Class 8 Maths Ganita Prakash Chapter 9 Solutions (NCERT 2026–27) – The Baudhayana-Pythagoras Theorem

These Class 8 Maths Ganita Prakash Chapter 9 solutions cover The Baudhayana-Pythagoras Theorem from the new NCF-2023 textbook (2026–27). This is Chapter 2 of Ganita Prakash Part II (the 9th chapter of the Class 8 course). Work through the Figure it Out exercises with original diagrams, construction steps and explanations. The page also covers the main Math Talk and Try This activities, including the lotus problem and the colour-box puzzle.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 9 (Part II Ch 2) Exercises: Figure it Out (4 sets) Session: 2026–27
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Chapter 9 Overview

This chapter, The Baudhayana-Pythagoras Theorem, traces one of geometry’s most famous results back to Baudhayana’s Śulba-Sūtra (c. 800 BCE). It begins with doubling and halving a square using its diagonal, then finds the hypotenuse of an isosceles right triangle (c2 = 2a2) and introduces √2 as a non-terminating, non-repeating number. By combining two different squares it builds up to the general theorem a2 + b2 = c2, explores Baudhayana (Pythagorean) triples, touches on Fermat’s Last Theorem, and ends with real-life applications such as a problem from Bhaskaracharya’s Līlāvatī. The Class 8 Maths Ganita Prakash Chapter 9 solutions below solve every exercise on this journey.

Key Concepts & Definitions

Right triangle: a triangle with one angle of 90°. The side opposite the right angle is the longest side.

Hypotenuse: the side opposite the right angle in a right triangle — always its longest side.

Baudhayana-Pythagoras Theorem: in a right triangle with legs a, b and hypotenuse c, a2 + b2 = c2. The chapter presents Baudhayana’s statement in the Śulba-Sūtra; the result is also called the Pythagorean Theorem.

Isosceles right triangle: a right triangle whose two legs are equal (a = b); then c2 = 2a2, so c = a√2.

Baudhayana (Pythagorean) triple: three positive integers (a, b, c) with a2 + b2 = c2, e.g. (3, 4, 5).

Primitive triple: a triple with no common factor greater than 1, e.g. (3, 4, 5). A scaled version (ka, kb, kc) is also a triple.

√2: the length of the diagonal of a unit square; it lies between 1.414 and 1.415 and cannot be written as a terminating decimal or a fraction.

A three-four-five right triangle has squares of areas nine and sixteen on its legs and twenty-five on its hypotenuse.
The theorem compares square areas. Identify the right angle before choosing the hypotenuse.

Important Formulas (Chapter 9)

Baudhayana-Pythagoras Theorem: a2 + b2 = c2  (c = hypotenuse). So c = √(a2 + b2) and a missing leg = √(c2 − b2).

Isosceles right triangle: c2 = 2a2, i.e. c = a√2 (hypotenuse), or a = c/√2 (each equal side).

Diagonal of a square of side s: d = s√2.

Scaling rule: if (a, b, c) is a triple, so is (ka, kb, kc) for every positive integer k.

Bound on a surd: if n2 < N < (n+1)2, then n < √N < n + 1 (refine digit by digit for one decimal place).

Figure it Out (Pages 39–40)

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. (Each square is cut along a diagonal into pieces 1 and 2, and 3 and 4.) Can you arrange these pieces to create a square with double the area of either square?

SOLUTION Yes. Cutting each unit square along a diagonal gives four right-angled isosceles triangles (pieces 1, 2, 3 and 4), each with two equal sides of length equal to the side of the original square. Place the four triangles with their right-angle corners meeting at a common centre and their hypotenuses pointing outwards. The four hypotenuses form the four sides of a new (tilted) square. Each small triangle has area = ½ of an original square, and there are 4 of them, so the new square has area 4 × ½ = 2 original squares — exactly double the area of either square. Its side is the diagonal of the original square.
Four equal right isosceles triangles meet at their right-angle vertices to form a diamond-shaped square with twice the area of either original paper square.
Page 39, Question 1: the four outer hypotenuses are equal and consecutive outer angles are 45° + 45° = 90°.

