NCERT Solutions Class 8 Maths Part 2 Ch 2 Baudhayana Theorem

Answers to every question in Part 2 Chapter 2, The Baudhayana-Pythagoras Theorem, of Ganita Prakash Grade 8 (first edition, December 2025, used in 2026-27): the questions in the text, every Figure it Out, Math Talk and Try This, and the Find the Colours puzzle. Page numbers follow the book (pages 33–54). Square roots that are not whole numbers are given to two decimal places. Short notes: Part 2 Chapter 2 notes.

Mind map of Chapter 2: doubling and halving, the number √2, isosceles right triangles, the theorem, Baudhayana triples, and beyond — The whole chapter on one page

Doubling and halving a square (pages 33–37)

Will doubling the side double the area? No: the new square has 2 × 2 = 4 times the area.

Why does the square on the diagonal have double the area? East-west and north-south lines through the corners cut the original square into 2 small right triangles and the square on the diagonal into 4, all congruent, so its area is double.

Doubling a square: the square on the diagonal is made of 4 identical triangles, the original of 2 — Doubling a square

Why do the extended sides pass through the corners of the dotted square? A diagonal of a square bisects its corner angle. The sides of the original square meet the diagonal at 45°, so they bisect the 90° corners of the dotted square, and the bisector of a corner of a square passes through the opposite corner.

Why are the small triangles congruent? Each is half of an identical small square, cut along a diagonal: all have two equal sides (half the diagonal) meeting at a right angle, so they are congruent by SAS.

Paper activity. Pieces 5, 6, 7 and 8, each a quarter of Square 2 cut along both diagonals, fit along the four sides of Square 1 to make a square with double the area.

Why is the inside square half the larger one? The lines through the midpoints cut the large square into 8 congruent triangles; the tilted square contains 4 of them.

Will a square with half the side have half the area? No, only a quarter: 4 such squares fill the original.

Why is PQRS a square with half the area? The four corner triangles are congruent (SAS: two halves of sides meeting at a right angle), so PQ = QR = RS = SP. Each corner triangle is isosceles and right-angled, so its other angles are 45°, and each angle of PQRS is 180° − 45° − 45° = 90°. Joining PR and QS cuts PQRS into four triangles congruent to the corner flaps, so the flaps exactly cover PQRS: it is half the paper.

Halving a square by folding: the corners fold onto square PQRS, which has half the area — Halving a square by folding

The hypotenuse and √2 (pages 37–41)

Hypotenuse of the 1-1 right triangle: √2, since the square on it has area 2. Bounds: 1 < √2 < 2; 1.4 < √2 < 1.5; 1.41 < √2 < 1.42; 1.414 < √2 < 1.415. A terminating decimal? No: its square would end in a non-zero digit. A fraction? No: 2n² = m² is impossible.

Figure it Out (pages 39–40)

1. Cut each square along one diagonal to get four right triangles. Put their right angles together at a centre point with the long sides outward: they form a square whose side is the diagonal, with double the area of either square.

Figure it Out 1: four triangles from two squares form one square with double the area — Figure it Out 1: four triangles

2. c² = 2a². (i) a = 3: c = √18 ≈ 4.24, so 4.2 < c < 4.3 (4.2² = 17.64, 4.3² = 18.49). (ii) a = 4: c = √32 ≈ 5.66, so 5.6 < c < 5.7. (iii) a = 6: c = √72 ≈ 8.49, so 8.4 < c < 8.5. (iv) a = 8: c = √128 ≈ 11.31, so 11.3 < c < 11.4. (v) a = 9: c = √162 ≈ 12.73, so 12.7 < c < 12.8.

3. The square on the hypotenuse has area 10² = 100, which is double the square on a side, so a² = 50 and each side is √50 ≈ 7.07 (between 7.0 and 7.1).

Combining two squares (pages 41–47)

Does the method agree for two equal squares? Yes: with equal shorter sides, the hypotenuse is the diagonal of the square, which is exactly the earlier doubling rule.

Why is the 4-sided figure a square? Its four sides are hypotenuses of congruent triangles, so they are equal. At each corner, the angles x and 90° − x of two of the triangles sit on a straight line (180°), leaving 180° − 90° = 90°. Four equal sides and four right angles make a square.

Two squares become one: two cuts and two moved triangles turn squares of sides a and b into a square of side c — Two squares become one

Figure it Out (page 47)

1. 5² + 12² = 169, so the hypotenuse is 13 cm.

2. 17² − 8² = 289 − 64 = 225, so the third side is 15 cm.

3. Triple: make a right triangle with shorter sides equal to the side a of the square and its diagonal a√2; the square on its hypotenuse has area a² + 2a² = 3a². Five times: use shorter sides a and 2a; the square on the hypotenuse has area a² + 4a² = 5a².

Three times and five times a square: right triangles with sides a and a√2, and a and 2a — Three times and five times a square

4. (i) c = √74 ≈ 8.60. (ii) c = √208 ≈ 14.42. (iii) b = √(225 − 81) = √144 = 12. (iv) c = √193 ≈ 13.89. (v) c = √(2.25 + 12.25) = √14.5 ≈ 3.81.

