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Class 8 Ganita Prakash · Part 2 · Chapter 2
Class 8 Ganita Prakash · Part 2 · Chapter 2
The Baudhฤyana-Pythagoras Theorem
Page-wise textbook solutions (pages 33–54, Part 2) · Every step · Figures · Common mistakes · Tips
๐️ Key ideas of this chapter
- The square on the diagonal of a square has double the area (Baudhฤyana, ลulba-Sลซtra 1.9).
- Isosceles right triangle: \(c^2 = 2a^2\), so \(c = a\sqrt{2}\). \(\sqrt{2} = 1.41421356\ldots\) — not a terminating decimal and not a fraction.
- Baudhฤyana–Pythagoras theorem: \(a^2 + b^2 = c^2\) (c = hypotenuse, the longest side).
- Triples: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25)… Scale by k; primitive = HCF 1.
- Fermat's Last Theorem: \(x^n + y^n = z^n\) has no solution for n > 2 (proved by Andrew Wiles, 1994).
๐ Textbook Page 33
๐ Page 33Think
If we double the side of a square, does the area double?
1Side a → area a². Side 2a → area (2a)² = 4a².
2Picture: a 2 × 2 grid of the original square.
✅ Answer: No — it becomes 4 times the area.
⚠️ Common mistake: Thinking 'double side = double area'.
๐ Textbook Page 34
๐ Page 34Why?
Why does the square drawn on the diagonal have double the area? Why are all the small triangles congruent?
1Draw the other diagonal and the east–west and north–south lines through the centre.
2The original square = 2 small triangles; the square on its diagonal = 4 small triangles.
3Each small triangle is a right isosceles triangle whose two equal sides are equal to the side of the original square, with a 90° angle between them → congruent by SAS.
✅ Answer: The new square holds 4 equal triangles, the old one 2 → double the area.
๐ Page 34Math Talk
Why do the extended sides of the original square pass through the vertices of the dotted square?
1The side of the original square makes 45° with the diagonal (a diagonal bisects the 90° corner).
2So at a vertex of the dotted square, the extended side cuts the 90° angle into 45° + 45° — it bisects the angle.
3In a square, the bisector of a corner angle is a diagonal, which passes through the opposite vertex.
✅ Answer: Because these lines are angle bisectors, i.e. diagonals of the dotted square.
๐ Textbook Page 35
๐ Page 35Sequence
Explain the sequence of squares made of 2, 4 and 8 small triangles.
1Each new square is drawn on the diagonal of the previous one, so its area doubles: 2 → 4 → 8 triangles.
Each square has double the area of the one before.
✅ Answer: Areas in the ratio 1 : 2 : 4.
๐ Page 35Activity
Cut Square 2 along both diagonals into pieces 5, 6, 7, 8 and place them around Square 1. What do you get?
1Each of 5, 6, 7, 8 is a quarter of a square (a right isosceles triangle).
2Put the long side of each piece on a side of Square 1, pointing outward.
3The four outer corners form a tilted square.
✅ Answer: A square with double the area of Square 1 (Square 1 + 4 quarters = 2 squares).
๐ Textbook Page 36
๐ Page 36Think
How do we make a square with half the area? Why is the tilted inside square half?
1Join the midpoints of the four sides.
2The east–west and north–south lines cut the big square into 8 equal triangles; the inner square contains 4 of them.
✅ Answer: Join the midpoints: the inner square has half the area.
๐ Page 36Think
Will a square with half the side length have half the area? How many such squares fill the original?
1Half side → area \((\tfrac{a}{2})^2 = \tfrac{a^2}{4}\).
✅ Answer: No — it has a quarter of the area. 4 such squares fill the original.
๐ Textbook Page 37
๐ Page 37Why?
Why is PQRS (made by folding corners to the centre) a square with half the area?
1Join QS and PR. They pass through the centre and are equal (each = side of the paper) and perpendicular.
2The four triangles around the centre are congruent (SAS: half-diagonals equal, 90° between them), so PQ = QR = RS = SP.
3Each angle at a vertex is 45° + 45° = 90°, so PQRS is a square.
