In-text — Scaling and Primitive Triples — The Baudhāyana–Pythagoras Theorem

1. 12 marksGanita Prakash Cl-8 Part 2, Math Talk, page 48

   Is (30, 40, 50) a Baudhayana triple? Is (300, 400, 500)? Is there an unending sequence of Baudhayana triples?

   Hint. Look for a common factor before you start squaring.

   Direct check for (30, 40, 50). 30² + 40² = 900 + 1600 = 2500, and 50² = 2500. They match, so yes.

   Direct check for (300, 400, 500). 300² + 400² = 90000 + 160000 = 250000, and 500² = 250000. They match, so yes again.

   The reason, without any squaring. Both are (3, 4, 5) scaled up — the first by 10, the second by 100. The chapter proves that if (a, b, c) is a Baudhayana triple then so is (ka, kb, kc), because (ka)² + (kb)² = k²a² + k²b² = k²(a² + b²) = k²c² = (kc)². So once you spot the common factor, the answer is immediate.

   An unending sequence. Since k can be any positive integer, (3k, 4k, 5k) gives (3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16, 20), (15, 20, 25), … and this list never stops. Therefore there are infinitely many Baudhayana triples.

   But notice what this does not give. Every triple in that list is a scaled copy of one small triple. It tells us there are infinitely many triples, but not that there are infinitely many genuinely different ones — which is why the chapter goes on to hunt for primitive triples.

   ✦ Yes to both — each is (3, 4, 5) scaled by 10 and by 100. And since (3k, 4k, 5k) is a triple for every positive integer k, there are infinitely many Baudhayana triples.

2. 23 marksGanita Prakash Cl-8 Part 2, Math Talk, page 48

   List down all the Baudhayana triples with numbers less than or equal to 20.

   Hint. The book's own list is not complete — search for triples that are not multiples of (3, 4, 5) as well.

   Search systematically. Take every pair a ≤ b with both at most 20, work out a² + b², and keep the pair only when the sum is a perfect square whose root is also at most 20.

   The multiples of (3, 4, 5). Scaling by k = 1, 2, 3, 4 gives (3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16, 20). Scaling by 5 would give (15, 20, 25), whose largest number exceeds 20, so the list of multiples stops here. These are the four the book prints.

   But two more exist. 5² + 12² = 25 + 144 = 169 = 13², so (5, 12, 13) qualifies — all three numbers are at most 20. 8² + 15² = 64 + 225 = 289 = 17², so (8, 15, 17) qualifies as well. Neither is a multiple of (3, 4, 5).

   Nothing else fits. Checking the remaining pairs turns up no further perfect squares within the limit — the next triples to appear are (20, 21, 29), (7, 24, 25) and (9, 40, 41), and each has a number above 20.

   The complete list — six triples: (3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15), (12, 16, 20).

   Why the question matters. The book deliberately says its four-triple list "contains" these triples rather than "is" these triples. Finding (5, 12, 13) and (8, 15, 17) is what shows that not every triple is a scaled (3, 4, 5) — and that is the observation the rest of the section is built on.

   ✦ There are six: (3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15) and (12, 16, 20).

3. 33 marksGanita Prakash Cl-8 Part 2, Math Talk, page 49

   Is (5, 12, 13) a primitive Baudhayana triple? What are the other primitive Baudhayana triples with numbers less than or equal to 20? Generate 5 scaled versions of each of these primitive triples. Are these scaled versions primitive?

   Hint. Primitive means the three numbers share no common factor bigger than 1.

   Is (5, 12, 13) primitive? The factors of 5 are 1 and 5; 5 does not divide 12; so the only number dividing all three is 1. Yes, it is primitive.

   The other primitives with all numbers at most 20. From the complete list of six triples found earlier: · (3, 4, 5) — common factor 1 → primitive · (5, 12, 13) — common factor 1 → primitive · (6, 8, 10) — common factor 2 → not primitive · (8, 15, 17) — common factor 1 → primitive · (9, 12, 15) — common factor 3 → not primitive · (12, 16, 20) — common factor 4 → not primitive So there are exactly three primitive triples in that range: (3, 4, 5), (5, 12, 13) and (8, 15, 17).

