Official NCERT 2026-27 Solutions
NCERT Solutions for Class 8 Maths Chapter 9: The Baudhayana-Pythagoras Theorem (Ganita Prakash Part II)
Authentic handwritten solutions in Topper Notebook style. Features geometric proofs, Baudhayana triples, the ancient Lilavati lotus stem problem, dot-grid integer square areas, and equilateral triangle altitudes from Pages 51 to 54 of Ganita Prakash Part II.
Quick Chapter Navigation & Blueprint
- Part A: Ancient Indian Geometry & Right Triangles (Lilavati & Q1–Q3) — The famous Lilavati swaying lotus problem (depth = 4), square diagonals (5√2), missing sides (9-40-41 triple), and rhombus side from diagonals 24 & 70.
- Part B: Triples, Rectangles & Geometric Justifications (Questions 4 to 6) — Proving hypotenuse is the strictly longest side, classification of Baudhayana triples, and 5 integer-diagonal rectangles.
- Part C: Grid Geometry & Sum of Two Squares (Questions 7 & 8) — Difference-of-squares construction (area 24) and which integer areas can be formed on dot grids (proving area 3 is impossible).
- Part D: Equilateral Altitude & Fermat’s Last Theorem (Question 9) — Altitude and exact area of equilateral triangle with side 6 (9√3), Andrew Wiles proof context, and the three-box logic puzzle.
Official Textbook PDF Reference: Download NCERT Class 8 Maths Chapter 9 (hegp202.pdf)
Part A: Lilavati Lotus & Core Right Triangle Problems (Pages 52–53)
Bhaskaracharya’s Lilavati Problem (Page 52):
“In a lake surrounded by birds, there is a lotus flower with the tip of its stem 1 unit above the water. When swayed by a gentle breeze, the tip touches the water 3 units away from its original vertical position. Find the depth of the lake.”
“In a lake surrounded by birds, there is a lotus flower with the tip of its stem 1 unit above the water. When swayed by a gentle breeze, the tip touches the water 3 units away from its original vertical position. Find the depth of the lake.”
Topper’s Right Triangle Model & Algebraic Solution:
- Let the depth of the lake be x units. This represents the vertical length of the stem submerged underwater.
- Since the tip rises 1 unit above water, the total length of the lotus stem is (x + 1) units.
- When swayed, the stem pivots about its rooted base at the lake bed, acting as the hypotenuse of length (x + 1).
- The horizontal displacement where it touches water is 3 units.
- Applying the Baudhayana-Pythagoras theorem:
Vertical Leg2 + Horizontal Leg2 = Hypotenuse2
x2 + 32 = (x + 1)2
x2 + 9 = x2 + 2x + 1 - Subtracting x2 from both sides:
9 = 2x + 1
2x = 8 → x = 4 units.
Final Answer: The depth of the lake is 4 units (and the total stem length is 5 units).
Question 1 (Page 52):
Find the diagonal of a square with sidelength 5 cm.
Find the diagonal of a square with sidelength 5 cm.
Step-by-Step Calculation:
- In a square of side s = 5 cm, the diagonal forms a right-angled isosceles triangle with two adjacent sides:
d2 = s2 + s2 = 52 + 52 = 25 + 25 = 50. - d = √50 = √(25 × 2) = 5√2 cm.
- Taking √2 ≈ 1.414: d ≈ 5 × 1.414 = 7.07 cm.
Final Answer: The diagonal of the square is 5√2 cm (approximately 7.07 cm).
Question 2 (Page 52):
Find the missing sidelengths in the following right triangles:
(i) Legs: 7 and 9. Hypotenuse = ?
(ii) Leg: 40, Hypotenuse: 41. Missing leg = ?
(iii) Legs: 27 and 36 (or 45). Find the third side.
Find the missing sidelengths in the following right triangles:
(i) Legs: 7 and 9. Hypotenuse = ?