4Each folded corner flap covers exactly one of the triangles, so PQRS = half the paper.
✅ Answer: PQRS is a square with half the area.
๐ Page 37Think
Find the hypotenuse of an isosceles right triangle with equal sides 1.
1Two such triangles make a 1 × 1 square PEAR.
2The square REST on the hypotenuse has area 2 × 1 = 2.
3c × c = 2 → c = √2.
✅ Answer: √2 units
๐ Textbook Page 38
๐ Page 38Bounds for √2
Is √2 more than 1 or 2? Find closer bounds.
11² = 1 < 2 and 2² = 4 > 2 → 1 < √2 < 2
21.4² = 1.96, 1.5² = 2.25 → 1.4 < √2 < 1.5
31.41² = 1.9881, 1.42² = 2.0164 → 1.41 < √2 < 1.42
41.414² = 1.999396, 1.415² = 2.002225 → 1.414 < √2 < 1.415
✅ Answer: √2 = 1.41421356…
๐ Page 38Think
Can a terminating decimal have square exactly 2?
1If the last non-zero digit is d, the square's last digit comes from d × d.
2d × d never ends in 0 when d is 1–9 (1, 4, 9, 6, 5, 6, 9, 4, 1).
3So the square has a non-zero digit after the decimal point — it can't be 2.000…
✅ Answer: No. √2 has a non-terminating decimal expansion.
๐ Textbook Page 39
๐ Page 39Try This
Can √2 be written as a fraction m/n?
1Suppose \(\sqrt{2} = \tfrac{m}{n}\). Then \(2n^2 = m^2\).
2In a square number, every prime appears an even number of times.
3In \(2n^2\) the prime 2 appears an odd number of times, in \(m^2\) an even number — impossible.
✅ Answer: No (Euclid's proof). √2 is not a fraction.
๐ Page 39Figure it Out · Q1
Two identical squares are each cut along a diagonal (pieces 1, 2 and 3, 4). Arrange them into a square of double the area.
1Each piece is a right isosceles triangle (half a square).
2Put the four right-angle corners together at one point.
3The four hypotenuses form the outside — a square with side = the diagonal.
✅ Answer: A tilted square made of all 4 triangles — double the area of either square.
๐ Page 39Figure it Out · Q2
Equal sides of an isosceles right triangle are given. Find the hypotenuse and bounds with one decimal place: (i) 3 (ii) 4 (iii) 6 (iv) 8 (v) 9
1\(c^2 = 2a^2\)
2(i) \(c = \sqrt{18}\): 4.2² = 17.64, 4.3² = 18.49 → 4.2 < c < 4.3
3(ii) \(\sqrt{32}\): 5.6² = 31.36, 5.7² = 32.49 → 5.6 < c < 5.7
4(iii) \(\sqrt{72}\): 8.4² = 70.56, 8.5² = 72.25 → 8.4 < c < 8.5
5(iv) \(\sqrt{128}\): 11.3² = 127.69, 11.4² = 129.96 → 11.3 < c < 11.4
6(v) \(\sqrt{162}\): 12.7² = 161.29, 12.8² = 163.84 → 12.7 < c < 12.8
✅ Answer: √18, √32, √72, √128, √162 (≈ 4.24, 5.66, 8.49, 11.31, 12.73)
๐ Textbook Page 40
๐ Page 40Figure it Out · Q3
The hypotenuse of an isosceles right triangle is 10. Find the other two sides.
1Square on the hypotenuse has area 100 = 2 × (square on a side).
2\(a^2 = 50\) → \(a = \sqrt{50}\)
37² = 49, 8² = 64 → between 7 and 8 (≈ 7.07)
✅ Answer: Each side is √50 ≈ 7.07
๐ Page 40Example 1
Hypotenuse of an isosceles right triangle with equal sides 12?
1\(c^2 = 2 \times 144 = 288\)
216² = 256, 17² = 289
✅ Answer: \(c = \sqrt{288}\), between 16 and 17 (≈ 16.97)
๐ Textbook Page 41
๐ Page 41Example 2
The hypotenuse is √72. Find the other two sides.
