NCERT Solutions for Class 8 Maths Ganith Prakash 2 Chapter 2 The Baudhayana - The Pythagoras Theorem

Chapter 2, The Baudhāyana-Pythagoras Theorem, explores the fascinating relationship between the sides of a right-angled triangle. Named after the ancient Indian mathematician Baudhāyana and the Greek philosopher Pythagoras, this theorem reveals that the square of the hypotenuse is equal to the sum of the squares of the other two sides ( $a^{2} + b^{2} = c^{2}$ ). By studying these geometric principles through historical sulba-sūtras and modern proofs, students develop a deep understanding of spatial relationships and numerical patterns.

Being able to master these things will allow students to approach real-life problems - whether finding out how far to travel based upon speed or figuring out how many people can fit inside an irrational shape. The NCERT Solutions for Class 8 provide students with a way of learning through examples and detailed solutions which had been lacking prior to today. Therefore, these solutions connect both ancient knowledge of mathematics to current education/board exam preparation within the realm of geometry.

1.0Download NCERT Class 8 Maths Ganith Prakash 2 Chapter 2 Solutions

Find out more about finding square areas and how to explore geometric connections using linear proofs with our large solution guide that includes proof methods utilizing Baudhāyana-Pythagorean theorem methods. The NCERT Solutions provided in a step by step manner can be downloaded from the link below:

Chapter 2 : The Baudhayana - The Pythagoras Theorem

2.0Key Concepts covered in Class 8 Maths : The Baudhayana - The Pythagoras Theorem

The chapter transitions from simple square constructions to the foundational rule governing right-angled triangles.

Doubling a Square: Understanding Baudhāyana’s observation that a square constructed on the diagonal of an original square has exactly double the area.
The Theorem Statement: In a right-angled triangle with sides a and b and hypotenuse c, the relationship is defined as $a^{2} + b^{2} = c^{2}$ .
Baudhāyana-Pythagoras Triples: Identifying sets of three positive integers (like 3, 4, 5) that satisfy the theorem.
Geometric Proofs: Visualizing the theorem through area-based proofs, including the "dissection" method where four identical triangles form a larger square.
Irrationality of $2$ : Exploring why the diagonal of a unit square cannot be expressed as a simple fraction or terminating decimal.

3.0NCERT Solutions for Class 8 Maths Chapter 2 : All Exercises

Exercise 2.1

Students have previously examined a geometric principle regarding increasing the square area twice, so they through the creation of their own square shapes at right angles from the diagonal of another square, have provided visual evidence as to the area relationship resulting in preliminary concepts of what squares are in geometry.

Exercise 2.2

The objective of this exercise is for you to become familiar with writing out formal theorem statements. There are several questions that will ask you to identify right triangles, find missing hypotenuse lengths, and discover how the three sides of any triangle relate to one another depending on the type of triangle.

Exercise 2.3

This section focuses on the identification of sets of integers according to the theorem. The student can do exercises that check whether there are already existing sets (for example, (3,4,5)) and use those sets to help them find other kinds of integer sets or groups.

Exercise 2.4

This exercise is all about putting the theorem to work. Students perform calculations using geometric values as they work through real-life examples showing how to use a ladder, find out the distance between telephone poles and how far to walk, in order to calculate their heights and straight-line distance between two objects in physical space.

Exercise 2.5

The focus of this exercise is on irrational numbers involving more advanced theorem proof of this concepts. Students analyze the length of the diagonals of a unit square, determine that the square root of any non-perfect square is a non-terminating decimal as well as etc., and use previously learned methods to prove the theory was proven.

4.0Detailed Class 8 Maths Chapter 2 The Baudhayana - NCERT Solutions

Figure it out-01

1. 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. Can you arrange these pieces to create a square with double the area of either square? Sol. Sliding all four right-angled triangles together, so that their right-angled corners all meet perfectly in the center, pointing inward as shown in the figure.

Thus, we have four identical triangles from two identical squares. Since, area of two triangles $=$ Area of the given square So, area of four triangles = double the area of the given square.

