The isosceles right triangle: why its hypotenuse must be a√2

What to assume they know

• Doubling a square: the diagonal is the construction — the square on a square's diagonal has twice the area, argued by counting congruent triangles
• Halving a square: the doubling construction run backwards — the same construction read the other way
• A square is made of two congruent right triangles, cut along a diagonal
• Area of a square is side × side, so a square of area A has side √A
• Squaring and un-squaring whole numbers: recognising 36, 256, 289 as squares
• Bounding a square root between consecutive whole numbers by squaring the candidates
• Substituting a number for a letter in a formula, and rearranging c² = 2a² in either direction

What they should be able to do

• Name the hypotenuse of a right triangle as the side facing the right angle
• Given an isosceles right triangle with legs of 1, find its hypotenuse by computing the area of the square built on that hypotenuse
• Follow the chapter's lettered figure and say which square has area 1 and which has area 2
• State that the hypotenuse of a unit isosceles right triangle is a length whose square is 2, and write it with the radical sign
• Generalise to legs of length a and derive c² = 2a² from the same figure
• Use the relation forwards, to get the hypotenuse from the legs, and backwards, to get the legs from the hypotenuse
• Trap an awkward hypotenuse between two whole numbers, and then between two one-decimal numbers, by squaring candidates
• Recognise a square's diagonal and an isosceles right triangle's hypotenuse as the same segment, so a diagonal question and a hypotenuse question are one question

Where it usually goes wrong

• "The hypotenuse is 1 + 1 = 2." Adding the legs is the commonest wrong move and it is easy to kill: 2 is the length of the bent path along the two legs, and the straight path is shorter. Draw both and let the student see which is longer.
• "The hypotenuse is the longest side, so it must be about 1.5." Guessing is reasonable; stopping at the guess is not. The chapter's answer is a length whose square is exactly 2, and the following topic shows that no terminating decimal and no fraction can be that length.
• "√2 means take half." The radical sign is unrelated to halving. If a student reads it as an operation on 2 they will produce 1. Insist on the definition: the length whose square is 2.
• **"c² = 2a² means c = 2a."** Square-rooting a product of two things means square-rooting both. This is the single most common algebraic slip in the section, and Example 1 is a good place to catch it: if c were 2a the answer for legs of 12 would be 24, not something between 16 and 17.
• "The hypotenuse is the sloping side." It is the side facing the right angle. In the exercise figures on Part II p.53 the right angle sits 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. A student who looks for "the slanted one" will misread the figure.
• **"This is the Pythagoras theorem, so just use a² + b² = c²."** Not yet. The chapter has not proved that; it reaches the isosceles case first, from the doubling construction, and the general theorem comes in §2.4. Getting this order right matters, because the isosceles result is the ancestor of the general one here and not a corollary of it.
• "Between 16 and 17 is not an answer." It is exactly the answer the chapter wants, and it is a real skill. A bound you can justify beats a decimal you have copied from a calculator.

Questions to check understanding

• Given the equal sides of an isosceles right triangle, state the hypotenuse using the radical sign, and bound it between two whole numbers with the squaring shown
• Given the hypotenuse, recover the equal sides
• Find the diagonal of a square of stated sidelength
• Given a figure, identify which marked side is the hypotenuse, with the right angle placed somewhere other than the bottom left
• Given a right triangle in a figure with one leg and the opposite side marked, decide which relation to use and whether the triangle is isosceles
• Tighten a bound: from two whole numbers to two tenths, showing the two squares that justify it
• Explain why the hypotenuse cannot equal the sum of the two legs
• Explain why c² = 2a² does not give c = 2a

Examples worth working on the board

Values marked derived are worked out here; where the chapter itself computes something, that is said.