(ii) Leg: 40, Hypotenuse: 41. Missing leg = ?
(iii) Legs: 27 and 36 (or 45). Find the third side.
Evaluation via Baudhayana-Pythagoras Theorem:
- (i) Legs 7 and 9:
Hypotenuse c = √(72 + 92) = √(49 + 81) = √130 ≈ 11.4 units. - (ii) Hypotenuse 41, Leg 40:
Missing leg b = √(412 − 402).
Using difference of squares: (41 − 40)(41 + 40) = 1 × 81 = 81.
b = √81 = 9 units. (Matches Baudhayana triple 9, 40, 41!). - (iii) Legs 27 and 36:
Hypotenuse = √(272 + 362) = √(729 + 1296) = √2025 = 45 units (Scaled 3-4-5 triple by 9!).
Missing Sidelengths: (i) √130 (≈ 11.4) | (ii) 9 units | (iii) 45 units.
Question 3 (Page 53):
Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.
Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.
Topper’s Rhombus Diagonal Property Derivation:
- Key Geometric Property: The diagonals of a rhombus bisect each other at right angles (90°).
- Let the diagonals be d1 = 24 and d2 = 70.
- Half-diagonal lengths = d1 / 2 = 12 units, and d2 / 2 = 35 units.
- The side of the rhombus s forms the hypotenuse of the right-angled triangle formed by the half-diagonals:
s2 = 122 + 352
= 144 + 1225 = 1369. - Taking square root:
s = √1369 = 37 units.
(Notice that 12, 35, 37 is a primitive Baudhayana triple: 144 + 1225 = 1369 ✓).
Final Answer: The sidelength of the rhombus is 37 units.
Part B: Properties, Proofs & Triples (Pages 53–54)
Question 4 (Page 53):
Is the hypotenuse the longest side of a right triangle? Justify your answer.
Is the hypotenuse the longest side of a right triangle? Justify your answer.
Two Independent Proofs (Geometric & Algebraic):
- Proof 1 (Angle-Side Inequality Theorem):
In any triangle, the sum of all interior angles is 180°.
In a right-angled triangle, one angle is 90°. Therefore, the sum of the other two acute angles is 90°.
Since each angle must be strictly positive, neither acute angle can exceed 90°.
Thus, the 90° right angle is strictly the largest angle in the triangle.
By the fundamental theorem of triangle geometry, the side opposite the largest angle is always strictly the longest side.
Therefore, the hypotenuse (opposite the 90° angle) is the longest side! - Proof 2 (Algebraic):
By the theorem: c2 = a2 + b2.
Since side lengths are positive (b > 0), c2 > a2 → c > a.
Similarly, since a > 0, c2 > b2 → c > b.
Hence, c is strictly greater than both legs.
Conclusion: Yes, the hypotenuse is always strictly the longest side of a right triangle.
Question 5 (Page 53):
True or False: Every Baudhayana triple is either a primitive triple or a scaled version of a primitive triple.
True or False: Every Baudhayana triple is either a primitive triple or a scaled version of a primitive triple.
Topper’s Number-Theoretic Proof:
- Let (a, b, c) be any set of positive integers satisfying a2 + b2 = c2.
- Let g = gcd(a, b, c) be their greatest common divisor.
- If g = 1, then (a, b, c) is by definition a primitive triple.
- If g > 1, then we can divide each term by g:
(a/g)2 + (b/g)2 = (c/g)2. - The resulting triple (a/g, b/g, c/g) has gcd = 1, making it a primitive triple, and the original triple is simply scaled by factor g.
Final Answer: TRUE. Every Baudhayana triple is either primitive or a positive integer multiple of a primitive triple.
Question 6 (Page 53):
Give 5 examples of rectangles whose sidelengths and diagonals are all integers.
Give 5 examples of rectangles whose sidelengths and diagonals are all integers.
