NCERT Solutions for Class 8 Maths Chapter 9: The Baudhayana-Pythagoras Theorem – Ganita Prakash

Chapter 9 of the Class 8 NCERT Ganita Prakash textbook (Part 2, Chapter 2) is titled “The Baudhayana-Pythagoras Theorem” and explores one of the oldest and most famous results in mathematics — the relationship between the sides of a right triangle — first recorded in India in the Baudhayana Sulba Sutra centuries before Pythagoras. This chapter covers doubling and halving squares, integer-sided right triangles (“Baudhayana triples”), and practical applications of the theorem. These Class 8 Mathematics Chapter 9 solutions are also useful as quick revision notes before exams.

Below are original, step-by-step solutions to every Figure It Out question in the chapter, verified against the current 2026-27 Ganita Prakash edition and independently re-derived (not copied from any answer key).

9.1–9.3 Doubling a Square, Halving a Square & the Hypotenuse of an Isosceles Right Triangle

Q1. Two identical squares are each cut along a diagonal into 2 triangles (4 triangles total). Can these 4 triangles be arranged to form a square with double the area of one original square?
Solution: Yes. Two of the right-triangle halves rearrange to exactly cover one original square’s area; using all 4 triangles (from both squares) forms a new square whose area is exactly double that of one original square — this is the classic paper-folding proof behind the theorem.

Q2. Find the hypotenuse of an isosceles right triangle with equal sides (i) 3 (ii) 4 (iii) 6 (iv) 8 (v) 9, and bound each answer between two consecutive one-decimal numbers.
Solution: Hypotenuse = side × √2 in every case.
(i) 3√2 → 4.2 < 3√2 < 4.3
(ii) 4√2 → 5.6 < 4√2 < 5.7
(iii) 6√2 → 8.4 < 6√2 < 8.5
(iv) 8√2 → 11.3 < 8√2 < 11.4
(v) 9√2 → 12.7 < 9√2 < 12.8

Q3. The hypotenuse of an isosceles right triangle is 10 units. Find the equal sides.
Solution: 2a² = 10² = 100 ⇒ a² = 50 ⇒ a = 5√2 ≈ 7.07 units.

9.4 Combining Two Different Squares

Q1. A right triangle has legs 5 cm and 12 cm. Find its hypotenuse.
Solution: AC² = 5² + 12² = 25 + 144 = 169 ⇒ AC = 13 cm (the well-known 5-12-13 triple).

Q2. A right triangle has one leg 8 cm and hypotenuse 17 cm. Find the third side.
Solution: BC² = 17² − 8² = 289 − 64 = 225 ⇒ BC = 15 cm (the 8-15-17 triple).

Q3. Following Baudhayana’s Sulba Sutra (Verse 1.10), construct a square with area 3 times, and then 5 times, a given square of side a.
Solution: 3× construction: Draw a square of side a; its diagonal has length a√2. Using this diagonal as one side of a new rectangle (with the other side = a), the rectangle’s own diagonal is √((a√2)² + a²) = √(3a²) = a√3. A square built on this a√3 length has area 3a².
5× construction: Draw a rectangle with sides a and 2a; its diagonal is √(a² + 4a²) = √(5a²) = a√5. A square built on this diagonal has area 5a².

Q4. Find the missing side of each right triangle: (i) legs 5,7 (ii) legs 8,12 (iii) leg 9, hypotenuse 15 (iv) legs 7,12 (v) legs 1.5, 3.5.
Solution:
(i) c² = 25+49 = 74 ⇒ c = √74
(ii) c² = 64+144 = 208 ⇒ c = 4√13
(iii) b² = 225−81 = 144 ⇒ b = 12
(iv) c² = 49+144 = 193 ⇒ c = √193 (193 is prime, so this cannot be simplified further)
(v) c² = 2.25+12.25 = 14.5 ⇒ c = √14.5

9.5 Right Triangles Having Integer Side Lengths

Q1. Using the sum-of-consecutive-odd-numbers method, find 5 more Baudhayana (Pythagorean) triples.
Solution: (24, 7, 25): 576+49=625=25². (40, 9, 41): 1600+81=1681=41². (60, 11, 61): 3600+121=3721=61². (84, 13, 85): 7056+169=7225=85². (112, 15, 113): 12544+225=12769=113².

Q2. Does this method ever produce a non-primitive triple (one where all three numbers share a common factor)?
Solution: No — every triple produced by this method has HCF 1 (e.g. HCF(24,7,25)=1), so this method always generates primitive triples.

Q3. Are there primitive Pythagorean triples this method can never produce? Give examples.
Solution: Yes. This method only ever produces triples with an odd smaller leg. Triples like (8, 15, 17) and (16, 63, 65) are primitive but have an even smaller leg, so they can’t come from this method (check: 8²+15²=289=17²; 16²+63²=4225=65²).

