Important Questions Class 8 Maths Part 2 Chapter 2 Baudhayana-Pythagoras Theorem

The Baudhayana-Pythagoras Theorem states that in a right triangle, the square on the hypotenuse equals the sum of the squares on the other two sides. Important Questions Class 8 Maths Part 2 Chapter 2 cover right triangles, hypotenuse, square diagonals, Baudhayana triples, missing side questions, and real-life applications.

A ladder against a wall, the diagonal of a square, and the shortest path across a rectangular field all use the same idea.

Class 8 Maths Chapter 2 Baudhayana Pythagoras Theorem teaches students how side lengths are connected in a right triangle. Once students know which side is the hypotenuse, questions on missing sides, triples, diagonals, and applications become easier to solve.

Key Takeaways

Class 8 Maths Part 2 Chapter List

Class 8 Maths Chapter 2 Baudhayana Pythagoras Theorem: Topics Covered

Before solving any question from this chapter, identify the right angle first. The side opposite the right angle is always the hypotenuse.

1. Doubling a square
2. Halving a square
3. Hypotenuse of an isosceles right triangle
4. Baudhayana-Pythagoras theorem
5. Finding missing side lengths
6. Baudhayana triples
7. Primitive and scaled triples
8. Diagonal of a square
9. Diagonal and side of a rhombus
10. Lotus-in-a-lake application
11. Real-life right triangle problems
12. Irrationality of √2

Important Questions for Class 8 Maths Chapter 2 with Answers

Important questions for Class 8 Maths Chapter 2 should be solved in this order: square diagonals, isosceles right triangles, missing sides, Baudhayana triples, and applications.

These baudhayana pythagoras theorem class 8 solutions help students practise formula use, reasoning, and real-life application together.

Baudhayana Pythagoras Theorem Class 8 Notes

Baudhayana pythagoras theorem class 8 notes begin with one core idea: a right triangle has one special side called the hypotenuse.

The theorem is written as:

a² + b² = c²

Here, a and b are the two shorter sides, and c is the hypotenuse.

Very Short Answer Important Questions Class 8 Maths Part 2 Chapter 2

Q1. What is a right triangle?
Ans. A right triangle is a triangle with one angle equal to 90°.

Q2. What is the hypotenuse in a right triangle?
Ans. The hypotenuse is the side opposite the right angle. It is the longest side of a right triangle.

Q3. State the Baudhayana-Pythagoras theorem.
Ans. In a right triangle, the square on the hypotenuse equals the sum of the squares on the other two sides.

Q4. Write the formula of the Baudhayana-Pythagoras theorem.
Ans. a² + b² = c², where c is the hypotenuse.

Q5. What is a Baudhayana triple?
Ans. A Baudhayana triple is a set of three positive integers a, b, and c that satisfy a² + b² = c².

Q6. Give one example of a Baudhayana triple.
Ans. 3, 4, 5 is a Baudhayana triple because 3² + 4² = 5².

Q7. What is a primitive Baudhayana triple?
Ans. A primitive Baudhayana triple has no common factor greater than 1 among its three numbers.

Q8. What is the diagonal of a square with side a?
Ans. The diagonal is a√2.

Doubling and Halving a Square Questions

Doubling a square uses the diagonal of the original square. If a square has side a, its diagonal becomes the side of a new square with double the area.

This idea prepares students for the theorem before they use algebraic formulas.

Ganita Prakash Class 8 Part 2 Chapter 2 Questions

Q1. A square is constructed on the diagonal of a given square. What is the area of the new square?
(a) Same as original
(b) Half the original
(c) Double the original
(d) Four times the original

Ans. (c) Double the original.

If the original square has side a, its diagonal is a√2.

Area of new square = (a√2)²
= 2a²

So, the new square has double the area.

Q2. If the side of a square is 5 cm, find the area of the square constructed on its diagonal.
Ans.

Diagonal = 5√2 cm

Area of new square = (5√2)²
= 25 × 2
= 50 cm²

Q3. Why is the smaller inside square formed by joining midpoints half the area of the larger square?
Ans.

Let the side of the larger square be a.

The smaller square is formed by joining the midpoints of the four sides.

Each side of the smaller square is the hypotenuse of a right triangle with legs a/2 and a/2.

Side of smaller square = √(a²/4 + a²/4)

= a/√2

Area = (a/√2)²
= a²/2

So, the smaller square has half the area of the larger square.

Q4. True or False: If the area of a square is doubled, its side also doubles.
Ans. False.

If the original side is a, original area = a².

Double area = 2a².

New side = √(2a²)
= a√2

The side becomes a√2, not 2a.

Isosceles Right Triangle Class 8 Questions

An isosceles right triangle has two equal shorter sides and one right angle. If each equal side is a, then the hypotenuse is a√2.