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

SOLUTION For an isosceles right triangle, c2 = 2a2, so c = a√2 = √(2a2). (i) a = 3: c2 = 2 × 9 = 18, so c = √18 = 3√2. Since 4.22 = 17.64 and 4.32 = 18.49, we get 4.2 < √18 < 4.3 (√18 ≈ 4.24). (ii) a = 4: c2 = 2 × 16 = 32, so c = √32 = 4√2. Since 5.62 = 31.36 and 5.72 = 32.49, we get 5.6 < √32 < 5.7 (√32 ≈ 5.66). (iii) a = 6: c2 = 2 × 36 = 72, so c = √72 = 6√2. Since 8.42 = 70.56 and 8.52 = 72.25, we get 8.4 < √72 < 8.5 (√72 ≈ 8.49). (iv) a = 8: c2 = 2 × 64 = 128, so c = √128 = 8√2. Since 11.32 = 127.69 and 11.42 = 129.96, we get 11.3 < √128 < 11.4 (√128 ≈ 11.31). (v) a = 9: c2 = 2 × 81 = 162, so c = √162 = 9√2. Since 12.72 = 161.29 and 12.82 = 163.84, we get 12.7 < √162 < 12.8 (√162 ≈ 12.73).

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.]

SOLUTION Let each equal side be a. Then c2 = 2a2 with c = 10. 102 = 2a2 ⇒ 100 = 2a2 ⇒ a2 = 50. So a = √50 = 5√2 ≈ 7.07. Each of the other two sides is √50 = 5√2 units (≈ 7.07).
A right isosceles triangle has hypotenuse ten and both legs five root two, approximately seven point zero seven.
Page 40, Question 3: choose the positive square root because a side length is positive.

Figure it Out (Page 47)

1. If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm, then what is the length of its hypotenuse? First draw the right-angled triangle with these sidelengths and measure the hypotenuse, then check your answer using Baudhayana’s Theorem.

SOLUTION By Baudhayana’s Theorem, c2 = 52 + 122 = 25 + 144 = 169. c = √169 = 13 cm. (On the drawn triangle the measured hypotenuse is also about 13 cm.)

2. If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhayana’s Theorem.

SOLUTION Let the third side be b. Then 82 + b2 = 172. 64 + b2 = 289 ⇒ b2 = 289 − 64 = 225. b = √225 = 15 cm. So (8, 15, 17) is a Baudhayana triple.
Two right triangles have sides five twelve thirteen centimetres and eight fifteen seventeen centimetres.
Page 47, Questions 1–2: diagrams preserve the side ratios. For a real 5 cm by 12 cm construction, use a ruler and set square on paper.

3. Using the constructions you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square? (Baudhayana’s Śulba-Sūtra, Verse 1.10)

SOLUTION Triple area: first make a square of double area on the diagonal of the given square (area 2). Now combine this double-area square with one more copy of the original square using Baudhayana’s combining method — make a right triangle whose legs are their sides (√2 and 1) and draw a square on its hypotenuse. Its area = 2 + 1 = 3 times the original. Five times the area: continue combining. Take the triple-area square (area 3) and the double-area square (area 2): a right triangle with legs √3 and √2 gives a hypotenuse-square of area 3 + 2 = 5. (Equivalently, a right triangle with legs of length 1 and 2 has hypotenuse-square 12 + 22 = 5 times the unit square.)
A right triangle with legs s and s root two yields hypotenuse s root three. A right triangle with legs s and two s yields hypotenuse s root five.
Page 47, Question 3: use the original square diagonal for s√2; a length 2s is obtained by placing two original side lengths end to end.