1\(72 = 2a^2\) → \(a^2 = 36\) → a = 6
✅ Answer: 6 and 6
๐ Textbook Page 42
๐ Page 42Think
Does Baudhฤyana's method agree with doubling when the two squares are the same size?
1Equal squares of side a → right triangle with legs a and a.
2Its hypotenuse is the diagonal of the square, and the square on it has area 2a² — the same as before.
✅ Answer: Yes — it is exactly the doubling construction.
๐ Textbook Page 44
๐ Page 44Why a square?
Explain why all angles of the new 4-sided figure (on the hypotenuse) are right angles.
1The four right triangles are congruent (legs a and b), so all four sides are c.
2In each triangle the two acute angles are x and 90° − x.
3At each corner of the new figure, a straight angle (180°) = x + (corner) + (90° − x) → corner = 90°.
✅ Answer: All sides equal and all angles 90° → it is a square of side c.
๐ Textbook Page 45
๐ Page 45Activity
Join squares a and b, make two cuts and rearrange the three pieces into one square. What does it show?
1Mark M on the base at distance a from the right end (b − a from the join).
2Cut from M to the top-left corner and from M to the top-right corner — both cuts have length c.
3Move the two triangles to the top: they fill the tilted square of side c.
✅ Answer: c² = a² + b² — the two squares together make the square on the hypotenuse.
๐ Textbook Page 47
๐ Page 47Example
Shorter sides 3 cm and 4 cm. Predict the hypotenuse.
1\(c^2 = 3^2 + 4^2 = 9 + 16 = 25\)
2c = 5
✅ Answer: 5 cm
๐ Page 47Figure it Out · Q1
Shorter sides 5 cm and 12 cm. Hypotenuse?
1\(c^2 = 25 + 144 = 169\)
✅ Answer: 13 cm
๐ Page 47Figure it Out · Q2
One short side 8 cm, hypotenuse 17 cm. Third side?
1\(b^2 = 17^2 - 8^2 = 289 - 64 = 225\)
✅ Answer: 15 cm
⚠️ Common mistake: Adding 17² + 8² — when the hypotenuse is known, SUBTRACT.
๐ Page 47Figure it Out · Q3
Construct a square with triple the area of a given square (side a). Five times?
1Triple: draw the diagonal d of the square (d² = 2a²). Make a right triangle with legs a and d. Its hypotenuse² = a² + 2a² = 3a².
2Five times: make a right triangle with legs a and 2a. Hypotenuse² = a² + 4a² = 5a².
✅ Answer: Square on the hypotenuse of (a, a√2) → 3a²; of (a, 2a) → 5a².
๐ Page 47Figure it Out · Q4
Find the missing side (c is the hypotenuse): (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
1(i) \(c^2 = 25 + 49 = 74\) → √74 ≈ 8.60
2(ii) \(c^2 = 64 + 144 = 208\) → √208 ≈ 14.42
3(iii) \(b^2 = 225 - 81 = 144\) → 12
4(iv) \(c^2 = 49 + 144 = 193\) → √193 ≈ 13.89
5(v) \(c^2 = 2.25 + 12.25 = 14.5\) → √14.5 ≈ 3.81
✅ Answer: √74, √208, 12, √193, √14.5
๐ Textbook Page 48
๐ Page 48Math Talk
List all Baudhฤyana triples with numbers up to 20. Are (30, 40, 50) and (300, 400, 500) triples?
1Check pairs with \(a^2 + b^2\) a perfect square.
2(3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16, 20), (5, 12, 13), (8, 15, 17)
330² + 40² = 2500 = 50²; 300² + 400² = 250000 = 500²
✅ Answer: Six triples up to 20; yes, both are triples (10 × and 100 × (3, 4, 5)).
๐ก Tip: The book lists only the (3, 4, 5) family first; (5, 12, 13) and (8, 15, 17) are also ≤ 20.
๐ Textbook Page 49
๐ Page 49Think
Is (5, 12, 13) primitive? Other primitive triples ≤ 20? Make 5 scaled versions. Are they primitive?