   Five scaled versions of each (using k = 2, 3, 4, 5, 6): · from (3, 4, 5): (6, 8, 10), (9, 12, 15), (12, 16, 20), (15, 20, 25), (18, 24, 30) · from (5, 12, 13): (10, 24, 26), (15, 36, 39), (20, 48, 52), (25, 60, 65), (30, 72, 78) · from (8, 15, 17): (16, 30, 34), (24, 45, 51), (32, 60, 68), (40, 75, 85), (48, 90, 102)

   Are the scaled versions primitive? No, and they never can be. Multiplying each member by k plants k as a common factor of all three, and k > 1 for every scaling. So every scaled version is non-primitive by construction — which is exactly what makes "primitive" a useful word: the primitive triples are the ones that are not copies of anything smaller.

   ✦ Yes, (5, 12, 13) is primitive; the other primitives with all numbers at most 20 are (3, 4, 5) and (8, 15, 17); and no scaled version is ever primitive, since the scaling factor k > 1 becomes a common factor of all three numbers.

4. 43 marksGanita Prakash Cl-8 Part 2, Math Talk, page 49

   If (a, b, c) is non-primitive and the integers have f — greater than 1 — as a common factor, then is (a/f, b/f, c/f) a Baudhayana triple? Check this statement for (9, 12, 15), and justify it.

   Hint. Divide the whole equation a² + b² = c² by f², and check that the three quotients are still whole numbers.

   Check on (9, 12, 15). All three are divisible by f = 3, giving (3, 4, 5). Test it: 3² + 4² = 9 + 16 = 25 = 5² ✓ So the divided triple is indeed a Baudhayana triple.

   Justify it in general. Since f divides a, b and c, the three quotients a/f, b/f and c/f are whole numbers — so the statement at least makes sense. Now start from what is given: a² + b² = c². Divide both sides by f², which is allowed since f ≠ 0: a²/f² + b²/f² = c²/f². Rewrite each term as a square: (a/f)² + (b/f)² = (c/f)². That is exactly the condition for (a/f, b/f, c/f) to be a Baudhayana triple. Therefore the statement is true for every non-primitive triple and every common factor f.

   Why this matters. This is the converse of the scaling rule proved just before it. Scaling turns a triple into a bigger triple; dividing by a common factor turns it back into a smaller one. Taking f to be the largest common factor strips a triple down as far as it can go, and what is left has no common factor — a primitive triple. This is what guarantees that every triple is a scaled copy of some primitive one, the fact asked about later in the chapter.

   ✦ Yes. Dividing a² + b² = c² by f² gives (a/f)² + (b/f)² = (c/f)², so (a/f, b/f, c/f) is again a Baudhayana triple — and for (9, 12, 15) with f = 3 it is (3, 4, 5).

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 2, Reprint 2026-27 (hegp202.pdf), where this is Chapter 2 (pages 33-54) — the ninth chapter of the Class 8 course. Like the rest of Part 2, this PDF carries NO printed answer key, so every answer here was derived from first principles and then independently recomputed in Python before being written — including all the √2-style one-decimal bounds, the six figure triangles, the rhombus side, the odd-square triple generator, and the complete list of Baudhāyana triples with all numbers at most 20. TWO POINTS WHERE THE BOOK'S OWN TEXT NEEDS CARE ARE FLAGGED IN PLACE: (1) on page 48 the book lists four triples with numbers at most 20 and says the list 'contains' them — the complete list has six, since (5, 12, 13) and (8, 15, 17) also qualify and are not multiples of (3, 4, 5); (2) the six right triangles in Figure it Out Q2 on pages 52-53 are labelled only in the printed figure, so each one's right-angle position was read directly off the rendered PDF page before solving, and the reading is stated in the solution so a student can check it against the book.. Questions are referenced from the NCERT textbook for identification.