4. Let a, b and c denote the length of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing sidelength in each of the following cases: (i) a = 5, b = 7   (ii) a = 8, b = 12   (iii) a = 9, c = 15 (iv) a = 7, b = 12   (v) a = 1.5, b = 3.5

SOLUTION Use a2 + b2 = c2; the missing side is found by adding (for c) or subtracting (for a leg). (i) a = 5, b = 7: c2 = 25 + 49 = 74, so c = √74 ≈ 8.60. (ii) a = 8, b = 12: c2 = 64 + 144 = 208, so c = √208 = 4√13 ≈ 14.42. (iii) a = 9, c = 15: b2 = 152 − 92 = 225 − 81 = 144, so b = √144 = 12. (Here (9, 12, 15) is a triple.) (iv) a = 7, b = 12: c2 = 49 + 144 = 193, so c = √193 ≈ 13.89. (v) a = 1.5, b = 3.5: c2 = 2.25 + 12.25 = 14.5, so c = √14.5 ≈ 3.81.

Figure it Out (Page 50)

1. Find 5 more Baudhayana triples using this idea. (Use an odd square number 2n − 1; then (n − 1)2 + (2n − 1) = n2 gives a triple.)

SOLUTIONThe textbook has already used odd squares 9 and 25 to obtain (3, 4, 5) and (5, 12, 13). For five more, choose 49, 81, 121, 169 and 225.For an odd number m > 1, put n = (m² + 1)/2. The triple is (m, n − 1, n).(7, 24, 25): 49 + 576 = 625.(9, 40, 41): 81 + 1600 = 1681.(11, 60, 61): 121 + 3600 = 3721.(13, 84, 85): 169 + 7056 = 7225.(15, 112, 113): 225 + 12544 = 12769.

2. Does this method yield non-primitive Baudhayana triples? [Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]

SOLUTION No — this method only gives primitive triples. In every triple it produces, the larger leg is (n − 1) and the hypotenuse is n, so they differ by exactly 1. Two whole numbers that differ by 1 are consecutive, so their only common factor is 1. Hence all three numbers share no common factor greater than 1, and the triple is primitive. (For example (3, 4, 5), (5, 12, 13), (7, 24, 25) are all primitive.)

3. Are there primitive triples that cannot be obtained through this method? If yes, give examples.

SOLUTION Yes. This method always makes the longer leg and the hypotenuse differ by 1, so it can never produce a primitive triple in which the two larger numbers differ by more than 1. Example: (8, 15, 17) is primitive (82 + 152 = 64 + 225 = 289 = 172), but here the hypotenuse 17 and the larger leg 15 differ by 2, so it is not generated by this method. Another example: (20, 21, 29) is primitive but the legs differ by 1 from each other (not from the hypotenuse), so it also cannot come from this method.

Figure it Out (Pages 52–54)

1. Find the diagonal of a square with sidelength 5 cm.

SOLUTION The diagonal splits the square into two isosceles right triangles with legs 5 and 5. d2 = 52 + 52 = 25 + 25 = 50, so d = √50 = 5√2 cm ≈ 7.07 cm.

2. Find the missing sidelengths in the following right triangles: (legs 7 and 9; right angle with legs 4 and 10; right angle with leg 40 and hypotenuse 41; leg 10 and hypotenuse √200; legs 10 and √150; leg 27 and hypotenuse 45).

SOLUTIONRead the six diagrams left to right, row by row. The side opposite the right-angle mark is the hypotenuse.(i) Legs 7 and 9: c = √(7² + 9²) = √130 ≈ 11.40.(ii) Legs 4 and 10: c = √(4² + 10²) = √116 = 2√29 ≈ 10.77.(iii) Hypotenuse 41, leg 40: other leg = √(41² − 40²) = √81 = 9.(iv) Hypotenuse √200, leg 10: other leg = √(200 − 100) = 10.(v) The right angle is between 10 and √150. They are both legs, so the missing hypotenuse is √(10² + (√150)²) = √250 = 5√10 ≈ 15.81.(vi) Hypotenuse 45, leg 27: other leg = √(45² − 27²) = √1296 = 36.
Three right triangles show legs seven and nine with hypotenuse root one hundred thirty; legs four and ten with hypotenuse root one hundred sixteen; legs nine and forty with hypotenuse forty-one.
Page 53, Question 2: the triangles are rotated into a common orientation for clarity; lengths and right-angle relationships are preserved.
Three right triangles show legs ten and ten with hypotenuse root two hundred; legs ten and root one hundred fifty with hypotenuse root two hundred fifty; legs twenty-seven and thirty-six with hypotenuse forty-five.
Page 53, Question 2: the triangles are rotated into a common orientation for clarity; lengths and right-angle relationships are preserved.

3. Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.

SOLUTION The diagonals of a rhombus bisect each other at right angles, so half-diagonals are 24 ÷ 2 = 12 and 70 ÷ 2 = 35. These are the legs of a right triangle whose hypotenuse is a side of the rhombus. side2 = 122 + 352 = 144 + 1225 = 1369, so side = √1369 = 37 units.
A rhombus has perpendicular bisecting diagonals of length twenty-four and seventy. Half-diagonals twelve and thirty-five form a right triangle with side thirty-seven as hypotenuse.
Page 53, Question 3: diagonals are not sides of the right triangle until they are halved.

4. Is the hypotenuse the longest side of a right triangle? Justify your answer.

SOLUTION Yes, the hypotenuse is always the longest side. Since c2 = a2 + b2, the value c2 is larger than both a2 and b2 (we add a positive amount). Therefore c > a and c > b, so the hypotenuse is greater than each leg. (It also lies opposite the largest angle, the right angle.)

5. True or False—Every Baudhayana triple is either a primitive triple or a scaled version of a primitive triple.

SOLUTION True. Take any Baudhayana triple (a, b, c) and let f be the greatest common factor of a, b, c. If f = 1, the triple is already primitive. If f > 1, then dividing through gives (a/f, b/f, c/f), which is a primitive triple, and the original triple is its scaled version (multiplied by f). So every triple falls into one of the two cases.

6. Give 5 examples of rectangles whose sidelengths and diagonals are all integers.

SOLUTION A rectangle’s diagonal is the hypotenuse of a right triangle whose legs are its two sides, so we just need Baudhayana triples (length, breadth, diagonal):
LengthBreadthDiagonalCheck (l2 + b2 = d2)
43516 + 9 = 25
12513144 + 25 = 169
861064 + 36 = 100
15817225 + 64 = 289
24725576 + 49 = 625

7. Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.

SOLUTIONRequired area = 7² − 5² = 24 square units; its side must be √24 = 2√6 ≈ 4.90 units.Construction: (1) Draw AB = 5 units. (2) Construct a perpendicular ray to AB at A. (3) With centre B and radius 7 units, draw an arc meeting that ray at C. (4) Construct a square on AC.Since ∠A = 90°, AC² = BC² − AB² = 49 − 25 = 24. Therefore the square on AC has the required area. This determines AC without first measuring a rounded √24.
A right triangle ABC has AB five, BC seven and AC root twenty-four. A square on a segment of length AC has area twenty-four.
Question 7: draw AB = 5, erect a perpendicular at A, and intersect it with a radius-7 arc centred at B to locate C. The separate square uses side AC.

8. (i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq. units, and (d) 5 sq. units? (ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

SOLUTION (i)(a) Area 2 — Yes: a tilted square on the diagonal of a 1×1 cell has side √2, so area (√2)2 = 2. (b) Area 3 — No: a grid (lattice) square’s area is always of the form a2 + b2 for whole numbers a, b (the side joins two dots a across and b up). The number 3 cannot be written as a sum of two squares, so no such square exists. (c) Area 4 — Yes: an upright 2×2 square has area 4 (here a = 2, b = 0). (d) Area 5 — Yes: a tilted square with side joining a dot 2 across and 1 up has side √5, so area = 5 (12 + 22 = 5). (ii) The possible integer areas are exactly those numbers that can be written as a2 + b2 with whole numbers a, b, not both zero — for example 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, … Numbers like 3, 6, 7, 11, 12, 14, 15 (which are not a sum of two squares) cannot be areas of such a grid square.
Three dot grids show squares of area two, four and five. Their side displacements are one and one, two and zero, and two and one.
Page 53, Question 8: a and b may be zero but cannot both be zero. The four vertices remain lattice points after a quarter-turn of the side vector.

9. Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.]

SOLUTIONDraw altitude AD from vertex A to opposite side BC. Triangles ABD and ACD are right triangles with equal hypotenuses AB = AC = 6 and common leg AD. They are congruent by RHS, so BD = DC = 3.AD² = 6² − 3² = 36 − 9 = 27, hence AD = 3√3 units.Area = ½ × BC × AD = ½ × 6 × 3√3 = 9√3 ≈ 15.59 square units.
An equilateral triangle of side six has altitude three root three, splitting its base into two segments of length three. Its area is nine root three square units.
Page 54, Question 9: the two right triangles are congruent by RHS, so the altitude bisects the base.

Math Talk & Try This Answers

Math Talk (Doubling a square, p. 34) Q. Why does the new dotted square (on the diagonal) have double the area of the original square? Why are the four small triangles congruent? Answer. Each diagonal cuts the original square into 2 congruent right-angled isosceles triangles, so the original square is made of 2 such triangles. The square built on the diagonal is made of 4 of the very same triangles. Since all four are right-angled isosceles triangles with the same leg length (the side of the original square), they are congruent (SAS). Therefore the new square = 4 triangles = 2 × (2 triangles) = double the original area.
Try This (Doubling/Halving using paper, pp. 35–37) Q. Will a square with half the sidelength have half the area? Why not? How many such squares fill the original? Why is PQRS (the inside tilted square) half the area? Answer. Halving a square’s side gives one-quarter of its area, so four such small squares fill the original. To obtain half the area, join the four side midpoints P, Q, R and S. Adjacent midpoint segments have equal length, and their angles are 45° + 45° = 90°, so PQRS is a square. Draw PR and QS: the inner square contains four small congruent right triangles; the four outer corners contain four more of the same size. Thus the inner area is 4/8 = 1/2 of the original.
A square joins the midpoints P Q R S of an outer square. Its diagonals divide it into four equal right triangles, each congruent to one corner triangle.
Pages 36–37: the outer square consists of eight equal small triangles; the inner square contains four of them.
Try This (Value of √2, p. 38) Q. Can √2 be written as a terminating decimal or as a fraction m/n? Answer. No. If √2 ended in a non-zero last digit, its square would also end in a non-zero digit and could not equal 2.000…, so its decimal never terminates. If √2 = m/n then 2n2 = m2; but in a perfect square each prime appears an even number of times, while the prime 2 would appear an odd number of times on the left and even on the right — impossible. So √2 = 1.41421356… is non-terminating and cannot be a fraction.
Math Talk (Baudhayana triples, p. 48) Q. List all Baudhayana triples with numbers ≤ 20. Is (30, 40, 50) a triple? Is (300, 400, 500)? Which are primitive? Answer. With a ≤ b and all entries ≤ 20, the complete list is (3,4,5), (6,8,10), (5,12,13), (9,12,15), (8,15,17), (12,16,20). There are six triples, without counting a swap of the legs again. The primitive triples are (3,4,5), (5,12,13), (8,15,17). Both (30,40,50) and (300,400,500) are triples, scaled by 10 and 100 from (3,4,5). In general, (ka)² + (kb)² = k²(a²+b²) = (kc)², so every positive integer scale k gives another triple.

