, OP, PM, PN, and OM.

MNOP is made up of 4 small triangles.
ABCD is made up of 8 small triangles.
Hence area of MNOP is half the area of square ABCD.

Activity: To make half of a square using paper.
1. Cut out a square sheet of paper (ABCD).

2. Mark the mid-points of sides as M, N, O, and P as shown.

3. Join MN, NO, OP, and PM.

4. Fold along MN, NO, OP, and PM as shown.

Then MNOP is a square, and the area of MNOP is half the area of square ABCD.

Hypotenuse of An Isosceles Right Triangle
We are given an isosceles right triangle with each equal side of length 1 unit.

We complete the square ABCD.

We observe that square ABCD is made of two isosceles right triangles with each equal side of 1 unit.
We draw a square on the diagonal AC.

We know that the square constructed on the diagonal of this square has twice the area of the original square.

Area of ACEF = 2 × Area of ABCD
= 2 × 1
= 2 sq. units.
If a is the hypotenuse AC, then, Area of ACEF = a × a = a²
a² = 2
⇒ a = √2
Hence, the hypotenuse is of length √2 units.

Decimal Representation of √2
Area of a square with sidelength 1 unit = 1 × 1 = 1 sq. unit.
Area of a square with sidelength √2 units = √2 × √2 = 2 sq. units.
Area square with sidelength 2 units = 2 × 2 = 4 sq. units.
Hence, 1 < √2 < 2
Now, 1.1² = 1.21
1.2² = 1.44
1.3² = 1.69
1.4² = 1.96
1.5² = 2.25
∴ 1.4 < √2 < 1.5
1.41² = 1.9881
1.42² = 2.0164
∴ 1.41 < √2 < 1.42
1.411² = 1.990921
1.412²= 1.993744
1.413² = 1.996569
1.414² = 1.999396
1.415² = 2.002225
∴ 1.414 < √2 < 1.415
This will go on forever.
Hence, √2 has a non-terminating decimal representation.

Can √2 be expressed as a fraction?
Euclid’s proof: Let √2 = \(\frac {p}{q}\)
Squaring both sides, we get,
2 = \(\frac{p^2}{q^2}\)
⇒ 2q² = p²
We know that in the prime factorization of a square number, each prime occurs an even number of times.
Thus, in the equation 2q² = p², the prime 2 would occur an odd number of times on the left side and an even number of times on the right side.
It is not possible.

General Solution
Let ‘a’ be the length of an equal side of an isosceles right triangle, and ‘c’ be the length of its hypotenuse.

Then c² = a² + a² = 2a²
This means c = a
Hence, √2 cannot be expressed as a fraction.
General Solution
Let ‘a’ be the length of an equal side of an isosceles right triangle, and ‘c’ be the length of its hypotenuse.
Then c2 = a² + a² = 2a²
This means c = a√2 or a = \(\frac{c}{\sqrt{2}}\)

Combining Two Different Squares
Verse 1.12 of the Sulba-Sutra: The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides.
We make a right-angled triangle whose perpendicular sides are the sidelengths of the two squares.
The square whose sidelength is the hypotenuse of this right-angled triangle has an area that is the sum of the areas of the original two squares. (Baudhayana’s theorem)

Proof of Baudhayana’s Theorem
1. Join the two squares.

2. Mark a rectangular portion of the larger square using a side of the smaller one, and draw its diagonal c.
By doing this, we get a right triangle with perpendicular sides a and b.

3. Colour this triangle.

4. Draw three more of such right triangles as shown.

5. We get

Triangles P, Q, T, and S are congruent right triangles.
Hence, they have equal hypotenuses.
The four sides of this figure obtained (P + Q + R) are the hypotenuses of the congruent right triangles P, Q, R, and T.
Also, all angles of the four-sided figure are right angles.
Hence, the 4-sided figure obtained is a square.
Area (P + Q + R) = Area (S + R + T)
Or
Area of square = Sum of the areas of the two smaller squares

c² = a² + b²

Activity: Create a new square using two squares of different sizes.
Method 1:
1. Cut out and join two differently sized squares as follows:

2. Make two cuts to make three pieces.

3. Rearrange the three pieces into a larger square as shown.

The two smaller squares have been combined into a larger square!
Area of square (side s) + area of square (side t) = area of square (side u)
s² + t² = u²

Method 2:
1. Make a right triangle using the two smaller squares.
Draw a square on the hypotenuse.

