Combining two different squares: a² + b² = c²

What to assume they know

• Doubling a square: the diagonal is the construction — combining two equal squares by building on the diagonal, and the triangle-counting proof
• The isosceles right triangle: why its hypotenuse must be a√2 — the isosceles case, and c² = 2a² from the same construction
• Angles on a straight line add to 180°, and the angles of a triangle add to 180°
• Congruent triangles have equal corresponding sides and angles, and equal area
• The area of a square is side × side, so a square of area A has side √A
• Cutting a figure into pieces and rearranging them leaves the area unchanged
• Substituting numbers into a² + b² = c², and reading it in both directions

What they should be able to do

• State Baudhāyana's theorem for a right-angled triangle with sides a, b and hypotenuse c
• Given two squares of different sizes, construct a single square whose area is their total, using the rectangle-and-diagonal recipe
• Identify, in the chapter's construction, the right triangle whose legs are the two given sidelengths
• Explain why the four-sided figure built on the four hypotenuses is a square, arguing both its sides and its angles
• Give the dissection proof: two arrangements of one middle piece and two copies of one triangle
• Check the construction against the equal-squares case of §2.1 and say what happens to the rectangle when the two squares match
• Predict the hypotenuse of a 3-and-4 right triangle before measuring it, and confirm by drawing
• Find a missing side of a right triangle when the hypotenuse is the unknown, and when the hypotenuse is one of the givens
• Attribute the theorem as the chapter attributes it, and explain why the chapter uses a joint name

Where it usually goes wrong

• **"a + b = c."** The straight side is shorter than the two-leg path. Show the 3, 4 and 5 triangle: the bent path is 7 and the straight one is 5.
• **"a² + b² = c², so a + b = c after square-rooting."** Square-rooting a sum is not adding the square roots. This is the same error one level up and it survives longer, because it is dressed as algebra. Put √(9 + 16) = 5 beside 3 + 4 = 7 on one screen.
• "The hypotenuse is the sloping side." It is the side facing the right angle. On Part II p.53 the right angle sits at the top in one drawing and at the far left in another, and in three of the six drawings the marked side is the hypotenuse rather than a leg — the triangles labelled 40/41, 10/√200 and 27/45. A student hunting for "the slanted one" will misread the figure and use the wrong relation.
• "The theorem gives the hypotenuse, so I always add." Only when the hypotenuse is the unknown. Item (iii) of Part II p.47 no.4 gives c and asks for a leg, and item no.2 does the same — those subtract. Sorting the givens before computing is the actual skill.
• "The recipe works because the picture looks right." The chapter goes to two full pages of dissection precisely because it does not. An explanation that shows the three-squares figure and moves to arithmetic has skipped the content of the section.
• "The four-sided figure is obviously a square." The chapter is careful to call it a four-sided figure until it has been proved. Four equal sides alone give a rhombus; the angle argument is what upgrades it. This caution is worth imitating — it models what a proof is for.
• "So this is a different theorem from §2.3's." It is the same one. Put b = a in a² + b² = c² and you have c² = 2a². The chapter asks the student to check that the general method agrees with the equal-squares method, and section 6 should answer it: when the two squares match, b − a is zero, the marked rectangle is the whole square, and its diagonal is the square's own diagonal.
• "The theorem tells you a triangle is right-angled." It does not, as stated. The chapter proves one direction — right angle, therefore the area relation — and never proves the reverse. See Notes; this matters from §2.5 onward.

Questions to check understanding

• Given the two shorter sides of a right triangle, find the hypotenuse
• Given the hypotenuse and one shorter side, find the third side
• Given three lengths, decide which could be a right triangle's sides and which must be the hypotenuse
• Given two squares, construct a single square of the combined area, and name the rectangle you marked
• Prove that the four-sided figure in the chapter's construction is a square
• Justify that the hypotenuse is the longest side
• Find the sidelength of a rhombus from its two diagonals
• Construct a square whose area is the difference of two given squares' areas
• Construct a square of three times, or five times, a given square's area
• Explain how the general theorem contains the isosceles result of §2.3

Examples worth working on the board

Where the chapter computes something, that is said; items marked derived are added here.