Five scaled versions of each primitive triple (page 49)

Original×2×3×4×5×6
3, 4, 56, 8, 109, 12, 1512, 16, 2015, 20, 2518, 24, 30
5, 12, 1310, 24, 2615, 36, 3920, 48, 5225, 60, 6530, 72, 78
8, 15, 1716, 30, 3424, 45, 5132, 60, 6840, 75, 8548, 90, 102

These scaled triples are not primitive: each has the common factor k > 1. Conversely, dividing a triple by a common factor f gives another integer triple, since (a/f)² + (b/f)² = (c/f)². Divide by the greatest common factor to obtain a primitive triple.

Why the combining construction makes a square (pages 42–46)

The four right triangles are congruent, so their hypotenuses are equal. Their acute angles are x and 90° − x. Each angle of the proposed square is 180° − x − (90° − x) = 90°. Thus its four equal sides form a square. In the textbook rearrangement, triangles T and U replace congruent triangles W and X while region V is shared. Therefore area(T + U + V) = area(W + X + V) = a² + b² = c².

The lotus problem (page 52)

Let the depth be x. The original stem is x + 1 units long. When its tip reaches the water 3 units away, the model gives a right triangle with legs x and 3 and hypotenuse x + 1. Hence x² + 9 = (x + 1)², giving 9 = 2x + 1 and x = 4 units. The unchanged stem length is 5 units.

A lotus stem stands one unit above water and x units below. Its tip moves three units horizontally to the water surface, forming a right triangle with legs x and three and hypotenuse x plus one. The depth is four units.
Page 52, Līlāvatī example: the model assumes a vertical original stem, a fixed base and unchanged straight stem length.

Find the Colours! (page 54)

Open the box labelled RED. Its contents must be blue or green because every label is wrong. If it contains blue, the GREEN-labelled box must contain red and the BLUE-labelled box green. If it contains green, the BLUE-labelled box must contain red and the GREEN-labelled box blue. One opened box determines all three contents.

Common Mistakes to Avoid

Watch out for these

  • Doubling the side to double the area — that multiplies the area by 4, not 2. To double area, build on the diagonal.
  • Writing √(a2 + b2) = a + b. The square root of a sum is not the sum of the roots; you must add the squares first, then take the root.
  • Adding when you should subtract: to find a leg, use leg = √(c2 − b2); only the hypotenuse uses c = √(a2 + b2).
  • Treating √2 as a fraction or a terminating decimal — it is non-terminating and non-repeating (irrational).
  • For a rhombus, using the full diagonals as legs. Use half of each diagonal, since the diagonals bisect each other.
  • Calling a triple primitive without checking the common factor — (6, 8, 10) and (9, 12, 15) are not primitive.
  • Leaving a surd un-simplified or giving wrong decimal bounds — check that the bound squared brackets the number (e.g. 8.42 < 72 < 8.52).

Practice MCQs & Assertion–Reason

1. In a right triangle, the side opposite the right angle is called the:

(a) base    (b) altitude    (c) hypotenuse    (d) median

2. The Baudhayana-Pythagoras Theorem states that for a right triangle with hypotenuse c:

(a) a + b = c    (b) a2 + b2 = c2    (c) a2 − b2 = c2    (d) a2 + b2 = 2c2

3. The hypotenuse of an isosceles right triangle with equal sides 5 is:

(a) 10    (b) √10    (c) 5√2    (d) 25

4. The diagonal of a square of side 1 unit has length:

(a) 1    (b) √2    (c) 2    (d) ½

5. Which of the following is a Baudhayana (Pythagorean) triple?

(a) (2, 3, 4)    (b) (5, 12, 13)    (c) (4, 5, 6)    (d) (6, 7, 8)

6. A square built on the diagonal of a given square has area that is … the original:

(a) the same as    (b) double    (c) four times    (d) half

7. The third side of a right triangle with hypotenuse 17 and one leg 8 is:

(a) 9    (b) 12    (c) 15    (d) 25

8. Which triple is primitive?

(a) (6, 8, 10)    (b) (9, 12, 15)    (c) (8, 15, 17)    (d) (12, 16, 20)