2. 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

Sol. If a is the length of two equal sides of an isosceles right triangle, then hypotenuse (i) $a = 3$

Hypotenuse $= a 2 = 32 = 3 \times 3 \times 2 = 18$ $18$ lies between $16$ and $25$ $16 < 18 < 25$ $\Rightarrow 4 < 32 < 5$ In one decimal point $4. 1^{2} = 16.81, 4. 2^{2} = 17.64, 4. 3^{2} = 18.49$ So, $4.2 < 32 < 4.3$ (ii) $a = 4$

Hypotenuse $= a 2 = 42 = 32$ $25 < 32 < 36$ $\Rightarrow 5 < 42 < 6$ In one decimal point $5. 1^{2} = 26.01, 5. 2^{2} = 27.04, 5. 3^{2} = 28.09$ , $5. 4^{2} = 29.16, 5. 5^{2} = 30.25, 5. 6^{2} = 31.36$ , $5. 7^{2} = 32.49$ So, $5.6 < 42 < 5.7$ (iii) $a = 6$

Hypotenuse $= a 2 = 62 = 6 \times 6 \times 2 = 72$ $64 < 72 < 81$ $\Rightarrow 8 < 62 < 9$ In one decimal point $8. 1^{2} = 65.61, 8. 2^{2} = 67.24, 8. 3^{2} = 68.89$ , $70.5 6^{2} = 8.5, = 72.25$ So, $8.4 < 62 < 8.5$ (iv) $a = 8$

Hypotenuse $= a 2 = 82 = 128$ $121 < 128 < 144$ $\Rightarrow 11 < 82 < 12$ In one decimal point $11. 1^{2} = 123.21, 11. 2^{2} = 125.44, 11. 3^{2} =$ $127.69, 11. 4^{2} = 129.96$ So, $11.3 < 82 < 11.4$ (v) $a = 9$

Hypotenuse $= a 2 = 92 = 162$ $144 < 162 < 169$ $\Rightarrow 12 < 92 < 13$ In one decimal point $12. 1^{2} = 146.41, 12. 2^{2} = 148.84$ , $12. 3^{2} = 151.29, 12. 4^{2} = 153.76$ , $12. 5^{2} = 156.25, 158.76$ , $12. 7^{2} = 161.29, 12. 8^{2} = 163.84$ So, $12.7 < 92 < 12.8$

3. 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.]

Sol.

If $a$ is the length of two equal sides of an isosceles right triangle, then $a^{2} + a^{2} = 1 0^{2}$ $\Rightarrow 2 a^{2} = 100$ $\Rightarrow a^{2} = 50$ $\Rightarrow a = 50$ $\Rightarrow a = 52$

Figure it out-02

1. 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 side lengths and measure the hypotenuse, then check your answer using Baudhāyana's Theorem.

Sol. Steps of construction: Step-1: Draw a line segment $AB = 5 cm$ Step-2: At a point A, draw a perpendicular and then cut an arc of length 12 cm such that AC $= 12 cm$ .

Step-3: Join B and C Measure the length of BC.

By measurement, $BC = 13 cm$ Now, using Baudhāyana's Theorem $AC^{2} + AB^{2} = BC^{2}$ $1 2^{2} + 5^{2} = BC^{2}$ $144 + 25 = BC^{2}$ $169 = BC^{2}$ $BC = 13 cm$ Therefore, the hypotenuse of the rightangled triangle is 13 cm , according to Baudhāyana's Theorem.

2. If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm . What is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhāyana's Theorem.

Sol. Step of construction: Step-1: Draw a line segment $PQ = 8 cm$ Step-2: Draw a line segment line passing through point P .

Step-3: With centre Q and radius 17 cm , draw an arc cutting perpendicular line at R. Join Q & R.

Step-4: Required $△ PQR$ is formed. Measure the length of PR.

By measurement PR=15 cm. Now, using Baudhāyana's theorem $PQ^{2} + PR^{2} = QR^{2}$ $8^{2} + PR^{2} = 1 7^{2}$ $PR^{2} = 289 - 64 = 225 = 1 5^{2}$ $∴ PR = 15 cm$ Therefore, the other side of the rightangled triangle is 15 cm , which satisfied Baudhāyana's theorem.

3. Using the constructions, you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square. (Baudhayāna's Ṡulba-Ṡutra, Verse 1.10)

Sol. (a) Steps of construction Step-1: Draw a square ABCD of side length $= 2 cm$ Then $AC = 22$ cm

Step-2: Draw $AC ⊥ CE$ such that $CE =$ 2 cm

By Baudhāyana Pythagoras theorem, $AE = 12 cm$ .