Generating Integer Rectangles from Primitive Triples:
- For a rectangle with sides a and b, the diagonal is d = √(a2 + b2). For d to be an integer, (a, b, d) must form a Baudhayana triple:
1. Sides 3 cm × 4 cm → Diagonal = √(9 + 16) = √25 = 5 cm.
2. Sides 5 cm × 12 cm → Diagonal = √(25 + 144) = √169 = 13 cm.
3. Sides 8 cm × 15 cm → Diagonal = √(64 + 225) = √289 = 17 cm.
4. Sides 7 cm × 24 cm → Diagonal = √(49 + 576) = √625 = 25 cm.
5. Sides 9 cm × 40 cm → Diagonal = √(81 + 1600) = √1681 = 41 cm.
(Bonus: 20 cm × 21 cm → Diagonal = 29 cm).
Five Rectangles (Sides → Diagonal):
(1) 3 by 4 → 5 | (2) 5 by 12 → 13 | (3) 8 by 15 → 17 | (4) 7 by 24 → 25 | (5) 9 by 40 → 41.
(1) 3 by 4 → 5 | (2) 5 by 12 → 13 | (3) 8 by 15 → 17 | (4) 7 by 24 → 25 | (5) 9 by 40 → 41.
Question 8 (Page 53 – Dot Grid Square Areas):
Using the dots of an integer grid as 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 sq. units?
What are all possible integer areas?
Using the dots of an integer grid as 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 sq. units?
What are all possible integer areas?
Topper’s Sum of Two Squares Theorem:
- On a square dot grid, let the vector between two adjacent vertices of a square have horizontal displacement Δx and vertical displacement Δy.
- By the theorem, the side length is s = √(Δx2 + Δy2).
- Therefore, the area of the square is Area = s2 = Δx2 + Δy2, which MUST be expressible as the sum of two integer squares!
- Testing each requested area:
• (a) Area = 2: 12 + 12 = 1 + 1 = 2 → YES (tilt by 1 step right, 1 step up).
• (b) Area = 3: The only squares ≤ 3 are 0 and 1. Sums possible: 0+0=0, 1+0=1, 1+1=2. The number 3 CANNOT be written as a2 + b2. → IMPOSSIBLE!
• (c) Area = 4: 22 + 02 = 4 + 0 = 4 → YES (horizontal square of side 2).
• (d) Area = 5: 22 + 12 = 4 + 1 = 5 → YES (tilt by 2 units along one axis, 1 along the other).
Answers:
(a) Yes (2 sq units) | (b) No, impossible (3 sq units) | (c) Yes (4 sq units) | (d) Yes (5 sq units).
Possible integer areas are precisely numbers that can be expressed as the sum of two squares: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 20, 25…
(a) Yes (2 sq units) | (b) No, impossible (3 sq units) | (c) Yes (4 sq units) | (d) Yes (5 sq units).
Possible integer areas are precisely numbers that can be expressed as the sum of two squares: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 20, 25…
Question 9 (Page 54):
Find the area of an equilateral triangle with sidelength 6 units.
Find the area of an equilateral triangle with sidelength 6 units.
Altitude Derivation & Area Calculation:
- Let ΔABC be an equilateral triangle with side s = 6 units.
- Draw altitude AD perpendicular to BC. By RHS congruence (ΔABD ≅ ΔACD), altitude AD bisects the base BC:
BD = CD = 6 / 2 = 3 units. - In right-angled ΔABD:
AD2 + BD2 = AB2
h2 + 32 = 62
h2 + 9 = 36 → h2 = 27 → h = √27 = 3√3 units. - Area of ΔABC = ½ × Base × Height
= ½ × 6 × 3√3 = 3 × 3√3 = 9√3 sq. units. - Substituting √3 ≈ 1.732: Area ≈ 9 × 1.732 = 15.588 sq. units.
Final Answer: The area of the equilateral triangle is 9√3 sq. units (approximately 15.59 sq. units).