9. The number √2 lies between:

(a) 1.313 and 1.314    (b) 1.414 and 1.415    (c) 1.515 and 1.516    (d) 2 and 3

10. The famous statement “an + bn = cn has no positive integer solution for integer n > 2”, proven in 1994, is:

(a) Euclid’s Theorem    (b) Baudhayana’s Theorem    (c) Fermat’s Last Theorem    (d) the Midpoint Theorem

Answer key: 1-(c), 2-(b), 3-(c), 4-(b), 5-(b), 6-(b), 7-(c), 8-(c), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: In any right triangle, the hypotenuse is the longest side.

Reason: Since c2 = a2 + b2, c2 is greater than both a2 and b2.

A-R 2. Assertion: √2 cannot be written as a fraction m/n of two counting numbers.

Reason: In a perfect square, each prime factor occurs an even number of times.

A-R 3. Assertion: (6, 8, 10) is a primitive Baudhayana triple.

Reason: (6, 8, 10) satisfies 62 + 82 = 102.

A-R 4. Assertion: A square built on the diagonal of a square has double the area of the original.

Reason: The original square is made of 2 congruent triangles and the new one of 4 such congruent triangles.

A-R 5. Assertion: A grid (lattice) square of area 3 sq. units can be drawn with its vertices on dots.

Reason: The area of a lattice square is always a2 + b2 for whole numbers a and b.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(D).

Quick Revision Summary

  • The Baudhayana-Pythagoras Theorem: for a right triangle, a2 + b2 = c2, where c is the hypotenuse (the longest side).
  • Building a square on the diagonal doubles a square’s area; the inside tilted square through the midpoints halves it.
  • Isosceles right triangle: c2 = 2a2, so c = a√2; the diagonal of a square of side s is s√2.
  • √2 lies between 1.414 and 1.415; it is non-terminating and cannot be written as a fraction (irrational).
  • Baudhayana (Pythagorean) triples: integers (a, b, c) with a2 + b2 = c2, e.g. (3, 4, 5), (5, 12, 13), (8, 15, 17). If (a, b, c) is a triple, so is (ka, kb, kc).
  • A primitive triple has no common factor > 1; every triple is primitive or a scaled version of a primitive triple.
  • Fermat’s Last Theorem: an + bn = cn has no positive-integer solution for integer n > 2 (proven by Andrew Wiles, 1994).

How to score full marks in this chapter

Always identify the hypotenuse first (it is opposite the right angle) and write a2 + b2 = c2 before substituting. Keep answers in surd form (e.g. √74, 5√2) unless a decimal or bound is asked, and when bounds are needed, show the two squares that bracket the number (n2 < N < (n+1)2). For rhombus and equilateral-triangle problems, draw the right triangle using half-diagonals or the bisected base first.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 9 about?

It is The Baudhayana-Pythagoras Theorem — Chapter 2 of Ganita Prakash Part II (the 9th chapter of the Class 8 course). It covers doubling/halving a square, the hypotenuse of an isosceles right triangle, √2, the theorem a2 + b2 = c2, Baudhayana (Pythagorean) triples and real-life applications.

What is the Baudhayana-Pythagoras Theorem?

For a right-angled triangle with legs a and b and hypotenuse c, a2 + b2 = c2. The textbook discusses its statement in Baudhayana’s Śulba-Sūtra (dated there to c. 800 BCE); it is also called the Pythagorean Theorem.

What is a Baudhayana (Pythagorean) triple?

A set of three positive integers (a, b, c) with a2 + b2 = c2, such as (3, 4, 5), (5, 12, 13) and (8, 15, 17). Scaling any triple by a positive integer gives another triple, so there are infinitely many.

Are these Class 8 Maths Ganita Prakash Chapter 9 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash Part II textbook for 2026–27, with every calculation verified.

Textbook reference: NCERT Ganita Prakash, Chapter 9 (official PDF). ClearStudy diagrams and worked explanations are original teaching aids.

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