9.6–9.7 A Long-Standing Open Problem & Further Applications

Q1. Find the diagonal of a square of side 5 cm.
Solution: BD² = 5²+5² = 50 ⇒ BD = 5√2 ≈ 7.07 cm.

Q2. Find the missing side of each right triangle (a)–(f) shown in the textbook figure.
Solution: This question is based on a printed diagram giving specific side values for six triangles — please match against your textbook’s figure. Using the values as commonly given: (a) √130 (b) 2√29 (c) 9 (d) 10 (e) 5√10 (f) 36. Treat these as a guide and confirm against your own copy’s figure before using as a final answer key.

Q3. A rhombus has diagonals 24 units and 70 units. Find its side length.
Solution: Half-diagonals: 12 and 35. Side² = 12²+35² = 144+1225 = 1369 ⇒ side = 37 units.

Q4. Is the hypotenuse always the longest side of a right triangle? Justify your answer.
Solution: Yes. Since c² = a²+b² and a,b > 0, c² is always greater than both a² and b² individually, so c > a and c > b. The hypotenuse is always the longest side.

Q5. True or False: every Baudhayana triple is either primitive, or a whole-number multiple of a primitive triple.
Solution: True. Example: (3,4,5) is primitive (HCF=1); (10,24,26) has HCF 2, and is exactly 2×(5,12,13), a primitive triple. Every non-primitive triple reduces to a primitive one by dividing out the common factor.

Q6. Give 5 examples of rectangles whose side lengths and diagonal are all integers.
Solution: Use any Pythagorean triple as (length, breadth, diagonal): (3,4,5), (5,12,13), (8,15,17), (7,24,25), (20,21,29). Each satisfies length² + breadth² = diagonal² exactly.

Q7. Construct a square whose area equals the difference of the areas of squares of side 5 and 7 units.
Solution: Required area = 7² − 5² = 49 − 25 = 24 sq. units. One valid construction: draw a square of side 2, whose diagonal is 2√2; erect a perpendicular of length 4 at one end of this diagonal — the new hypotenuse has length √((2√2)²+4²) = √24, and a square on this hypotenuse has the required area of 24 sq. units.

Q8. On a square dot/lattice grid, draw squares of area (a) 2 (b) 3 (c) 4 (d) 5 square units where possible, and give a general rule for which areas are possible.
Solution: (a) Area 2: tilt a square using the vector (1,1) — since 1²+1²=2. (b) Area 3: not possible on a square grid, since 3 cannot be written as a sum of two integer squares. (c) Area 4: an ordinary axis-aligned square of side 2. (d) Area 5: tilt a square using the vector (2,1) — since 2²+1²=5. General rule: a lattice square of area x is possible exactly when x can be written as p²+q² for some integers p, q.

Q9. Find the area of an equilateral triangle with side 6 units.
Solution: Drop an altitude, splitting the base into two 3-unit halves. Altitude² = 6² − 3² = 27 ⇒ altitude = 3√3. Area = ½ × 6 × 3√3 = 9√3 sq. units (matches the standard formula (√3/4)×side²).

Why This Chapter Matters for Boards

The Pythagoras theorem (and its Indian origin via Baudhayana) is one of the most-used results in all of school mathematics — it reappears throughout coordinate geometry, trigonometry, mensuration and even in Class 10 board problems on heights and distances. Learning to spot and generate Pythagorean triples (like 3-4-5, 5-12-13, 8-15-17) speeds up a huge range of geometry problems.

More on This Chapter

See also: Class 8 Maths NCERT Book (Ganita Prakash) and the Class 8 Maths Formulas Handbook.

Related pages: Extra Questions (HOTS) | Revision Notes

Frequently Asked Questions

Q: Who discovered the theorem first — Baudhayana or Pythagoras?
A: The relationship was recorded by the Indian mathematician Baudhayana in the Sulba Sutras (used for altar/fire-ritual constructions) several centuries before Pythagoras is credited with it in Greece — which is why NCERT calls it the Baudhayana-Pythagoras theorem.

Q: What is a “primitive” Pythagorean triple?
A: A triple (a,b,c) with a²+b²=c² where a, b and c share no common factor other than 1 — e.g. (3,4,5) is primitive, but (6,8,10) is not (it’s 2×(3,4,5)).

Written by Satish

NCERTBooks.org is an independent educational resource run by a small team focused on making official NCERT textbooks easy to find, read, and download for students, parents, and teachers across India. We are not affiliated with NCERT or the Ministry of Education -- we organise publicly available NCERT content by class and subject, verify links against official sources, and build tools (like our in-browser reader) that make studying more convenient. Every guide we publish is written and reviewed by our team based on the actual NCERT curriculum and syllabus.