In-text — Doubling, Halving and Combining Squares — The Baudhāyana–Pythagoras Theorem

1. 13 marksGanita Prakash Cl-8 Part 2, Math Talk, page 34

   Why does the new dotted square — the square drawn on the diagonal of the original square — have double the area of the original square?

   Hint. Draw the horizontal and vertical lines through the corner of the original square that sits at the centre of the dotted square.

   Set the picture up with coordinates so nothing is left to the eye. Let the original square be A(0, 0), B(1, 0), C(1, 1), D(0, 1). Its diagonal DB runs from D(0, 1) to B(1, 0). The square built on that diagonal has vertices D(0, 1), B(1, 0), (2, 1) and (1, 2).

   The centre of the dotted square is the corner C. The two diagonals of the dotted square join D(0, 1) to (2, 1), and B(1, 0) to (1, 2). They cross at (1, 1), which is exactly the corner C of the original square. So the horizontal line y = 1 and the vertical line x = 1 — the two sides of the original square that meet at C, extended — are precisely the diagonals of the dotted square.

   Now count triangles. Those two diagonals cut the dotted square into four triangles, each with its right angle at C and each with two legs of length 1: · D(0,1) – B(1,0) – C · B(1,0) – (2,1) – C · (2,1) – (1,2) – C · (1,2) – D(0,1) – C Meanwhile the diagonal DB cuts the original square into two triangles, D–B–A and D–B–C, and each of those also has two legs of length 1 with a right angle between them.

   All six triangles are congruent by SAS — same two legs, same included right angle — so they all have the same area, call it t.

   Area of original square = 2t. Area of dotted square = 4t. Therefore the dotted square has exactly double the area.

   ✦ Because the original square is made of exactly 2 of these small congruent triangles while the square on its diagonal is made of 4 of the very same triangles, the square on the diagonal has double the area.

2. 23 marksGanita Prakash Cl-8 Part 2, Math Talk, page 34

   Why should the extension of the vertical and horizontal sides of the original square pass through the vertices of the dotted square? [Hint: the line that bisects an angle of a square passes through the opposite vertex.]

   Hint. Show first that the original square's side bisects an angle of the dotted square; the diagonal property then does the rest.

   What has to be shown. Take the corner B of the original square, where the dotted square also has a vertex. The vertical side of the original square through B, extended upward, must be shown to hit the opposite vertex of the dotted square.

   Step 1 — the sides of the dotted square make 45° with the sides of the original square. At B, one side of the dotted square runs along BD, the diagonal of the original square. A diagonal of a square bisects the 90° corner angle, so BD makes 45° with the side BA and 45° with the side BC.

   Step 2 — the other dotted side at B is perpendicular to the first. The dotted figure is a square, so its two sides at B meet at 90°. Since one of them is 45° away from the vertical, the other must be 45° away from the vertical on the opposite side.

   Step 3 — so the vertical bisects the dotted square's angle at B. The two dotted sides at B sit 45° on either side of the vertical line through B. That vertical line is exactly the side BC of the original square, extended. Therefore the extended side of the original square bisects the angle of the dotted square at B.

   Step 4 — apply the hint. In a square, the line that bisects a corner angle is the diagonal, and a diagonal ends at the opposite vertex. So this bisector passes through the vertex of the dotted square opposite to B.

   The same argument at the neighbouring corner shows the extended horizontal side passes through the other pair of opposite vertices. This is why the two extended sides turn out to be the two diagonals of the dotted square — which is what made the triangle-counting in the previous question work.

   ✦ Because a diagonal of a square bisects its corner angles, the sides of the dotted square lie 45° on either side of the original square's extended sides; those extended sides therefore bisect the dotted square's angles, and an angle bisector of a square is a diagonal, which ends at the opposite vertex.

3. 32 marksGanita Prakash Cl-8 Part 2, Math Talk, page 36

   Will a square having half the sidelength have half the area? Why not? How many such squares will fill the original square?

   Hint. Halving a length halves it twice over when you build a square on it — once in each direction.

   Test it with a number. Take a square of side 8, whose area is 64. Half the sidelength is 4, and the square on it has area 4 × 4 = 16 — which is a quarter of 64, not half.

   Why length and area do not halve together. Area is a product of two lengths. Halving the side halves both factors at once, so the area is multiplied by ½ × ½ = ¼. In symbols, a square of side L has area L², and a square of side L/2 has area (L/2)² = L²/4.

   How many fit? Since each small square is a quarter of the original, exactly four of them fill it. You can see this without any arithmetic: cut the original square with its horizontal midline and its vertical midline, and you get four squares of side L/2 tiling it perfectly.