This is the most direct use of the theorem in the early part of the chapter.

Hypotenuse Class 8 Maths Questions

Q1. Find the hypotenuse of an isosceles right triangle with equal sides of 3 units.
Ans.

c² = 3² + 3²
= 9 + 9
= 18

c = √18
= 3√2

Since 4.2² = 17.64 and 4.3² = 18.49,

4.2 < 3√2 < 4.3

Q2. Find the hypotenuse of an isosceles right triangle with equal sides of 4 units.
Ans.

c² = 4² + 4²
= 16 + 16
= 32

c = √32
= 4√2

Since 5.6² = 31.36 and 5.7² = 32.49,

5.6 < 4√2 < 5.7

Q3. Find the hypotenuse of an isosceles right triangle with equal sides of 6 units.
Ans.

c² = 6² + 6²
= 36 + 36
= 72

c = √72
= 6√2

Since 8.4² = 70.56 and 8.5² = 72.25,

8.4 < 6√2 < 8.5

Q4. Find the hypotenuse of an isosceles right triangle with equal sides of 8 units.
Ans.

c² = 8² + 8²
= 64 + 64
= 128

c = √128
= 8√2

Since 11.3² = 127.69 and 11.4² = 129.96,

11.3 < 8√2 < 11.4

Q5. The hypotenuse of an isosceles right triangle is 10 units. Find the other two sides.
Ans.

Let each equal side be a.

c² = 2a²

10² = 2a²

100 = 2a²

a² = 50

a = √50
= 5√2

Each equal side is 5√2 units.

Q6. If the hypotenuse of an isosceles right triangle is √72, find the other two sides.
Ans.

Let each equal side be a.

c² = 2a²

(√72)² = 2a²

72 = 2a²

a² = 36

a = 6

Each equal side is 6 units.

Pythagoras Theorem Class 8 Questions on Missing Sides

Pythagoras theorem class 8 questions usually ask students to find the missing side of a right triangle.

First identify the hypotenuse. Then decide whether to add squares or subtract squares.

Right Triangle Class 8 Maths Questions

Q1. A right triangle has shorter sides 5 cm and 12 cm. Find the hypotenuse.
Ans.

c² = 5² + 12²
= 25 + 144
= 169

c = √169
= 13 cm

Q2. A right triangle has one shorter side 8 cm and hypotenuse 17 cm. Find the third side.
Ans.

8² + b² = 17²

64 + b² = 289

b² = 225

b = 15 cm

Q3. Find the missing side if a = 5 and b = 7.
Ans.

c² = 5² + 7²
= 25 + 49
= 74

c = √74 units

Q4. Find the missing side if a = 8 and b = 12.
Ans.

c² = 8² + 12²
= 64 + 144
= 208

c = √208
= 4√13 units

Q5. Find the missing side if a = 9 and c = 15.
Ans.

9² + b² = 15²

81 + b² = 225

b² = 144

b = 12 units

Q6. Find the missing side if a = 7 and b = 12.
Ans.

c² = 7² + 12²
= 49 + 144
= 193

c = √193 units

Q7. Find the missing side if a = 1.5 and b = 3.5.
Ans.

c² = 1.5² + 3.5²

= 2.25 + 12.25

= 14.5

c = √14.5 units

Q8. One side of a right triangle is 40 and the hypotenuse is 41. Find the missing side.
Ans.

40² + b² = 41²

1600 + b² = 1681

b² = 81

b = 9 units

Q9. One side is 27 and the hypotenuse is 45. Find the missing side.
Ans.

27² + b² = 45²

729 + b² = 2025

b² = 1296

b = 36 units

Baudhayana Triples Class 8 Questions

Baudhayana triples class 8 questions ask students to verify whether three whole numbers can be sides of a right triangle.

Always square the two smaller numbers and compare their sum with the square of the largest number.

Primitive Baudhayana Triple Class 8 Questions

Q1. List Baudhayana triples with numbers less than or equal to 20.
Ans.

(3, 4, 5): 3² + 4² = 5²
(6, 8, 10): 6² + 8² = 10²
(5, 12, 13): 5² + 12² = 13²
(9, 12, 15): 9² + 12² = 15²
(8, 15, 17): 8² + 15² = 17²
(12, 16, 20): 12² + 16² = 20²

Q2. Which of the following is a Baudhayana triple?
(a) 4, 5, 6
(b) 6, 8, 10
(c) 5, 6, 7
(d) 8, 10, 12

Ans. (b) 6, 8, 10

6² + 8² = 36 + 64
= 100
= 10²

Q3. Which of the following is a primitive Baudhayana triple?
(a) 6, 8, 10
(b) 9, 12, 15
(c) 3, 4, 5
(d) 12, 16, 20

Ans. (c) 3, 4, 5

The numbers 3, 4, and 5 have no common factor greater than 1.