1HCF(5, 12, 13) = 1 → primitive.
2Primitive ≤ 20: (3, 4, 5), (5, 12, 13), (8, 15, 17).
3(5, 12, 13) × 2, 3, 4, 5, 6: (10, 24, 26), (15, 36, 39), (20, 48, 52), (25, 60, 65), (30, 72, 78)
4(8, 15, 17) × 2…6: (16, 30, 34), (24, 45, 51), (32, 60, 68), (40, 75, 85), (48, 90, 102)
✅ Answer: Scaled versions are never primitive (they have the common factor k).
๐ Page 49Think
If a triple has a common factor f, is (a/f, b/f, c/f) a triple? Check (9, 12, 15).
1\((\tfrac{a}{f})^2 + (\tfrac{b}{f})^2 = \tfrac{a^2 + b^2}{f^2} = \tfrac{c^2}{f^2} = (\tfrac{c}{f})^2\)
2(9, 12, 15) ÷ 3 = (3, 4, 5) ✔
✅ Answer: Yes — dividing by a common factor keeps it a triple.
๐ Textbook Page 50
๐ Page 50Think
What is the sum of the first (n − 1) odd numbers? How do odd squares give triples?
1Sum of first (n − 1) odd numbers = (n − 1)².
2So \((n-1)^2 + (2n - 1) = n^2\).
3If 2n − 1 is itself a square k², we get a triple: \((n-1)^2 + k^2 = n^2\). E.g. 9 → n = 5: 4² + 3² = 5².
✅ Answer: (n − 1)²
๐ Page 50Figure it Out · Q1
Find 5 more Baudhฤyana triples using odd squares.
149 = 2 × 25 − 1 → n = 25 → (7, 24, 25)
281 → n = 41 → (9, 40, 41)
3121 → n = 61 → (11, 60, 61)
4169 → n = 85 → (13, 84, 85)
5225 → n = 113 → (15, 112, 113)
✅ Answer:
| Odd square | n | Triple |
|---|---|---|
| 49 (7²) | 25 | (7, 24, 25) |
| 81 (9²) | 41 | (9, 40, 41) |
| 121 (11²) | 61 | (11, 60, 61) |
| 169 (13²) | 85 | (13, 84, 85) |
| 225 (15²) | 113 | (15, 112, 113) |
๐ก Tip: Shortcut: for odd k, the triple is \((k, \tfrac{k^2-1}{2}, \tfrac{k^2+1}{2})\).
๐ Page 50Figure it Out · Q2
Does this method give non-primitive triples?
1The method gives b = n − 1 and c = n — consecutive numbers.
2Any common factor of b and c divides c − b = 1.
3So the HCF is 1.
✅ Answer: No — all triples from this method are primitive.
๐ Page 50Figure it Out · Q3
Are there primitive triples this method cannot give?
1This method always has c − b = 1.
2(8, 15, 17): 17 − 15 = 2, 17 − 8 = 9 → not from this method.
3(20, 21, 29), (12, 35, 37), (28, 45, 53) are other examples.
✅ Answer: Yes, e.g. (8, 15, 17) and (20, 21, 29).
๐ Textbook Page 52
๐ Page 52Lฤซlฤvatฤซ lotus
A lotus tip is 1 unit above water; the wind bends it to touch the water 3 units away. Depth of the lake?
1Depth x, stem x + 1 (assume the stem is vertical at first).
2\(3^2 + x^2 = (x+1)^2\) → \(9 + x^2 = x^2 + 2x + 1\)
39 = 2x + 1 → x = 4
✅ Answer: 4 units
๐ Page 52Figure it Out · Q1
Diagonal of a square with side 5 cm?