Step-3: Now, draw a square AEGH of side length $12 cm$ .

Proof: Here, ar (AEGH) = $12 cm^{2}$ $ar (ABCD) = (2)^{2} = 4 cm^{2}$

From (i) and (ii), we conclude, $ar (AEGH) = 3 \times ar (ABCD)$

(b) Steps of construction:

Step-1: Draw two identical square of side lengths 'a' on keeping side by side.

Step-2: Join D and E to make a rightangled triangle ADE.

Step-3: Draw a square on the hypotenuse DE.

The area of square DEGH formed is 5 times the area of square ABCD .

Proof: The area of square $A BC D$ of side length ' a ' $= a^{2}$

In $△ ADE, AD = a$ and $AE = 2 a$ $DE = a^{2} + (2 a)^{2} = 5 a^{2}$ Area of square DEGH $= 5 a^{2}$ $= 5$ times the area of square ABCD .

4. 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 side length 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) $a = 1.5, b = 3.5$

Sol. Here $c^{2} = a^{2} + b^{2}$ (i) Now, $c^{2} = 5^{2} + 7^{2} = 25 + 49 = 74$ $\Rightarrow c = 74$ (ii) Now, $c^{2} = 8^{2} + 1 2^{2} = 64 + 144 = 208$

$\Rightarrow c = 208 = 2 \times 2 \times 2 \times 2 \times 13 = 413$

(iii) Here, $1 5^{2} = 9^{2} + b^{2} = b^{2} = 1 5^{2} + 9^{2}$

$= 225 - 81 = 144 \Rightarrow b = 144$ $= 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 12$

(iv) $c^{2} = 7^{2} + 1 2^{2} = 49 + 144 = 193$ $\Rightarrow c = 193$ (v) $c^{2} = 1. 5^{2} + 3. 5^{2} = 2.25 + 12.25 = 14.5$ $\Rightarrow c = 14.5$

Figure it out-03

1. Find 5 more Baudhāyana triples using this idea.

Sol. (i) $(1 + 3 + 5 + \dots + 47) + 49 = 2 5^{2}$

$2 4^{2} + 7^{2} = 2 5^{2} (24, 7, 25)$

(ii) $(1 + 3 + 5 + \dots + 79) + 81 = 4 1^{2}$

$4 0^{2} + 9^{2} = 4 1^{2} (40, 9, 41)$

(iii) $(1 + 3 + 5 + \dots + 119) + 121 = 6 1^{2}$

$6 0^{2} + 1 1^{2} = 6 1^{2} (60, 11, 61)$

(iv) $(1 + 3 + 5 + \dots + 167) + 169 = 8 5^{2}$

$842 + 132 = 852 (84, 13, 85)$

(v) $(1 + 3 + 5 + \dots + 223) + 225 = 11 3^{2}$

$11 2^{2} + 1 5^{2} = 11 3^{2} (112, 15, 113)$

Does this method yield non-primitive Baudhāyana triples? [Hint: Observe that among the triples generated, one of the smaller side-lengths is one less than the hypotenuse.) Sol. No, it never yields non-primitive triples. Let's look at the generated triplesexample (7,24,25)

The hypotenuse (n) and one of the legs ( $n - 1$ ) are consecutive integers (like 24 and 25).
Consecutive integers never share a common factor other than 1 .
Because two of the three numbers share no common factors, it is impossible for all three to share a common factor. Therefore, the triple is always primitive.

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

Sol. Yes, there are. The method on this page strictly produces triples where the hypotenuse is exactly 1 unit larger than one of the legs (because they are n and $n - 1$ ).

However, there are primitive triples which don't follow this logic Example:

Some examples are (8, 15, 17), (20, 99, 101), (120, 209, 241)

Thus, not all primitive triples can be generated by this method.

Figure it out-04

1. Find the diagonal of a square with side length 5 cm.

Sol.

BD^{2} = 5^{2} + 5^{2} = 25 + 25 = 50

BD^{2} = 50 = 5 \times 5 \times 2

= 52 = 5 \times 1.414 = 7.07 (\sim 7.1)

Hence, the length of the diagonal is

52

( 7.1 cm ) approx.

