Baudhāyana's construction: turning a rectangle into a square of matching area

What to assume they know

• From rectangle to parallelogram: area survives rearrangement — area of a rectangle, and that cut-and-move preserves area
• The Baudhāyana–Pythagoras theorem, in the form leg² = hypotenuse² − other leg²
• The identity for the difference of two squares
• Constructing a midpoint, a perpendicular, a square on a given segment, and an arc of given centre and radius
• That all radii of one circle are equal
• Reading a construction written as an ordered list of instructions

What they should be able to do

• State what "squaring a shape" asked for in ancient practice, and what the target is for a rectangle of sides a and b
• State the identity the construction realises, and verify it algebraically
• Carry out Baudhāyana's construction step by step from the printed instructions
• Explain why the square drawn on the first stage has side equal to the average of a and b
• Explain why the segment from the rectangle's top side up to the square's top corner is half the difference of a and b
• Identify the right-angled triangle the argument uses, name its right angle, and supply the step the printed proof leaves out
• Complete the algebra to show the final square's area is ab
• Explain why the construction can never fail, using the fact that the geometric mean of two lengths never exceeds their average
• Describe how to square a triangle, and say why squaring a circle is a different kind of problem

Where it usually goes wrong

• "Squaring means multiplying by itself." Here it means building a square of the same area. Both senses are live in this chapter within two pages of each other, and the chapter's own gloss on p. 140 is the fix.
• "The construction is a recipe to memorise." It is one algebraic identity drawn. Once a student sees that (a + b)/2 is the arc's radius and (a − b)/2 is the gap above the rectangle, the six instructions stop being arbitrary.
• **"AF is half of AD."** It is half of AE + AD, because F halves the leftover ED rather than the whole of AD. Getting this wrong makes every later step come out wrong and it looks superficially plausible.
• **"HP² = HK² − BH² is a mistake in the book."** It is a correct line with an unstated step. The triangle's legs are HP and KP, and KP = BH because BHPK is a rectangle. Show the rectangle, and the printed line becomes obviously right.
• **"The arc might miss BC."** It cannot: the arc's radius is bigger than the distance from its centre to the line, always, because (a + b)/2 > (a − b)/2 whenever b is positive.
• **"√(ab) could come out bigger than the square AFGH's side, and then P would fall off the end."** It cannot, and the reason is the average-versus-geometric-mean inequality. Worth one beat, because it turns a worry into a theorem.
• "If a rectangle can be squared, so can a circle — just take a fine enough approximation." An approximation is not a construction. The circle case is genuinely impossible with these tools, and the chapter, having raised the circle on p. 140, never says so.

Questions to check understanding

• Carry out the construction on a rectangle of given side lengths, and measure the result against the computed √(ab)
• Verify the identity for the difference of two squares algebraically, and identify each term in the figure
• Given the rectangle's sides, state the arc's radius and the length BH before drawing anything
• Fill in the missing justification in a proof of the construction
• Describe a procedure for squaring a triangle, and justify each step
• Explain why the construction is impossible to break, whatever a and b are, provided a > b > 0
• Reasoning question: state what changes in the construction when a = b

Examples worth working on the board

Inputs, not answers, except where the chapter itself prints the result. Values marked Verified are worked out here; the chapter prints no answers and this volume has no appended answer key.