1\(d^2 = 25 + 25 = 50\)
✅ Answer: \(5\sqrt{2} = \sqrt{50}\) ≈ 7.07 cm
๐ Textbook Page 53
๐ Page 53Figure it Out · Q2
Find the missing sides: (a) legs 7, 9 (b) legs 4, 10 (c) leg 40, hypotenuse 41 (d) leg 10, hypotenuse √200 (e) legs 10, √150 (f) leg 27, hypotenuse 45
1(a) \(c^2 = 49 + 81 = 130\) → √130 ≈ 11.40
2(b) \(c^2 = 16 + 100 = 116\) → √116 ≈ 10.77
3(c) \(b^2 = 1681 - 1600 = 81\) → 9
4(d) \(b^2 = 200 - 100 = 100\) → 10
5(e) \(c^2 = 100 + 150 = 250\) → √250 = 5√10 ≈ 15.81
6(f) \(b^2 = 2025 - 729 = 1296\) → 36
✅ Answer: √130, √116, 9, 10, √250, 36
๐ก Tip: First find the right angle: the side opposite it is the hypotenuse (the longest side).
๐ Page 53Figure it Out · Q3
Side of a rhombus with diagonals 24 and 70?
1Diagonals bisect each other at 90° → half-diagonals 12 and 35.
2\(s^2 = 144 + 1225 = 1369\)
✅ Answer: 37 units
๐ Page 53Figure it Out · Q4
Is the hypotenuse the longest side? Justify.
1\(c^2 = a^2 + b^2\), and \(b^2 \gt 0\), so \(c^2 \gt a^2\) → c > a.
2Similarly c > b.
✅ Answer: Yes, always.
๐ Page 53Figure it Out · Q5
True or false: every Baudhฤyana triple is primitive or a scaled version of a primitive triple.
1Divide the triple by the HCF of its numbers.
2The result is still a triple (shown on p49) and has HCF 1 → primitive.
3So the original = HCF × primitive triple.
✅ Answer: True
๐ Page 53Figure it Out · Q6
Give 5 rectangles with integer sides and diagonals.
1Use triples: sides a, b and diagonal c.
✅ Answer: 3 × 4 (5), 6 × 8 (10), 5 × 12 (13), 8 × 15 (17), 7 × 24 (25)
๐ Page 53Figure it Out · Q7
Construct a square with area = 7² − 5².
1Area needed = 49 − 25 = 24 → side √24.
2Draw a 5-unit line, a perpendicular at one end, and an arc of radius 7 from the other end to meet it.
3The perpendicular side is √(7² − 5²) = √24. Build the square on it.
✅ Answer: A square of side √24 (the leg of a right triangle with hypotenuse 7 and leg 5).
๐ Page 53Figure it Out · Q8
On a dot grid, make squares of area 2, 3, 4, 5. Which integer areas are possible on an endless grid?
1A tilted square with side 'across p, up q' has area \(p^2 + q^2\).
22 = 1² + 1² ✔; 4 = 2² + 0² ✔; 5 = 2² + 1² ✔
33 is not a sum of two squares ✘
✅ Answer: (i) 2, 4, 5 possible; 3 impossible. (ii) Exactly the numbers that are sums of two squares: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, …
๐ Textbook Page 54
๐ Page 54Figure it Out · Q9
Area of an equilateral triangle of side 6.
1The altitude makes two congruent right triangles (RHS), so it bisects the base: 3 and 3.
2\(h^2 = 6^2 - 3^2 = 27\) → \(h = \sqrt{27} = 3\sqrt{3}\)
3Area = \(\tfrac{1}{2} \times 6 \times 3\sqrt{3} = 9\sqrt{3}\)
✅ Answer: \(9\sqrt{3}\) ≈ 15.59 sq units
๐ Page 54Puzzle: Find the Colours!
Three boxes labelled RED, BLUE, GREEN are ALL wrongly labelled. Open only one box to fix all labels.
1Open the box labelled RED (any box works). It is not red — say it has blue balls.
2The box labelled GREEN is not green, and blue is taken → it must be red.
3The box labelled BLUE is then green.
✅ Answer: Open one box; since every label is wrong, the other two boxes are fixed automatically.
๐ Where do we use this?
- Builders check right angles with a 3-4-5 rope, ladders against walls, TV screen sizes (diagonal), distance on maps.
๐ Link to higher classes
- Class 9: irrational numbers, Heron's formula. Class 10: distance formula, trigonometry. Class 11: 3D distances.
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