• The equal-squares recap (Part II, §2.4, p.41, upper figure). Two identical squares with a plus sign between them, an arrow, and a larger square drawn with a broken outline. The chapter states that the new square's side is the diagonal of either of the small ones.
• The unequal-squares question (Part II, §2.4, p.41, lower figure). A small square, a plus sign, a visibly larger square, an arrow, and a question mark where the answer should be. Hold this frame; it is the section's whole premise.
• Verse 1.12, in a tinted box (Part II, §2.4, p.41). Its content: the square built on the diagonal has the combined area of the squares built on the two sides. The chapter restates it operationally on the next page — build a right triangle whose two perpendicular sides are the two given sidelengths, and the square on its hypotenuse is the square you want.
• The three-squares figure (Part II, §2.4, p.42). A right triangle with a square on each of its three sides, the smaller leg square labelled Area A by a leader line, the larger leg square labelled Area B, and the square on the hypotenuse labelled with their total. Beside it, the two loose squares with a plus and an arrow. This is the picture the whole section is out to justify.
• Verse 2.1, in a tinted box (Part II, §2.4, p.42). Its content: inside the larger square, mark off a rectangle whose width is the smaller square's side; that rectangle's diagonal is the side of the square you want.
• The construction, in four numbered stages (Part II, §2.4, pp.42–44).
  • Joining. The small square, side a, and the large square, side b, are set side by side on one baseline, so the baseline reads a then b.
  • Stage 1. The rectangle of width a is marked at the far end of the larger square and its diagonal drawn, so the baseline now reads a, then b − a, then a. The right triangle produced is shaded on the page; its two perpendicular sides are a and b.
  • Stage 2. A second copy of that triangle is shaded in, at the upper left.
  • Stage 3. A third copy, rotated up out of the figure.
  • Stage 4. The fourth, completing a four-sided figure standing on the four hypotenuses.
• The angle figure (Part II, §2.4, p.44). The drawing carries **four x and four 90 − x, but not one of each at every vertex — verified on the printed page of the two that differ. At the top and bottom** vertices of the new four-sided figure the pair is x with 90 − x, and each of those vertices sits on a straight line. At the right vertex the pair is ***x* with x, and at the left vertex it is 90 − x with 90 − *x*** — and both of those sit at a right-angled corner of the outline, where the ambient angle is 90°, not 180°. Derived, and this is the argument the chapter asks the student to complete: each triangle's two non-right angles add to 90°, so at every vertex the two angles abutting the figure are the two acute angles of congruent right triangles and therefore add to 90°. What is left over is the figure's own angle: at the top and bottom vertices 180 − x − (90 − x) = 90°, and at the left and right vertices the ambient right angle less the two marked angles, again 90°. State both cases. An explanation that runs the straight-line subtraction at all four vertices teaches an invalid step at two of them, even though the conclusion — four right angles, four sides that are hypotenuses of congruent triangles, so the figure is a square — is correct.
• The two arrangements (Part II, §2.4, pp.44–45). The chapter names five pieces. The square on the hypotenuse is T + U + V. The two given squares together are V + W + X. T, U, W and X are four congruent copies of the right triangle with legs a and b, and V is the same piece in both pictures. So T + U equals W + X, and therefore the square on the hypotenuse has the same area as the two given squares. The chapter states the chain of equalities and prints the two arrangements side by side at the top of Part II p.45.
• The paper version (Part II, §2.4, pp.45–46). A red square of side a joined to a yellow square of side b on one baseline. Two cuts are made, giving three pieces; both cuts are marked c on the figure, and the baseline again reads a, b − a, a. The three pieces are rearranged into one tilted square, its sides marked c. Then a right triangle with legs a and b is drawn with a square on each side, and the same three pieces are laid into the square on the hypotenuse, leaving the two leg squares as empty outlines. Derived: both cuts have length c because each is the hypotenuse of a right triangle with legs a and b — that is why exactly two cuts suffice.
• The theorem, in a tinted box (Part II, §2.4, p.46). For a right-angled triangle with sides a, b and hypotenuse c: a² + b² = c².
• Attribution (Part II, §2.4, p.47). Baudhāyana stated it first in this general and essentially modern form. Pythagoras, dated about 500 BCE, lived a couple of centuries later and also studied and admired it. The chapter calls the joint name transitional and uses it so that readers know which theorem is meant.
• The first prediction, worked on the page (Part II, p.47). Draw a right triangle with the shorter sides 3 cm and 4 cm; measuring the hypotenuse gives about 5 cm. The theorem gives it exactly: 3² + 4² = 9 + 16 = 25, so c = 5 cm. The margin figure marks 3 cm on the vertical side, 4 cm on the horizontal, and a question mark on the hypotenuse.
• Figure it Out, Part II p.47. No.1: legs of 5 cm and 12 cm, hypotenuse wanted, by drawing and then by the theorem. No.2: one short side 8 cm and hypotenuse 17 cm, third side wanted, again both ways. No.3: construct a square of three times, and of five times, a given square's area, attributed to Śulba-Sūtra Verse 1.10. No.4 supplies two of the three sides in five cases and wants the third; the five givens, with an added answer after each since none is printed, are a = 5 with b = 7, giving √74; a = 8 with b = 12, giving √208; a = 9 with c = 15, giving 12; a = 7 with b = 12, giving √193; and a = 1.5 with b = 3.5, giving √14.5. Derived for nos.1 and 2 as well: 13, then 15. Note that the third of the five cases is the only one supplying the hypotenuse, and the fifth the only one with decimals.
• Later items that are this section's (Part II, p.53). No.3: a rhombus with diagonals 24 and 70, sidelength wanted. No.4: whether the hypotenuse is always the longest side, with justification. No.7: a square holding the area left when a square of side 5 is taken away from one of side 7. Derived: 37 for the rhombus, from half-diagonals 12 and 35 — and (12, 35, 37) is one of the triples Baudhāyana himself lists on Part II p.48, which the chapter never points out; c > a for no.4, because c² exceeds a² by b²; and side √24 for no.7.