Q4. Is 30, 40, 50 a Baudhayana triple? Is it primitive?
Ans.

30² + 40² = 900 + 1600
= 2500
= 50²

So, it is a Baudhayana triple.

HCF of 30, 40, and 50 is 10. So, it is not primitive.

It is 10 × (3, 4, 5).

Q5. If (a, b, c) is a Baudhayana triple, prove that (ka, kb, kc) is also a Baudhayana triple.
Ans.

Given:

a² + b² = c²

Multiply both sides by k²:

k²a² + k²b² = k²c²

So,

(ka)² + (kb)² = (kc)²

Therefore, (ka, kb, kc) is also a Baudhayana triple.

Q6. Check whether 9, 12, 15 is a primitive Baudhayana triple.
Ans.

9² + 12² = 81 + 144
= 225
= 15²

So, it is a Baudhayana triple.

HCF of 9, 12, and 15 is 3. So, it is not primitive.

It is 3 × (3, 4, 5).

Q7. Find five Baudhayana triples using the odd number method.
Ans.

Using known triples:

7, 24, 25
9, 40, 41
11, 60, 61
13, 84, 85
15, 112, 113

Each set satisfies a² + b² = c².

Baudhayana Pythagoras Theorem Question Answer: Real-Life Applications

Baudhayana pythagoras theorem question answer problems often use ladders, diagonals, rhombus sides, fields, and the lotus-in-a-lake problem.

Draw the right triangle before substituting values.

Application Questions from Ganita Prakash Class 8 Part 2 Chapter 2

Q1. Find the diagonal of a square with side length 5 cm.
Ans.

The diagonal is the hypotenuse of a right triangle with both legs 5 cm.

d² = 5² + 5²
= 25 + 25
= 50

d = √50
= 5√2 cm

Q2. Find the side length of a rhombus whose diagonals are 24 units and 70 units.
Ans.

Diagonals of a rhombus bisect each other at right angles.

Half-diagonals are 12 and 35.

Side² = 12² + 35²
= 144 + 1225
= 1369

Side = √1369
= 37 units

Q3. A lotus stem rises 1 unit above water. When swayed by wind, its tip touches water 3 units away. Find the depth of the lake.
Ans.

Let depth = d.

Stem length = d + 1

When the tip touches water 3 units away, a right triangle forms.

Hypotenuse = d + 1
One leg = d
Other leg = 3

Using the theorem:

(d + 1)² = d² + 3²

d² + 2d + 1 = d² + 9

2d + 1 = 9

2d = 8

d = 4 units

So, the depth of the lake is 4 units.

Q4. How can a square with triple the area of a given square be constructed?
Ans.

Let the side of the given square be a.

Its area is a².

First construct a length a√2 using the diagonal of the square.

Now make a right triangle with legs a√2 and a.

Hypotenuse² = (a√2)² + a²
= 2a² + a²
= 3a²

The square on this hypotenuse has area 3a².

Q5. How can a square with five times the area of a given square be constructed?
Ans.

Let the side of the given square be a.

Make a rectangle with sides a and 2a.

Its diagonal satisfies:

d² = a² + (2a)²
= a² + 4a²
= 5a²

A square built on this diagonal has area 5a².

Q6. Give five examples of rectangles whose side lengths and diagonals are all integers.
Ans.

Use Baudhayana triples:

1. 3 × 4 rectangle, diagonal 5
2. 5 × 12 rectangle, diagonal 13
3. 8 × 15 rectangle, diagonal 17
4. 7 × 24 rectangle, diagonal 25
5. 20 × 21 rectangle, diagonal 29

Q7. Is the hypotenuse always the longest side of a right triangle?
Ans. Yes.

In a right triangle:

a² + b² = c²

Since c² is greater than a² and b² individually, c is greater than a and b.

So, the hypotenuse is always the longest side.

Baudhayana Pythagoras Theorem Class 8 Worksheet

A baudhayana pythagoras theorem class 8 worksheet should include direct theorem questions, missing side questions, triples, and applications.

Use this mixed practice set after completing the solved examples.

Maths Questions for Class 8 with Answers

Q1. Check whether 7, 24, 25 is a Baudhayana triple.
Ans.

7² + 24² = 49 + 576
= 625
= 25²

So, it is a Baudhayana triple.

Q2. Find the hypotenuse of a right triangle with legs 9 cm and 12 cm.
Ans.

c² = 9² + 12²
= 81 + 144
= 225

c = 15 cm

Q3. Find the missing side if one side is 10 and the hypotenuse is √200.
Ans.

10² + b² = (√200)²

100 + b² = 200

b² = 100

b = 10 units

Q4. Check whether 4, 5, 6 is a Baudhayana triple.
Ans.

