Applying the theorem: the lotus-in-the-lake problem from the Līlāvatī

What to assume they know

• Combining two different squares: a² + b² = c² — the theorem a² + b² = c², and how to find any one of the three sides from the other two
• The isosceles right triangle: why its hypotenuse must be a√2 — the isosceles case, needed for the diagonal item
• Expanding (x + 1)² correctly as x² + 2x + 1
• Solving a linear equation in one unknown by subtracting the same thing from both sides
• That a rhombus's diagonals cut each other in half at right angles
• Reading a right angle out of a figure wherever it happens to be drawn, rather than assuming it is at the bottom left
• Area of a triangle as half the base times the height

What they should be able to do

• Turn a problem stated in words and a picture into a labelled right triangle
• Choose the unknown so that a second quantity in the problem becomes an expression in it, rather than a second unknown
• State the assumptions the solution rests on, including the one the chapter leaves unspoken
• Set up 3² + x² = (x + 1)², expand it, and see the x² terms cancel
• Solve the resulting linear equation and read the answer back into the picture
• Notice that the finished triangle is 3, 4, 5, and connect it to §2.5's list
• Apply the theorem to a square's diagonal, and to a rhombus given its two diagonals
• Justify that a right triangle's longest side is always its hypotenuse
• Use the theorem on a non-right triangle by dropping an altitude, as the equilateral-triangle item requires

Where it usually goes wrong

• "There isn't enough information." The chapter says out loud that it looks that way, which makes this the rare misconception the textbook itself raises. The answer is that one unknown covers two quantities, so one equation is enough.
• **"Call the depth x and the stem y."** Two unknowns and one equation, and the problem stalls. Watching it stall is instructive; the fix is noticing that the stem is 1 more than the depth.
• **"(x + 1)² is x² + 1."** The commonest algebraic wreck in this problem, and it destroys it — the middle term 2x is the only place the unknown survives the cancellation. Expand it slowly.
• **"There's an x² in it, so this needs methods we don't have."** The x² appears on both sides and leaves. Point this out before doing it, so the student is watching for it.
• "3 is the hypotenuse — it's the horizontal one." The hypotenuse is x + 1, the swayed stem, because the right angle is between the water surface and the upright stem. Getting this wrong gives 3² + (x + 1)² = x², i.e. 9 = −1 − 2x, so x = −5 — a negative depth, which is a useful wrong answer to show.
• "The stem gets longer when it bends." Nothing in the printed problem says the stem keeps its length, and the whole solution depends on it. Say the assumption out loud; the chapter states only the perpendicularity one.
• "The hypotenuse is the slanted side in the drawing." On Part II p.53 the right angle appears at the top, at the bottom left and at the far left in different drawings, and in three of the six the labelled side is the hypotenuse rather than a leg — those labelled 40/41, 10/√200 and 27/45. Teach the habit of finding the right-angle mark first.
• "You can only use the theorem on right triangles." You can use it on any figure you can cut into right triangles: a rhombus split by its diagonals, an equilateral triangle split by an altitude. That is what section 11 is for.
• "Every whole number is a possible grid-square area." 3 is not. Let the class try and fail before the reason is given.

Questions to check understanding

• Set up and solve a problem where one unknown appears in two of the three sides — a reed in a river, a ladder against a wall, a broken bamboo
• Explain why the problem is solvable despite appearing to lack data
• State the assumptions a stated word problem depends on, including any left unsaid
• Find the diagonal of a square from its side, and the side from its diagonal
• Find a rhombus's sidelength from its diagonals, and the diagonals from a side and one diagonal
• Given a figure with a right-angle mark and two labels, find the third side — with the right angle placed away from the bottom left
• Justify that the hypotenuse is the longest side
• Find the height and area of an equilateral triangle of given side
• Draw squares of stated areas on a dot grid, and say which small whole numbers cannot be done

Examples worth working on the board

The lotus problem the chapter solves in full; everything else in the exercise set it leaves open, and those results are marked derived.