• The setting (p. 140). In ancient practice, to square a figure was to build a square of the same area, and the figure could be a rectangle, a triangle or a circle. The chapter then shows how Baudhāyana squared a rectangle, taking the construction from the Śhulbasūtra, dated 800 BCE, and says it is presenting a slightly simplified form of it. It requires a > b.
• The target (pp. 140–141). Rectangle with AD = a and AB = b, so the square to be built has area ab and therefore side √(ab).
• The identity, as the chapter prints it (p. 141). The square on (a + b)/2 minus the square on (a − b)/2 equals ab. Verified: the two expanded squares are (a² + 2ab + b²)/4 and (a² − 2ab + b²)/4, and the difference is 4ab/4.
• Fig. 6.30 (p. 141), read off the printed page. The rectangle sits with A at the bottom left, D at the bottom right, B above A and C above D, so AD = a runs along the bottom and AB = b up the left. E is marked on AD and F between E and D. Square AFGH stands on AF with G top right and H top left, H sitting above B on AB produced. A dashed arc runs from A up to G, crossing the top edge BC of the rectangle at K, and P is on HG directly above K. The square HPQS hangs below HP, with S on AB between A and B and Q directly below P. The printed figure shades four regions in different colours; the shading is decoration and the labels are the content.
• The construction, as the six printed instructions run (p. 141). Start from the rectangle. Mark E on AD so that AE equals AB. Take F as the midpoint of ED. Build the square AFGH, with H on AB produced. Draw the arc from A to G about centre H, and let it meet BC at K. Through K draw a line parallel to AH, meeting HG at P. The square on HP, namely HPQS, is the answer.
• **Why AF is the average (p. 141).** AE = b and AD = a, and F halves ED. Verified: AF = b + (a − b)/2 = (a + b)/2, which is what the chapter gets by averaging AE and AD.
• **Why HG and HK are equal (p. 141).** HG is a side of square AFGH, so it is (a + b)/2; HK is a radius of the same arc as HA and HG, so it is that too.
• **Why BH is half the difference (p. 141).** AH is the square's side, (a + b)/2, and AB is b. Verified: BH = (a + b)/2 − b = (a − b)/2.
• The step the printed proof leaves out, and it is the one line. The chapter says to consider the right-angled triangle HKP and then writes HP² = HK² − BH². But the right angle in that triangle is at P, so what Baudhāyana–Pythagoras gives is HP² = HK² − KP². The substitution of BH for KP is correct and unstated: B and K both lie on the line BC, H and P both lie on the line HG, and those two lines are parallel, so BHPK is a rectangle and its two vertical sides are equal. Verified. Say this; it is one sentence and without it the printed line looks like an error.
• The algebra closing (p. 141). Verified: HP² = ((a + b)/2)² − ((a − b)/2)² = ab, so square HPQS has area ab, equal to the rectangle's.
• A numerical instance, not in the book. a = 8, b = 2. Verified: AE = 2; ED = 6 so EF = 3 and AF = 5; the arc has radius 5; BH = 5 − 2 = 3; HP² = 25 − 9 = 16, so HP = 4; and 4² = 16 = 8 × 2. Choose numbers giving a Pythagorean triple, so the construction lands on a whole number and the arithmetic does not distract from the geometry. A second pass with a = 9, b = 4 gives HP = 6, and a third with a = 5, b = 3 gives HP = √15, which is the honest general case.
• Why it cannot fail, which the chapter does not discuss. Verified: the arc has radius (a + b)/2 and the line BC sits (a − b)/2 away from H, which is smaller, so the arc must cross BC and K exists. And HP = √(ab) never exceeds HG = (a + b)/2, so P really does land on the segment HG and not beyond G. That is all this inequality gives, and it is all the construction needs — resist adding that S therefore lands between A and B, which is a different and weaker claim. S sits on HA at distance √(ab) from H while HB = (a − b)/2, so S falls inside AB only when a/b ≤ 3 + 2√2 ≈ 5.83; at a = 100, b = 1 the distances are 10 against 49.5 and the square rises clear of the side AB altogether. Nothing breaks when it does. The inequality √(ab) ≤ (a + b)/2 is therefore not a curiosity here — it is what makes the drawing possible, with equality exactly when the rectangle was already a square.
• Think and Reflect (p. 142). How would you square a given triangle? The chapter asks and does not answer. Verified route: a triangle of base b and height h has area bh/2, which is the area of a rectangle with sides b and h/2; build that rectangle, then square it by this construction. Two steps, both already available.
• The shape the chapter drops (p. 140). It names the circle as one of the figures the ancients wanted to square, squares a rectangle, and never returns to the circle — checked on the printed pages 140, 141 and 142. Section 10 should close the loop: with straight edge and compasses a circle cannot be squared, and the reason is a stronger property of π than the irrationality §6.3 established. See Notes.

Figures to have open

• Fig. 6.30 (p. 141), redrawn and fully able to be shown moving, with each of the six instructions appearing as its line is drawn. **The printed figure's labels — A, B, C, D, E, F, G, H, K, P, Q, S and the letters a and b — all sit inside the artwork.** Rebuild the figure; do not lift it.
• A separate panel isolating the right-angled triangle HKP, with rectangle BHPK shaded, to carry section 7. Not in the book, and the topic's most important addition.
• Two side-by-side runs of the construction with numbers on it: a = 8, b = 2 giving a whole-number answer, then a = 5, b = 3 giving a surd. Standard schematic.
• A chain for section 10: triangle → rectangle of the same area → square, then a circle with the final arrow crossed out. Not in the book.
• No portrait and no manuscript image is needed. The Śhulbasūtra can be named.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics (NCF-SE 2023), Chapter 6, §6.9 "Squaring a Rectangle" (pp. 140–142), comprising the opening paragraph on p. 140, the construction and its figure on p. 141, the starred subsection asking why the method works on p. 141, and the Think and Reflect on p. 142.
• Fig. 6.30 (p. 141), with all its labels.
• The boxed panel on special cases and generalisation (p. 139) sits immediately before this section and supplies the vocabulary section 1 uses.
• The Baudhāyana–Pythagoras theorem is used here and in §6.8.1's Examples 3, 4 and 5 (pp. 135–136), handled in Heron's formula: area from the three sides alone.
• π's irrationality, needed for the closing beat, is §6.3 (pp. 123–124), handled in π is irrational, and what that rules out.
• The Chapter Summary (p. 154) does not mention this construction.