Baudhayana triples (pages 48–50)

Triples up to 20: (3, 4, 5), (6, 8, 10), (5, 12, 13), (9, 12, 15), (8, 15, 17) and (12, 16, 20). (30, 40, 50) and (300, 400, 500) are triples: 900 + 1600 = 2500 and 90,000 + 1,60,000 = 2,50,000. Pattern: (6, 8, 10), (9, 12, 15) and (12, 16, 20) are (3, 4, 5) times 2, 3 and 4. (5, 12, 13) is primitive; the other primitive triples up to 20 are (3, 4, 5) and (8, 15, 17). Their scaled versions are below; none is primitive, since k divides all three numbers.

Primitive triples (3, 4, 5), (5, 12, 13) and (8, 15, 17) with five scaled versions each — Primitive triples and their multiples

Is (a/f, b/f, c/f) a triple? Yes: dividing a² + b² = c² by f² gives (a/f)² + (b/f)² = (c/f)². For (9, 12, 15) with f = 3: (3, 4, 5), and 9 + 16 = 25. The nth odd number is 2n − 1, and the first n − 1 odd numbers add up to (n − 1)².

Figure it Out (page 50)

1. From the odd squares 49, 81, 121, 169 and 225: (7, 24, 25), (9, 40, 41), (11, 60, 61), (13, 84, 85) and (15, 112, 113).

Triples from odd squares: 9, 25, 49 and more give (3, 4, 5), (5, 12, 13), (7, 24, 25) and others — Triples from odd squares

2. No. One shorter side, n − 1, is one less than the hypotenuse n, and consecutive numbers have no common factor greater than 1, so every triple it makes is primitive.

3. Yes, for example (8, 15, 17), (20, 21, 29), (12, 35, 37) and (28, 45, 53): in these, no side is one less than the hypotenuse.

Applications (pages 52–54)

Lilavati’s lotus: 3² + x² = (x + 1)² gives x = 4, so the lake is 4 units deep.

Bhaskaracharya’s lotus: a right triangle with sides x and 3 and hypotenuse x + 1 shows the lake is 4 units deep — Bhaskaracharya’s lotus

Figure it Out (pages 52–54)

1. 5√2 = √50 ≈ 7.07 cm.

2. Sides 7 and 9 round the right angle: hypotenuse √130 ≈ 11.40. Sides 4 and 10: hypotenuse √116 ≈ 10.77. Side 40, hypotenuse 41: third side √(1681 − 1600) = 9. Side 10, hypotenuse √200: third side √(200 − 100) = 10. Sides 10 and √150: hypotenuse √250 ≈ 15.81. Side 27, hypotenuse 45: third side √(2025 − 729) = 36.

3. The diagonals bisect each other at right angles, giving right triangles with sides 12 and 35: side = √(144 + 1225) = √1369 = 37 units.

Question 3: the rhombus’s half-diagonals 12 and 35 give a side of 37 — Question 3: a rhombus

4. Yes. c² = a² + b² is greater than a² and greater than b², so c is greater than both a and b.

5. True: divide any non-primitive triple by the greatest common factor of its numbers to get a primitive triple, of which it is a scaled version.

6. For example 3 × 4 (diagonal 5), 6 × 8 (10), 5 × 12 (13), 8 × 15 (17) and 7 × 24 (25).

7. The area needed is 7² − 5² = 49 − 25 = 24. Draw a right triangle with hypotenuse 7 and one side 5: draw a 5-unit segment, a perpendicular at one end, and an arc of radius 7 from the other end to cut it. The perpendicular side is √24, and the square on it has area 24.

8. (i) (a) Yes: a tilted square whose side joins dots 1 across and 1 up (side² = 1 + 1 = 2). (b) No: 3 is not a sum of two squares. (c) Yes: a 2 × 2 square. (d) Yes: a tilted square with sides going 1 across and 2 up (side² = 1 + 4 = 5). (ii) Exactly the numbers that are sums of two squares p² + q²: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, and so on.

Question 8: squares of area 1, 2, 4 and 5 on a dot grid — Question 8: squares on a dot grid

9. The altitude splits the triangle into two congruent right triangles (RHS: the hypotenuses are equal sides of 6 and the altitude is shared), so it bisects the base into 3 and 3. Then h² = 6² − 3² = 27, h = 3√3 ≈ 5.20, and the area is ½ × 6 × 3√3 = 9√3 ≈ 15.59 square units.

Question 9: the altitude of an equilateral triangle of side 6 is √27, and its area is 9√3 — Question 9: an equilateral triangle

Find the Colours (page 54)

Open the box labelled RED. Its label is wrong, so it holds blue or green balls. If it holds blue, then the box labelled BLUE cannot hold blue (wrong label, and blue is taken), so it holds green, and the box labelled GREEN holds red. If it holds green, then the box labelled GREEN cannot hold green, so it holds blue, and the box labelled BLUE holds red. One box is enough.

Short notes and more practice are on the topic pages: the theorem, Pythagorean triples and the square root of 2.

NCERT Solutions Class 8 Maths Part 2 Chapter 2: The Baudhayana-Pythagoras Theorem