4² + 5² = 16 + 25
= 41

6² = 36

Since 41 ≠ 36, it is not a Baudhayana triple.

Q5. Find the diagonal of a square with side 7 units.
Ans.

d² = 7² + 7²
= 49 + 49
= 98

d = √98
= 7√2 units

Important MCQs for Class 8 Maths Chapter 2

MCQs in this chapter usually test triples, hypotenuse, diagonal, and theorem meaning.

Check whether the largest number is placed as the hypotenuse before verifying a triple.

Q1. According to Baudhāyana, a square with double the area of a given square can be constructed on its:
(a) Side
(b) Perimeter
(c) Diagonal
(d) Angle bisector

Ans. (c) Diagonal

Q2. Which of the following is not a Baudhayana triple?
(a) 3, 4, 5
(b) 5, 12, 13
(c) 8, 15, 17
(d) 4, 5, 6

Ans. (d) 4, 5, 6

4² + 5² = 41, but 6² = 36.

Q3. The triple 10, 24, 26 is:
(a) Primitive Baudhayana triple
(b) Not a Baudhayana triple
(c) Scaled Baudhayana triple
(d) None of these

Ans. (c) Scaled Baudhayana triple

10² + 24² = 100 + 576
= 676
= 26²

It is 2 × (5, 12, 13).

Q4. Which equation represents Fermat’s Last Theorem?
(a) a² + b² = c²
(b) a³ + b³ = c³ has solutions
(c) aⁿ + bⁿ = cⁿ for n > 2 has no positive integer solutions
(d) a + b = c

Ans. (c) aⁿ + bⁿ = cⁿ for n > 2 has no positive integer solutions

Q5. A triangle with sides 7 cm, 24 cm, and 25 cm:
(a) Is not a right triangle
(b) Satisfies the Baudhayana-Pythagoras relation
(c) Is equilateral
(d) None of these

Ans. (b) Satisfies the Baudhayana-Pythagoras relation

7² + 24² = 49 + 576
= 625
= 25²

Q6. A Baudhayana triple with no common factor is called:
(a) Scaled triple
(b) Perfect triple
(c) Primitive triple
(d) Integer triple

Ans. (c) Primitive triple

Q7. The diagonal of a square with side 7 units is:
(a) 7 units
(b) 14 units
(c) 7√2 units
(d) 49 units

Ans. (c) 7√2 units

Class 8 Maths Extra Questions Chapter 2

Class 8 maths extra questions chapter 2 test whether students can use the theorem beyond direct substitution.

Use diagrams wherever possible.

HOTS and Application-Based Questions

Q1. True or False: Every Baudhayana triple is either primitive or a scaled version of a primitive triple.
Ans. True.

Any Baudhayana triple can be divided by the HCF of its three numbers.

The reduced triple has no common factor greater than 1. So, the original triple is either primitive or a scaled version of one.

Q2. Using grid dots, create squares with areas 2, 4, and 5 sq. units.
Ans.

Area 2: Draw a tilted square with side √2.
Area 4: Draw a square with side 2.
Area 5: Draw a tilted square with side √5.

Each side length can be formed using right triangles on the grid.

Q3. Find the area of an equilateral triangle with side length 6 units.
Ans.

The altitude bisects the base.

Half-base = 3

Using the theorem:

h² + 3² = 6²

h² = 36 - 9
= 27

h = 3√3

Area = 1/2 × base × height
= 1/2 × 6 × 3√3
= 9√3 sq. units

Q4. A right triangle has one short side 6 cm and hypotenuse 10 cm. Find the third side and verify.
Ans.

6² + b² = 10²

36 + b² = 100

b² = 64

b = 8 cm

Verification:

6² + 8² = 36 + 64
= 100
= 10²

Ganita Prakash Class 8 Part 2 Solutions: Chapter 2 Practice

Ganita Prakash Class 8 Part 2 solutions for this chapter focus on reasoning through right triangles, not only memorising a² + b² = c².

Students should learn when to add squares, when to subtract squares, and how to identify triples quickly.

Ganita Prakash Class 8 Part 2 Solutions Questions

Q1. When do we add squares in the theorem?
Ans. Add squares when both shorter sides are known and the hypotenuse is missing.

Q2. When do we subtract squares in the theorem?
Ans. Subtract squares when the hypotenuse and one shorter side are known.

Q3. How can students identify the hypotenuse?
Ans. The hypotenuse is opposite the right angle.

It is always the longest side in a right triangle.

Q4. Why are triples useful?
Ans. Triples help students recognise right triangles quickly without long calculations.

For example, 5, 12, 13 immediately shows a right triangle.

All Key Formulas from Class 8 Maths Chapter 2