Figures to have open

• The chapter's four-stage construction, step by step: the joined squares, the marked rectangle with its diagonal, and the three further triangles (Part II, §2.4, pp.42–44). Sections 6 to 8 cannot be told without it. Redraw as a schematic; the baseline labels a, b − a, a must be kept, because they are what makes the marked rectangle findable.
• The two arrangements of pieces T, U, V, W, X, with V drawn in a distinct colour in both so a student can see it is the same piece (Part II, §2.4, pp.44–45). This is the proof and it must be one continuous movement, not two stills.
• The angle figure (Part II, §2.4, p.44), redrawn as a schematic with the marks as printed: x with 90 − x at the top and bottom vertices, x with x at the right vertex, and 90 − x with 90 − x at the left. The two corner vertices must be distinguishable from the two straight-line vertices, because the argument differs at them.
• The three-squares-on-a-right-triangle figure (Part II, §2.4, p.42), which the explanation will want again in section 11 with the pieces laid into the hypotenuse square (Part II, p.46).
• The 3-4-5 triangle with its measured and predicted hypotenuse (Part II, p.47). Standard schematic.
• Optional: the rhombus with diagonals 24 and 70, split into four right triangles, for the assessment section.
• No photograph is needed. The chapter's coloured paper figures can be redrawn flat.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part II, printed Chapter 2, "The Baudhāyana-Pythagoras Theorem", §2.4 "Combining Two Different Squares", Part II pp.41–47. §2.4 prints two bold unnumbered subheadings, nameable but not citable by number: "Combining Two Squares Using Paper" (Part II p.45) and "Using Baudhāyana's Theorem" (Part II p.47).
• The two Śulba-Sūtra verses the section rests on, both in tinted boxes: Verse 1.12 at Part II p.41 and Verse 2.1 at Part II p.42. Verse 1.10, cited for the area-triple construction, is at Part II p.47.
• The theorem's own statement box is at Part II p.46; the attribution paragraph naming Pythagoras and the joint name is at Part II p.47.
• Exercises: Part II p.47, Figure it Out nos.1–4; Part II p.53, Figure it Out nos.3, 4 and 7.
• Backward dependencies inside the chapter: §2.1 at Part II pp.33–35 and §2.3 at Part II pp.37, 40–41.
• The chapter's SUMMARY at Part II p.54 states the theorem and calls it one of geometry's most fundamental.