• The problem's data (Part II, §2.7, p.52). A lake with chakra and krauñcha birds around it. A lotus rises above the waterline, its stem tip standing 1 unit clear. A breeze tips it over until that tip meets the water 3 units from where it had been. The depth of the lake is wanted. Those two numbers, 1 and 3, are the entire numerical input.
• The artwork (Part II, §2.7, p.52, right of the text). A cutaway view of a lake: green water in a shallow basin, reeds at both ends, a small green frog on a pad near the surface. One pink lotus stands upright with its tip above the waterline, and the gap between the tip and the surface is marked with a short vertical tick and the numeral 1. A second pink lotus leans left with its flower at the surface. Below the waterline the depth is marked x, and the two stem positions are drawn as two straight lines meeting at a common point on the lake bed. Both the 1 and the x are lettering inside the illustration and were read from the printed page.
• The set-up, worked on the page (Part II, §2.7, p.52). Let x be the length of stem under water, which is the depth asked for. Since 1 unit stands clear, the whole stem is x + 1. The chapter states the assumption that the resting stem is perpendicular to the water surface, and calls it a very reasonable one. That gives a right triangle with sides 3, x and x + 1.
• The solution, worked on the page (Part II, §2.7, p.52). 3² + x² = (x + 1)²; expanding the right side gives 9 + x² = x² + 2x + 1. Take the squared term off each side and what is left is 9 = 2x + 1, so x = 4, a lake 4 units deep. Note that the printed working steps straight from 9 = 2x + 1 to x = 4 without an intermediate line.
• The triple hiding in the answer (derived). The finished triangle has sides 3, 4 and 5, the first triple on Baudhāyana's list at Part II p.48. The chapter does not remark on this. It is the best single beat available in this topic, because it ties a 12th-century word problem (this chapter prints no date, but Part I p.144 dates the Līlāvatī to 1150 CE) to an 800 BCE list with nothing in between.
• Figure it Out no.1 (Part II, p.52). The diagonal of a square of side 5 cm. Derived: √50 cm, which lies between 7 and 8, and between 7.0 and 7.1 — and it is the isosceles case of §2.3 asked as a diagonal.
• Figure it Out no.2 (Part II, p.53). Six right triangles drawn at varied orientations, missing sides wanted. The labels are inside the artwork, and the right-angle mark is what tells you which labelled side is the hypotenuse. As drawn:
  • two legs 7 and 9, hypotenuse unlabelled — derived: √130
  • two legs 4 and 10, hypotenuse unlabelled — derived: √116
  • one leg 40 and the hypotenuse 41, the short leg unlabelled — derived: 9, giving the triple (9, 40, 41), which is one of the odd-square machine's outputs from §2.5
  • one leg 10 and the hypotenuse √200 — derived: the other leg is 10, so this one is isosceles
  • two legs 10 and √150, hypotenuse unlabelled — derived: √250
  • one leg 27 and the hypotenuse 45 — derived: 36, giving (27, 36, 45), which is nine times (3, 4, 5).
• Figure it Out no.3 (Part II, p.53). A rhombus whose diagonals are 24 and 70; the sidelength is wanted. Derived: the diagonals bisect each other at right angles, so each side is the hypotenuse of a right triangle with legs 12 and 35, giving 37 — and (12, 35, 37) is on Baudhāyana's list at Part II p.48. A second uncredited closure, and worth showing beside the first.
• Figure it Out no.4 (Part II, p.53). Whether the hypotenuse is always a right triangle's longest side, with justification wanted. Derived: c² is a² plus something positive, so c² exceeds both a² and b², so c exceeds both. One line, and it is a proof rather than an observation.
• Figure it Out no.7 (Part II, p.53). A square holding the area left over when a square of side 5 is taken away from one of side 7. Derived: 49 − 25 = 24, so the side is √24, between 4 and 5. Constructively: make a right triangle with hypotenuse 7 and one leg 5, and build on the third side.
• Figure it Out no.8 (Part II, p.53, marked Math Talk and Try This). Part (i): using grid dots as corners, draw squares of area 2, 3, 4 and 5 square units; the figure shows a 5-by-5 array of dots with faint gridlines. Part (ii): on a grid extending indefinitely, which whole-number areas are reachable? Derived: a square whose corner-to-corner step across the grid is m right and n up has area m² + n², so 2 is reachable (step 1 and 1), 4 is (step 2 and 0), 5 is (step 2 and 1) — and 3 is not. The reachable areas are exactly the numbers that are a total of two squares: 1, 2, 4, 5, 8, 9, 10, 13, 16 and so on, missing 3, 6, 7, 11, 12. Treat the impossibility of 3 as something the class discovers by failing to draw it; a full proof is beyond this class and the item is marked exploratory.
• Figure it Out no.9 (Part II, p.54, marked Try This). The area of an equilateral triangle of side 6, with a hint to show first that an altitude cuts the opposite side in half. Derived: the altitude makes a right triangle with hypotenuse 6 and one leg 3, so the height squared is 36 − 9 = 27 and the height is √27; the area is half of 6 times √27. This is the only item in the chapter that produces a length involving √3 — the altitude is √27, which is 3√3 — rather than √2. No length in the item has square 3.
• The SUMMARY and what follows it (Part II, p.54). The chapter's summary box sits above a puzzle page headed "Find the Colours!", which the spine marks as carrying no exposition and which has no brief.

Figures to have open

• The lake cutaway with the lotus in both positions, the 1 above the surface and the depth x marked, and the right triangle overlaid on it in a second pass (Part II, §2.7, p.52). This is worth showing as a movement: the stem swaying is the whole argument, because it is what shows the length is unchanged. Redraw as original artwork; the chapter's illustration is not to be reproduced.
• The six exercise triangles at their printed orientations, with the right-angle mark emphasised and the given labels placed on the sides they label (Part II, p.53, no.2). Redraw as schematics. Orientation matters here — straightening them all up would remove the point of the item.
• A rhombus with its two diagonals drawn and one of the four right triangles picked out, legs 12 and 35 (Part II, p.53, no.3). Standard schematic.
• An equilateral triangle of side 6 with the altitude dropped and the half-base marked (Part II, p.54, no.9). Standard schematic.
• A dot grid, at least 5 by 5, with tilted and upright squares drawn on it (Part II, p.53, no.8). Standard schematic, and it should extend beyond the printed 5 by 5 for part (ii).
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part II, printed Chapter 2, "The Baudhāyana-Pythagoras Theorem", §2.7 "Further Applications of the Baudhāyana-Pythagoras Theorem", Part II pp.52–54. §2.7 prints one subheading in colour, nameable but not citable by number: "A Problem from Bhāskarāchārya's Līlāvatī" (Part II p.52).
• The chapter's closing Figure it Out, nine items, runs from Part II p.52 (nos.1–2) across Part II p.53 (the six triangles of no.2 and then nos.3–8) to Part II p.54 (no.9). Items 8 and 9 carry Math Talk and Try This markers.
• The chapter's SUMMARY is at Part II p.54, and the puzzle page "Find the Colours!" occupies the lower half of the same page. The spine marks that puzzle video: no, so it has no brief; its content is a three-box logic problem.
• Backward dependencies inside the chapter: the theorem at Part II p.46, the isosceles relation at Part II pp.37 and 40, and the triple list at Part II p.48.
• Attribution printed in the chapter: the problem is given as a translation from Bhāskarāchārya, also named Bhāskara II, from the Līlāvatī (Part II p.52). No date is printed for either the author or the text.