2. Find the missing side lengths in the following right triangles:

(a)

(b)

(f)

(d)

(e)

Sol. (a) $a^{2} = 7^{2} + 9^{2} = 49 + 81 = 130$

$\Rightarrow a = 130$

(b) $b^{2} = 4^{2} + 1 0^{2} = 16 + 100 = 116$

$\Rightarrow b = \Rightarrow b = 116 2 \times 2 \times 29 = 229 = 116$

(c) $4 0^{2} = c^{2} + 4 1^{2}$ $\Rightarrow 1600 + c^{2} = 1681$ $\Rightarrow c^{2} = 1681 - 1600$ $\Rightarrow c^{2} = 81$ $\Rightarrow c = 81$ $\Rightarrow c = 9$ (d) $d^{2} + 1 0^{2} = (200)^{2}$ $d^{2} + 100 = 200$ $d^{2} = 200 - 100$ $d^{2} = 100$ d = 10 (e) $e^{2} = 1 0^{2} + (150)^{2}$ $\Rightarrow e^{2} = 100 + 150$ $\Rightarrow e^{2} = 250$ $\Rightarrow e = 250$ $\Rightarrow e = 5 \times 5 \times 5 \times 2$ $\Rightarrow d = 510$ (f) $2 7^{2} = f^{2} = 4 5^{2}$ $\Rightarrow 729 = f^{2} = 2025$ $\Rightarrow f^{2} = 2025 - 729$ $\Rightarrow f^{2} = 1296$ $\Rightarrow f = 1296$ $= 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$ $= 2 \times 2 \times 3 \times 3 = 36$

3. Find the side length of a rhombus whose diagonals are of length 24 units and 70 units. Sol.

Since diagonals of Rhombus bisect each other at $9 0^{\circ}$ . $OA = \frac{1}{2} \times 24 = 12$ OD $= \frac{1}{2} \times 70 = 35$ In $△ AOD, a^{2} = 1 2^{2} + 3 5^{2}$ $= 144 + 1225 = 1369$ $a = 1369 = 37$ $∴$ The side of the rhombus is 37 units.

4. Is the hypotenuse the longest side of a right triangle? Justify your answer. Sol. $c^{2} = a^{2} + b^{2}$ $∴ c^{2} > a^{2}$ and $c^{2} > b^{2}$ Or $c > a$ and $c > b$ Hence, 'c' is the longest side of the right triangle.

5. True or False: Every Buadhayana triple is either a primitive triple or a scaled version of a primitive triple. Sol. Let $(a, b, c)$ be a Baudhayana triple. Then $HCF (a, b, c) 1$ or $HCF (a, b, c) \neq = 1$ $(3, 4, 5) : HCF$ is 1 $(10, 24, 26) : HCF = 2$ If HCF is 1 , then the triple is primitive If HCF is a number other than 1 , then the triple is scaled.

6. Give 5 examples of rectangles whose side lengths and diagonals are all integers.
Sol. (i)

(ii)

(iii)

(iv)

(v)

7. Construct a square whose area is equal to the difference of the areas of squares of side-lengths 5 units and 7 units.

Sol. To form a square whose area should be $\Rightarrow 7^{2} - 5^{2} = 49 - 25 = 24$ sq. units. Now, draw a right angles triangle XYZ such that $XY = 5 cm$ and $YZ = 7 cm$ .

Here, $XZ^{2} - XY^{2} = ZY^{2}$

$X Z^{2} = 7^{2} - 5^{2} = 49 - 25 = 24 XZ = 24 cm$

Then draw a square XZUV of side length $24 cm$ as shown.

Hence, the square XZUN is formed whose area is equal to difference of area of squares of side lengths 5 units and 7 units.

8. (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, (d) 5 unit (ii) Suppose the grid extends indefinitely. What are the possible integer-values areas of squares you can create in this manner?

Sol. (i) Since, one side of square formed will be hypotenuse 'c' of a right angled triangle formed by the grid lines.

The legs of this triangle will be 'a' units and 'b' units.

Now, $c^{2} = a^{2} + b^{2}$ Because, the dots are on a grid, a and b must be whole number ( $0, 1, 2, 3, \dots$ ). This means that the area of a square on a dot grid must be equal to sum of two perfect square. (a) Area of 2 square units

Since $2 = (1)^{2} + (1)^{2}$ So, $AP = PD = 1$ units

$\Rightarrow AD = AP^{2} + PD^{2} = 1 - 11 = 2 units$

Area of square $ABCD = 2 sq$ . units

Tilted square ABCD is formed.

