Exercise 2.1

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The Baudhayana-Pythagoras Theorem - Exercise 2.1 Class 8 Maths - Chapter 2 NCERT Solutions

Class 8 Maths Ganita Prakash 2026–27 Term 2 Chapter 2, The Baudhayana-Pythagoras Theorem - Exercise 2.1 detailed solutions, curated by ALLEN’s expert faculty members, are available on this page. These solutions follow the updated NCERT syllabus to help students with their exam preparation. You can also download the chapter-wise PDF of the detailed solutions provided below.The very first exercise covers the Baudhāyana-Pythagoras Theorem for right-angled triangles, including how to find the hypotenuse when two shorter sides are given and how to find a missing side when the hypotenuse is known. It also includes questions on isosceles right triangles, Baudhāyana triples, and constructions of squares with three times and five times the area of a given square.

1.0Download Class 8 Maths Chapter 2 Ex 2.1 NCERT Solutions PDF

Download the NCERT Class 8 Maths The Baudhayana-Pythagoras Theorem - Exercise 2.1 PDF for learning and practising the questions at your convenience.

Chapter 2 : The Baudhayana - The Pythagoras Theorem Exercise 2.1

2.0Detailed NCERT Solutions on The Baudhayana-Pythagoras Theorem Exercise 2.1 Class 8 Maths (Ganita Prakash Term 2)

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 $a^{2} + a^{2} = 1 0^{2}$ $\Rightarrow 2 a^{2} = 100$ $\Rightarrow a^{2} = 50$ $\Rightarrow a = 50$ $\Rightarrow a = 52$

3.0Key Concepts of Chapter 2 - The Baudhayana-Pythagoras Theorem Exercise 2.1

Applying Baudhāyana-Pythagoras Theorem to find the hypotenuse of a right-angled triangle.
Finding a missing side when one side and the hypotenuse are given.
Finding the hypotenuse of an isosceles right triangle and estimating its value to one decimal place.
Constructing squares with three times and five times the area of a given square using right-angled triangles.
Using the theorem to calculate unknown side lengths in right triangles with different given measurements.
Understanding Baudhāyana triples and identifying primitive and scaled triples.
Applying the theorem to practical constructions involving squares and right-angled triangles.

4.0Benefits of NCERT Solutions Class 8 Maths Chapter 2 Exercise 2.1

Explains how to apply Baudhāyana-Pythagoras Theorem to find unknown sides of right-angled triangles.
Shows the steps for finding the hypotenuse when the two shorter sides are known.
Helps understand how to find the missing side when the hypotenuse and one side are given.
Makes isosceles right triangle problems easier by showing how to find the equal sides or hypotenuse.
Provides clear steps for constructing squares with three times and five times the area of a given square.
Helps practise calculations involving different side lengths, including whole numbers and decimals.
Explains how Baudhāyana-Pythagoras Theorem is used in geometric constructions and practical problems.

NCERT Class 8 Maths Ch. 2 The Baudhayana - The Pythagoras Theorem Other Exercises:-

Exercise 2.1

Exercise 2.2

Exercise 2.3

Exercise 2.4

NCERT Solutions Class 8 Maths All Chapters:-

Chapter 1 : Fractions in Disguise

Chapter 2 : The Baudhayana - The Pythagoras Theorem

Chapter 3 : Proportional Reasoning - 2

Chapter 4 : Exploring Some Geometric Themes

Chapter 5 : Tales by Dots and Lines

Chapter 6 : Algebra Play

Chapter 7 : Area