   The lesson for this chapter. This is precisely why halving a square is not as easy as it looks, and why Baudhāyana's answer is a tilted square rather than a shrunken one. To get half the area you need a side of L/√2, not L/2 — and L/√2 cannot be marked off with a ruler as easily as a midpoint can, so the fold-through-the-midpoints construction is used instead.

   ✦ No — a square of half the sidelength has one quarter of the area, because both factors in length × length are halved; exactly four such squares fill the original.

4. 44 marksGanita Prakash Cl-8 Part 2, Math Talk, page 37

   In the paper-folding construction, the square paper is folded inward so that the crease lines pass through the midpoints of the sides, giving the figure PQRS. Why is PQRS a square? Why is its area half that of the original paper? Explain by joining QS and PR, finding the angles formed, and using triangle congruence.

   Hint. The four pieces folded away are congruent corner triangles; that gives the four sides, and the 45° angles give the four right angles.

   Name the pieces. Let the paper be square ABCD of side L, and let P, Q, R, S be the midpoints of AB, BC, CD, DA. Folding along PQ, QR, RS and SP turns the four corners inward, so PQRS is what remains.

   Why the four sides are equal. The four corner triangles are APS, BQP, CRQ and DSR. In each one, the two legs are halves of sides of the square, so each leg is L/2, and the angle between them is a corner of the square, 90°. By SAS the four triangles are congruent, and therefore their hypotenuses are equal: SP = PQ = QR = RS.

   Why the four angles are right angles. Each corner triangle is right-angled with two equal legs, so its two base angles are each (180° − 90°) ÷ 2 = 45°. Look at the point P on side AB. Three angles sit along that straight line: ∠APS from one triangle, ∠SPQ from the figure, and ∠QPB from the next triangle. So ∠SPQ = 180° − 45° − 45° = 90°. The same holds at Q, R and S, so all four angles of PQRS are right angles. Four equal sides plus four right angles make PQRS a square.

   Why the area is exactly half. Each corner triangle has area ½ × (L/2) × (L/2) = L²/8. Four of them together have area 4 × L²/8 = L²/2, which is half the paper. What is left is the other half: area of PQRS = L² − L²/2 = L²/2.

   A second way to see it. Join PR and QS. Since P and R are midpoints of opposite sides, PR is parallel to a side and has length L; likewise QS has length L, and the two cross at right angles at the centre. They are the diagonals of PQRS and they cut it into four right triangles with legs L/2 and L/2, of total area 4 × ½ × (L/2)(L/2) = L²/2 ✓. This also shows the side of PQRS is the length whose square is half of L² — the halving construction is the doubling construction read backwards.

   ✦ PQRS is a square because the four congruent corner triangles give it four equal sides and their 45° base angles leave 90° at each vertex; its area is L²/2, exactly half the original paper, since the four discarded corners also total L²/2.

5. 53 marksGanita Prakash Cl-8 Part 2, Math Talk, pages 41 and 44

   In Baudhayana's construction for combining two different squares, four congruent right triangles are drawn on the hypotenuse to make a four-sided figure. Explain why all the angles of that figure are right angles, so that it is a square. Also check that the method agrees with the earlier doubling method when the two squares are the same size.

   Hint. Give the acute angles of the right triangle the names x and 90 − x, then add up the three angles that sit on each straight edge.

   Name the angles once. Each of the four triangles is right-angled with legs a and b. Its three angles are 90°, and two acute ones which must add to 90° — call them x and 90° − x. Every triangle in the figure is congruent to every other, so every triangle carries the same three angles.

   Look at one vertex of the new figure. The four triangles are placed so that at each vertex of the four-sided figure, the leg of one triangle continues in a straight line into the leg of the next. Along that straight line three angles sit side by side: the acute angle x from one triangle, the angle of the new figure, and the acute angle 90° − x from the neighbouring triangle. Angles on a straight line add to 180°, so angle of the new figure = 180° − x − (90° − x) = 180° − 90° = 90°.

   The x cancels, which is the whole point — the result does not depend on the shape of the triangle at all. The same computation applies at all four vertices, so all four angles are right angles.

   Why the four sides are equal. Each side of the new figure is the hypotenuse of one of the four congruent triangles, so all four sides have the same length c. Four equal sides and four right angles make it a square of side c, and its area is therefore the sum of the two original squares — which is Baudhayana's claim.

   The same-size case. Put b = a. The right triangle becomes isosceles, its hypotenuse is the diagonal of a square of side a, and the new square has area c² = a² + a² = 2a². That is exactly the doubling result of §2.1: the square on the diagonal has double the area. So the general method contains the doubling method as its special case, and the two agree.

   ✦ Because the acute angles of the congruent triangles are x and 90° − x, each vertex angle of the new figure is 180° − x − (90° − x) = 90°; with four equal hypotenuses as sides it is a square, and setting b = a reduces it to c² = 2a², the earlier doubling result.

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.