Class 8 Maths Chapter 2 The Baudhayana-Pythagoras Theorem

Understanding the Baudhayana-Pythagoras theorem in Class 8 lays a strong foundation in geometry. This chapter introduces the core theorem, explains the properties of right-angled triangles, and demonstrates how to apply these concepts to real-life mathematical problems. Mastering this chapter helps students build logical thinking, making future problem-solving much easier and more enjoyable.

The Baudhayana-Pythagoras theorem applies specifically to right-angled triangles (where one angle is exactly 90 degrees). In a right-angled triangle with base a, perpendicular b, and hypotenuse c (the longest side opposite the 90-degree angle), the theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Class 8 Maths Chapter 2 The Baudhayana-Pythagoras Theorem

The mathematical equation is written as: a² + b² = c²

Detailed Questions & Answers

Q1. A ladder 15 m long reaches a window which is 12 m above the ground on one side of a street. Find the distance of the foot of the ladder from the wall.

When a ladder is placed against a wall, it forms a right-angled triangle where the wall and the ground are perpendicular. Hypotenuse (ladder) = 15 m; Perpendicular (window height) = 12 m; Base (distance from wall) = x m.

  • x² + 12² = 15²
  • x² + 144 = 225
  • x² = 225 − 144 = 81
  • x = √81 = 9

Conclusion: The distance of the foot of the ladder from the wall is 9 m.

Q2. A tree is broken by the wind at a height of 5 m from the ground. Its top touches the ground at a distance of 12 m from the base. Find the original height of the tree.

The upright part, the ground, and the broken fallen part form a right-angled triangle. Perpendicular (upright) = 5 m; Base = 12 m; Hypotenuse (broken part) = h.

  • h² = 5² + 12² = 25 + 144 = 169
  • h = √169 = 13
  • Original Height = 5 + 13 = 18 m

Conclusion: The original height of the tree was 18 m.

Q3. The length of a rectangle is 40 cm and one of its diagonals is 41 cm. Find the perimeter of the rectangle.

A rectangle's diagonal divides it into two right-angled triangles. Hypotenuse (diagonal) = 41 cm; Base (length) = 40 cm; Perpendicular (width) = w.

  • w² + 40² = 41²
  • w² + 1600 = 1681
  • w² = 81, so w = 9 cm
  • Perimeter = 2 × (40 + 9) = 2 × 49 = 98

Conclusion: The perimeter of the rectangle is 98 cm.

Q4. Two poles of heights 6 m and 11 m stand vertically on plane ground. If the distance between their feet is 12 m, find the distance between their tops.

A horizontal line from the top of the shorter pole creates a right-angled triangle. Base = 12 m; Perpendicular (difference in heights) = 11 − 6 = 5 m; Hypotenuse = d.

  • d² = 12² + 5² = 144 + 25 = 169
  • d = √169 = 13

Conclusion: The distance between the tops of the two poles is 13 m.

Q5. Check whether the numbers 7, 24, and 25 form a Baudhayana (Pythagorean) triplet. Explain your reasoning.

For a Pythagorean triplet, the square of the largest number must equal the sum of the squares of the other two. Largest = 25; others = 7 and 24.

  • LHS = 7² + 24² = 49 + 576 = 625
  • RHS = 25² = 625
  • LHS = RHS (625 = 625)

Conclusion: Yes, (7, 24, 25) forms a Baudhayana triplet.

Q6. The diagonals of a rhombus measure 16 cm and 30 cm. Find the perimeter of the rhombus.

The diagonals of a rhombus bisect each other at right angles, forming four identical right triangles. Base = 8 cm (half of 16); Perpendicular = 15 cm (half of 30); Hypotenuse (side) = s.

  • s² = 8² + 15² = 64 + 225 = 289
  • s = √289 = 17 cm
  • Perimeter = 4 × 17 = 68

Conclusion: The perimeter of the rhombus is 68 cm.

Q7. Two ships leave a port at the same time. One travels North at 24 km/hr, the other East at 10 km/hr. How far apart are they after 1 hour?

North and East are perpendicular, forming a right-angled triangle. Perpendicular = 24 km; Base = 10 km; Hypotenuse = D.

  • D² = 24² + 10² = 576 + 100 = 676
  • D = √676 = 26

Conclusion: After 1 hour, the ships are 26 km apart.

Q8. Find the altitude (height) of an equilateral triangle whose side length is 10 cm.

The altitude divides the equilateral triangle into two identical right triangles and bisects the base. Hypotenuse (side) = 10 cm; Base = 5 cm (half of 10); Perpendicular (altitude) = h.

  • h² + 5² = 10²
  • h² + 25 = 100, so h² = 75
  • h = √75 = √(25 × 3) = 5√3

Conclusion: The altitude of the equilateral triangle is 5√3 cm.

Q9. A wire is stretched from the top of a 24 m high vertical pole to a peg in the ground 7 m away from the base. Find the length of the wire.

The pole and ground meet at 90°; the wire is the hypotenuse. Perpendicular = 24 m; Base = 7 m; Hypotenuse (wire) = w.

  • w² = 24² + 7² = 576 + 49 = 625
  • w = √625 = 25

Conclusion: The length of the wire is 25 m.

Q10. The sides of a triangle are 9 cm, 40 cm, and 41 cm. Determine whether it is a right-angled triangle.

By the converse of the theorem, if the square of the longest side equals the sum of squares of the other two, the triangle is right-angled. Longest = 41; others = 9 and 40.

  • Sum = 9² + 40² = 81 + 1600 = 1681
  • 41² = 1681
  • 1681 = 1681

Conclusion: Yes, since the sides satisfy the theorem, it is a right-angled triangle.

Frequently Asked Questions (FAQs)

1. What types of important questions are asked from this chapter in school exams? Class 8 exams usually feature MCQs, short-answer questions, long-answer derivations, and case-based word problems. Focus on the main formula, real-life applications, and finding unknown sides.

2. How should I write answers for this chapter to score full marks? Always start with a clear statement of the theorem or formula. Draw neatly labeled diagrams. Show every step of your calculation. Underline important keywords and write the proper units (cm or m) in your final answer.

3. What are the highest-weightage subtopics in this chapter? The most critical topics are the statement and proof of the theorem, finding unknown sides of a right-angled triangle using the formula, and solving word problems on heights, distances, and areas.

4. Are diagrams necessary for scoring in geometry questions? Yes. Clearly labeled diagrams are essential. Drawing the right-angled triangle and marking known and unknown sides helps you gain step marks and makes your solution easier to evaluate.

5. How can I quickly revise this chapter before a test? Review all your solved important questions, practice past MCQs, summarize the core formulas on a single page, and review your notes on high-weightage word problems.

6. Do I get step marks if my final calculation is wrong? Yes. CBSE and most boards follow step marking. Even if the final answer is incorrect, you get partial marks for the correct formula, an accurate diagram, and correct substitution of values before the error.