• The opening triangle (Part II, §2.3, p.37, margin figure). A right triangle with the right angle at the bottom left, the vertical side marked 1, the horizontal side marked 1, and the slanting side marked with a question mark.
• The lettered figure (Part II, §2.3, p.37). A square PEAR with P at the bottom left, E at the bottom right, A at the upper right and R at the upper left, all four sides marked 1, right-angle marks at P and at A. Its diagonal ER is drawn. A larger square REST is drawn with a broken outline on that diagonal, standing on a corner, with T at the top and S at the right, so that R, E, S, T run round it. All the lettering sits inside the artwork and was read from the printed page.
• The chapter's own computation (Part II, §2.3, p.37). Area of REST is twice the area of PEAR, which is 2 × 1 = 2 square units. Writing the hypotenuse ER as c, the area of REST is c × c = c². So c² = 2 and c = √2. The chapter states all of this.
• The two small closing figures (Part II, §2.3, p.37, foot of page). A right triangle with both legs marked 1 and the hypotenuse marked √2; beside it a square with right-angle marks at all four corners, the left side marked 1 and the bottom side marked 1, the top and right sides unlabelled, and its diagonal — drawn from the top-left corner to the bottom-right — marked √2. The second figure is the point of section 11.
• The unit convention (Part II, p.38, first line). From here on the chapter takes all lengths in one fixed unit unless it says otherwise.
• The general figure (Part II, §2.3, p.40, under "General Solution"). A square PQRS with P at the bottom left, Q at the bottom right, R at the upper right — which falls at the centre of the broken-line square SQVU — and S at the upper left, the left side marked a, the bottom marked a, a right-angle mark at P, and the diagonal SQ drawn and marked c. A broken-line square SQVU is drawn on SQ with U at the top and V at the right. The chapter writes: area of SQVU is twice the area of PQRS, so c² = 2a².
• Example 1, worked on the page (Part II, p.40). Legs of 12. Then c = √(2 × 12²) = √288. The chapter supplies 16² = 256 and 17² = 289, and concludes that √288 sits between 16 and 17. Note what it does not do: it never gives a decimal for √288.
• Example 2, worked on the page (Part II, pp.40–41). Hypotenuse √72. Then 72 = 2a², so a² = 36 and a = 6. The chapter then tells the student to use the formula to check the answers to the earlier exercise.
• Figure it Out no.2 (Part II, p.39). The equal sides are given as (i) 3, (ii) 4, (iii) 6, (iv) 8, (v) 9, and the task is the hypotenuse together with bounds carrying at least one decimal place. Derived, since the chapter prints no answers: the hypotenuses are √18, √32, √72, √128 and √162. Bounds to one decimal: 4.2 and 4.3 (from 4.2² = 17.64, 4.3² = 18.49); 5.6 and 5.7 (31.36, 32.49); 8.4 and 8.5 (70.56, 72.25); 11.3 and 11.4 (127.69, 129.96); 12.7 and 12.8 (161.29, 163.84).
• Figure it Out no.3 (Part II, p.40). The hypotenuse is 10; the legs are wanted. The chapter's hint is to find the area of the square made from two such triangles. Derived: 100 = 2a², so a² = 50 and each leg is √50, which sits between 7 and 8 because 7² = 49 and 8² = 64, and between 7.0 and 7.1 because 7.0² = 49 and 7.1² = 50.41.
• The diagonal item (Part II, p.52, Figure it Out no.1). A square of side 5 cm; its diagonal is wanted. Derived: √50 cm, the same number as the previous item — worth putting the two side by side, because one is asked as a hypotenuse and one as a diagonal and they are the same computation.
• The isosceles triangle hidden in the exercise figures (Part II, p.53, no.2, the middle-right triangle). Right angle at the top vertex; one leg marked 10; the side opposite the right angle marked √200. Derived: the other leg is 10, since 200 − 100 = 100 — so this drawing is an isosceles right triangle and the relation from this section settles it in one line.
• The chapter's SUMMARY line (Part II, p.54). It records the relation as a² + a² = 2a² = c², so c = a√2. Read the caution in Notes before showing that line as printed.

Figures to have open

• The chapter's lettered figure: square PEAR with side 1, diagonal ER, and square REST built on ER with a broken outline (Part II, §2.3, p.37). Sections 4 to 6 cannot be told without it. Redraw as a schematic, keeping the printed letters.
• The general version: square PQRS of side a, diagonal SQ marked c, square SQVU on it (Part II, §2.3, p.40). Same treatment, same letters.
• The square from Part II, §2.3, p.37, foot: right-angle marks at all four corners, the left and bottom sides marked 1 (the top and right unlabelled, as printed), and the diagonal from top-left to bottom-right marked √2 — sitting beside the bare isosceles right triangle with legs 1 and hypotenuse √2 from the same foot of page. The point of section 11 is that these are one figure.
• A number line for sections 8 and 10, marked with whole numbers and then re-marked in tenths, with the squaring candidates shown above it. Standard schematic; not printed in the chapter.
• Optional: the middle-right triangle from Part II p.53 no.2, with its right angle at the top, to make the point about not hunting for the sloping side.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part II, printed Chapter 2, "The Baudhāyana-Pythagoras Theorem", §2.3 "Hypotenuse of an Isosceles Right Triangle", the opening of the section at Part II p.37 and the "General Solution" subheading with its two worked examples at Part II pp.40–41. §2.3 also contains the subheading "Decimal Representation of √2" (Part II pp.38–39), which belongs to Why √2 is neither a terminating decimal nor a fraction, not here.
• Exercises: Part II p.39, Figure it Out no.2; Part II p.40, Figure it Out no.3, which begins and ends at the top of that page; Part II p.52, Figure it Out no.1; Part II p.53, Figure it Out no.2, the triangle whose marked sides are 10 and √200.
• Backward dependency inside the chapter: the doubling construction and its triangle-counting proof, §2.1, Part II pp.33–35.
• The chapter's SUMMARY at Part II p.54 carries the isosceles relation.