Figure it Out — Using Baudhāyana's Theorem — The Baudhāyana–Pythagoras Theorem

1. 12 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 47

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

   Hint. Square both legs, add, and look for a perfect square.

   By drawing. Draw a 5 cm segment, put a 12 cm segment perpendicular to it at one end, and join the two free ends. Measuring that joining segment gives about 13 cm.

   By the theorem. With a = 5 and b = 12, a² + b² = c² 5² + 12² = c² 25 + 144 = c² 169 = c² Since 13 × 13 = 169, we get c = 13.

   The measurement and the calculation agree, which is the point of the exercise — but the calculation gives exactly 13, while the ruler only ever gives "about 13". This is why the theorem is worth having.

   Worth remembering. (5, 12, 13) is one of the triples Baudhayana himself listed in the Sulba-Sutra, and it comes up constantly. Its multiples work too: (10, 24, 26), (15, 36, 39), (25, 60, 65).

   ✦ The hypotenuse is exactly 13 cm.

2. 22 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 47

   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? Try drawing and measuring, then check using Baudhayana's Theorem.

   Hint. Here the unknown is a leg, not the hypotenuse — so subtract instead of adding.

   Careful reading first. The 17 cm side is the hypotenuse, not a leg, so it goes on the right-hand side of a² + b² = c². Putting it in the wrong place is the single commonest error in this chapter.

   By the theorem. With a = 8 and c = 17, a² + b² = c² 8² + b² = 17² 64 + b² = 289 b² = 289 − 64 = 225 Since 15 × 15 = 225, we get b = 15.

   By drawing. Draw an 8 cm segment, erect a perpendicular at one end, and from the other end swing a compass arc of radius 17 cm to cut that perpendicular. The perpendicular piece measures about 15 cm, agreeing with the calculation.

   Worth remembering. (8, 15, 17) is another triple from Baudhayana's list. Notice it is not a multiple of (3, 4, 5) or of (5, 12, 13) — it is a genuinely new one, which is why the chapter goes on to ask how many such triples exist.

   ✦ The third side is exactly 15 cm.

3. 34 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 47

   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 Sulba-Sutra, Verse 1.10)

   Hint. You can now combine any two squares. So ask: which two squares add up to three times the given one?

   The tool available. Baudhayana's rule lets us combine any two squares: build a right triangle whose legs are the two sides, and the square on its hypotenuse has area equal to the sum. In symbols, legs p and q give a square of area p² + q².

   Tripling. We want area 3s² from a square of side s. Split 3s² as 3s² = s² + 2s². The first part is the given square, of side s. The second part is the square of double the area, which §2.1 already showed how to build — its side is the diagonal d of the given square. So:

   1. Draw the given square and its diagonal d.
   2. Build a right triangle with legs s and d.
   3. The square on its hypotenuse has area s² + d² = s² + 2s² = 3s².

   Five times. Split 5s² as 5s² = s² + 4s². A square of area 4s² is easy — it is simply the square of side 2s, since (2s)² = 4s². So:

   1. Mark off a length 2s (twice the given side).
   2. Build a right triangle with legs s and 2s.
   3. The square on its hypotenuse has area s² + 4s² = 5s².

   Alternatively for five times, combine the tripled square (side √3 s) with the doubled square (side √2 s), since 3s² + 2s² = 5s². Both routes are correct; the s and 2s route needs no earlier construction at all.

   The general idea behind Verse 1.10. Once you can double, you can add the given square repeatedly: the hypotenuse of a right triangle with legs s and (the side of the n-times square) gives the (n + 1)-times square. Doing this over and over produces a square of any whole-number multiple of the original area — the spiral of squares sometimes drawn as the "spiral of Theodorus".

   ✦ For triple: build a right triangle with legs s and the diagonal of the given square (legs s and s√2) — the square on its hypotenuse has area 3s². For five times: build a right triangle with legs s and 2s — the square on its hypotenuse has area 5s².

4. 45 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 47

   Let a, b and c denote the lengths of the sides of a right triangle, with c the length of the hypotenuse. Find the missing sidelength in each case: (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

   Hint. When two legs are given you add the squares; when a leg and the hypotenuse are given you subtract. Give bounds where the answer is not a whole number.

   Which operation to use. In (i), (ii), (iv) and (v) the two given lengths are both legs, so c² = a² + b². In (iii) one of the given lengths is the hypotenuse, so the missing leg comes from b² = c² − a².

   (i) a = 5, b = 7 c² = 25 + 49 = 74. Since 8² = 64 and 9² = 81, the answer is not a whole number. 8.6² = 73.96 < 74 and 8.7² = 75.69 > 74, so c = √74 ≈ 8.60, with 8.6 < c < 8.7.

   (ii) a = 8, b = 12 c² = 64 + 144 = 208. Since 14² = 196 and 15² = 225, again not a whole number. 14.4² = 207.36 and 14.5² = 210.25, so c = √208 = 4√13 ≈ 14.42, with 14.4 < c < 14.5.

   (iii) a = 9, c = 15 b² = 15² − 9² = 225 − 81 = 144, therefore b = 12 exactly. This is the triple (9, 12, 15), which is 3 × (3, 4, 5).

   (iv) a = 7, b = 12 c² = 49 + 144 = 193. 13.8² = 190.44 and 13.9² = 193.21, so c = √193 ≈ 13.89, with 13.8 < c < 13.9.

   (v) a = 1.5, b = 3.5 c² = 2.25 + 12.25 = 14.5. 3.8² = 14.44 and 3.9² = 15.21, so c = √14.5 ≈ 3.81, with 3.8 < c < 3.9. (Decimals cause no trouble: the theorem never asked the sides to be whole numbers.)

   A sanity check that costs nothing. In every case the hypotenuse came out larger than both legs — 8.60 > 7, 14.42 > 12, 15 > 12, 13.89 > 12, 3.81 > 3.5. If it ever comes out smaller, a leg and the hypotenuse have been swapped.

   ✦ (i) √74 ≈ 8.60 (ii) √208 = 4√13 ≈ 14.42 (iii) b = 12 exactly (iv) √193 ≈ 13.89 (v) √14.5 ≈ 3.81

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.