Figure it Out — Doubling a Square and √2 — The Baudhāyana–Pythagoras Theorem

1. 13 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 39

   Earlier we saw a method to create a square with double the area of a given square paper. There is another method, in which two identical square papers are each cut along a diagonal to give four pieces (1, 2, 3 and 4). Can you arrange these pieces to create a square with double the area of either square?

   Hint. Work out the total area of the four pieces first. Then ask which edge of a piece is long enough to become a side of the new square.

   Step 1 — settle the area before touching the paper. Let each square have side 1, so each has area 1. Cutting a square along a diagonal gives two congruent right-angled isosceles triangles, each of area ½. Two squares therefore give four such triangles, of total area 4 × ½ = 2. So any square built out of all four pieces must have area 2 — already double one original square. The only thing left to check is whether the pieces actually fit into a square.

   Step 2 — decide which edge becomes a side of the new square. Each triangle has two legs of length 1 and one hypotenuse, which was the diagonal of the original square. The new square has area 2, so its side is longer than 1. Since the hypotenuse is the only edge longer than 1, the four hypotenuses must be the four sides of the new square, and the legs must all end up on the inside.

   Step 3 — the arrangement. Put the four triangles down with their right-angle corners meeting at one central point, turning each a quarter-turn from the one before. The four right angles fill 4 × 90° = 360° around that point, so they close up with no gap and no overlap. The four hypotenuses then face outward and form a closed four-sided figure.

   Why the figure is a square. All four hypotenuses are equal, because the four triangles are congruent — so all four sides are equal. At each outer corner, a leg from one triangle meets a leg from its neighbour; each leg makes 45° with its own hypotenuse, so the corner angle is 45° + 45° = 90°. Four equal sides and four right angles make it a square.

   This is the same picture the chapter drew in §2.1 — the original square is two of the triangles, the new square is all four — reached by cutting instead of by drawing.

   ✦ Yes. Cut each square along a diagonal to get four congruent right isosceles triangles, then set their right-angle corners together at one point so the four hypotenuses form the outside. The result is a square of area 2 — double either original square.

2. 25 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 39

   The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse, and find bounds on it that have at least one digit after the decimal point. (i) 3 (ii) 4 (iii) 6 (iv) 8 (v) 9

   Hint. Two copies of the triangle make a square of side a, and the square on the hypotenuse has twice that area — so c² = 2a².

   The relation to use. Two copies of an isosceles right triangle with equal sides a fit together into a square of side a, whose area is a². The square built on the hypotenuse has double that area, so c² = 2a², and therefore c = a√2. For the bounds, square one-decimal numbers until one lands below c² and the next lands above it.

   (i) a = 3 → c² = 2 × 9 = 18, so c = √18 = 3√2 ≈ 4.2426 4.2² = 17.64 < 18 and 4.3² = 18.49 > 18, therefore 4.2 < c < 4.3

   (ii) a = 4 → c² = 2 × 16 = 32, so c = √32 = 4√2 ≈ 5.6569 5.6² = 31.36 and 5.7² = 32.49, therefore 5.6 < c < 5.7

   (iii) a = 6 → c² = 2 × 36 = 72, so c = √72 = 6√2 ≈ 8.4853 8.4² = 70.56 and 8.5² = 72.25, therefore 8.4 < c < 8.5

   (iv) a = 8 → c² = 2 × 64 = 128, so c = √128 = 8√2 ≈ 11.3137 11.3² = 127.69 and 11.4² = 129.96, therefore 11.3 < c < 11.4

   (v) a = 9 → c² = 2 × 81 = 162, so c = √162 = 9√2 ≈ 12.7279 12.7² = 161.29 and 12.8² = 163.84, therefore 12.7 < c < 12.8

   A pattern worth noticing. Every answer is the equal side multiplied by √2 ≈ 1.414. Once you know √2 to three decimals you can predict all five at a glance — 3 × 1.414 = 4.24, 4 × 1.414 = 5.66, 6 × 1.414 = 8.49, and so on. The multiplying is what finds the answer; the squaring is what proves the bounds.

   ✦ (i) √18 = 3√2, 4.2 < c < 4.3 (ii) √32 = 4√2, 5.6 < c < 5.7 (iii) √72 = 6√2, 8.4 < c < 8.5 (iv) √128 = 8√2, 11.3 < c < 11.4 (v) √162 = 9√2, 12.7 < c < 12.8

3. 33 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 40

   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.]

   Hint. This time you know the big square and want the small one, so divide by 2 instead of multiplying.

   Work backwards from the square on the hypotenuse. The square built on the hypotenuse has area 10 × 10 = 100. That square is double the square built on an equal side, so the square on an equal side has area 100 ÷ 2 = 50.

   If a is the length of each equal side, then a² = 50, and therefore a = √50 = 5√2 ≈ 7.07.

   Bounds. 7.0² = 49 < 50 and 7.1² = 50.41 > 50, so 7.0 < a < 7.1.

   Check with the formula. a² + a² = 50 + 50 = 100 = 10² ✓

   Notice this is the previous question run in reverse: there we multiplied the equal side by √2, here we divide the hypotenuse by it, since 10 ÷ √2 = 10√2 ÷ 2 = 5√2.

   ✦ Both equal sides have length √50 = 5√2 ≈ 7.07 units, which lies between 7.0 and 7.1.

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.