2. Extend DC to meet the side of the square at L.
Draw LM and PQ parallel to AC.

3. Cut out the pieces of the squares and cover the square on the hypotenuse using them.
The pieces cover the square on the hypotenuse completely.

Hence, if a right triangle has shorter sides of length a and b, and a hypotenuse of length c, then the areas of the two smaller squares, a² and b², add up to the area of the larger square, c².
Or a² + b² = c²

Baudhayana’s Theorem on Right-angled Triangles:
If a right-angled triangle has sidelengths a, b, and c, where c is the length of the hypotenuse, then a² + b² = c²

Right-Triangles Having Integer Sidelengths
In Sulba-Sutra (Verse 1.13), Baudhayana lists integer triples (a, b, c) that form the sidelengths of a right triangle and therefore satisfy a² + b² = c²
These include (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (9, 40, 41), (11, 60, 61), (12, 35, 37), and (15, 36, 39).
Such triples (a, b, c) are called Baudhayana triples or Baudhayana-Pythagoras triples or right-angled triangle triples.
1. If (a, b, c) is a Baudhayana triple, then (ka, kb, kc) is also a Baudhayana triple, where k is a counting number greater than 1.
(5, 12, 13) is a Baudhayana triple. Then (10, 24, 26) is also a Baudhayana triple.

2. If (a, b, c) is a Baudhayana triple, then \(\left(\frac{a}{k}, \frac{b}{k}, \frac{c}{k}\right)\) is also a Baudhayana triple, where k is a
counting numbers other than 1.
(7, 24, 25) is a Baudhayana triple. Then, (0.7, 2.4, 2.5) is also a Baudhayana triple.

3. Let ‘n’ be an odd number. Then \(\left(n, \frac{\left(n^2-1\right)}{2}, \frac{\left(n^2+1\right)}{2}\right)\) is a Baudhayana triple.
Let n = 7, then \(\left(7, \frac{\left(7^2-1\right)}{2}, \frac{\left(7^2+1\right)}{2}\right)\) or (7, 24, 25) is a Baudhayana triple.

4. Let ‘n’ be an even number. Then \(\left(n,\left(\frac{n}{2}\right)^2-1,\left(\frac{n}{2}\right)^2+1\right)\) is a Baudhayana triple.
Let n = 4, then 4, \(\left(\frac{4}{2}\right)^2-1,\left(\frac{4}{2}\right)^2+1\) or (4, 3, 5) is a Baudhayana triple.

5. For every counting number greater than 2, we can generate a Baudhayana triple.

6. The number of Baudhayana triples is uncountable.

7. If numbers in a Baudhayana triple are co-prime, it is called a primitive Baudhayana triple.
(3, 4, 5), (5, 12, 13), (7, 24, 25) are primitive Baudhayana triples.

8. If numbers in a Baudhayana triple are not co-prime, it is called a scaled Baudhayana triple.
Baudhayana triples of type (ka, kb, kc) and \(\left(\frac{k}{a}, \frac{k}{b}, \frac{k}{c}\right)\) are called scaled Baudhayana triples.
(6, 8, 10), (1.5, 3.6, 3.9) are scaled Baudhayana triples.

9. For any primitive triple, any number of scaled triples can be generated.

A Long-Standing Open Problem
This problem was proposed by Fermat, a French mathematician of the 17th century.
The problem: Is there a solution to the equation: xn + yn = zn,
where x, y, and z are natural numbers, and n > 2?
He could neither find a solution to the problem nor could he prove that a solution does not exist.
After more than 200 years, in 1994, it was finally proved that the solution does not exist.
Now it is known as ‘Fermat’s Last Theorem’.
Theorem: The equation xn + yn = zn has no solution in positive integers when n > 2.

Further Applications of the Baudhayana-Pythagoras Theorem
The Baudhayana-Pythagoras theorem is applied to find unknown sides of right-angled triangles. It is widely used in construction, navigation, engineering, and physics to calculate distances, heights, and diagonals accurately.

Example: Manisha started from a point A. She went 10 km north and then 24 km East and reached point C. How far is she from the starting point?
Solution:

We know, AC² = AB² + BC²
= 10²+ 24²
= 100 + 576
= 676
= 26²
⇒ AC = 26
Hence, she is 26 km away from the starting point.

The post The Baudhayana Pythagoras Theorem Class 8 Notes Maths Part 2 Chapter 2 appeared first on Learn CBSE.

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