(b) Area of 3 sq. units

There is no combination of two perfect square that add up to 3 .

It is impossible to draw a square with an area of 3 sq. units on a standard integer dot grid. (c) Area of 4 sq. units $∵ 4 = (2)^{2} + 0^{2} = 4 + 0 = 4$

Non tilted square ABCD is formed.

(d) Area of 5 sq. units

5 = (2)^{2} + (1)^{2}

Area of square

ABCD = 5 sq

. units Tilted square ABCD is formed.

(ii) If the grid extends indefinitely, the possible integer area are exactly the numbers that can be written as the sum of two perfect squares.

$\Rightarrow \Rightarrow 0^{2} + 1^{2} = 1 1^{2} + 1^{2} = 2 2^{2} + 0^{2} = 4 2^{2} + 1^{2} = 5 2^{2} + 2^{2} = 8 3^{2} + 0 = 9 3^{2} + 1^{2} = 10 3^{2} + 2^{2} = 13 4^{2} + 0^{2} = 16 4^{2} + 1^{2} = 17$

... and so on Here, 3 , 6 , 7 , 11 , 12 , 14 , 15 are shifted as it is impossible to draw a square on a standard integer dot grid.

9. Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.] Sol. Let $△ ABC$ be an equilateral triangle. $AB = BC = CA = 6 cm$

Let AD be perpendicular to BC . Then $∠1 = ∠2$ (each $9 0^{\circ}$ ) $AB = AC$ (each 6 cm ) $AD = AD$ (common) $△ ADB ≅ △ ADC$ (by RHS congruency) $BD = DC$ (CPCT)

$∴ BD = DC = \frac{1}{2} \times 6 cm = 3 cm$

In $△ ADC, h^{2} + 3^{2}$

$\Rightarrow \Rightarrow h = 6^{2} (Baudhayana’s triple) h^{2} = 36 - 9 = 27 = ar (Δ ABC) = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times 6 \times 33 sq. units = 93 sq. units$

5.0Key Features and Benefits of Class 8 Maths Ganith Prakash 2 Chapter 2

Integration of Historical Heritage: The chapter uniquely combines the Indian mathematical tradition (Baudhāyana’s Sulba-Sūtras) with global mathematical history, fostering pride and a broader perspective on geometry.
Strong Visual Foundation: By starting with the physical construction of squares and diagonals, the solutions help students "see" the math before calculating it, which is vital for spatial reasoning.
Rigorous Conceptual Clarity: The material clearly explains the difference between rational and irrational numbers using the length of the hypotenuse, simplifying a traditionally difficult topic for Grade 8 students.
Problem-Solving Versatility: The student will be able to determine the missing side lengths using geometry as well as valid triple proofs and the theorem of applied use in "height and distance" real-world applications as they prepare for advanced trigonometry.
Alignment with CBSE Standards: The solution(s) follow the new CBSE methodology developed on experiential learning and critical thinking versus traditional rote memorization.
Edge in Competitive Exams: Understanding Baudhāyana-Pythagoras triples and their uses will give you an edge in Olympiads and NTSE where geometry is a key element of these contests.

Frequently Asked Questions

What is taught in NCERT Solutions for Class 8 Maths Chapter 2 The Baudhayana – The Pythagoras Theorem?

How do NCERT Solutions for Class 8 help students learn the Pythagoras Theorem?

Why is the Pythagoras Theorem important for Class 8 students?

What do students learn about Baudhayana in this chapter?

Can NCERT Solutions for Class 8 help students prepare for exams?

Are solved examples included in NCERT Solutions for Class 8?

How can students improve their understanding of theorem-based questions?

Is this chapter connected to real-life applications?

Do NCERT Solutions for Class 8 cover all textbook exercises?

Are these NCERT Solutions for Class 8 based on the latest CBSE syllabus?

Join ALLEN!

NCERT Solutions for Class 8 Maths Ganith Prakash 2 Chapter 2 The Baudhayana - The Pythagoras Theorem

1.0Download NCERT Class 8 Maths Ganith Prakash 2 Chapter 2 Solutions

2.0Key Concepts covered in Class 8 Maths : The Baudhayana - The Pythagoras Theorem

3.0NCERT Solutions for Class 8 Maths Chapter 2 : All Exercises