Chapter 11.6 — Exercise 11.5 — Geometric Representations
Geometrical representations of algebraic identities. This is Lesson 6 of 6 in Chapter 11: Algebraic Expressions.
Turning an Algebraic Identity Into an Area
Every identity from the previous lesson can be seen, not just calculated. A square of side (a + b) has area (a + b)² by definition — and that same square, cut along two lines parallel to its sides, splits cleanly into four smaller pieces: a square of side a, a square of side b, and two identical rectangles of sides a and b.
Four Pieces, One Square
Area of whole square = (a+b)²
Sum of four pieces = a² + ab + ab + b² = a² + 2ab + b²
∴ (a+b)² ≡ a² + 2ab + b²The identity isn't a separate fact needing its own proof — it's a direct consequence of the fact that the whole square's area must equal the sum of its parts, however that square happens to be sliced.
Verifying (a + b)² With Actual Measurements
Substituting real numbers into the picture above confirms the identity numerically, not just visually:
| a, b | L.H.S. = (a+b)² | R.H.S. = a² + 2ab + b² |
|---|---|---|
| a=2, b=4 | 6² = 36 sq.u. | 4 + 16 + 16 = 36 sq.u. |
| a=3, b=1 | 4² = 16 sq.u. | 9 + 6 + 1 = 16 sq.u. |
| a=5, b=2 | 7² = 49 sq.u. | 25 + 20 + 4 = 49 sq.u. |
In every row, the R.H.S. is written out as four separate areas before adding — the square of side a, the two a-by-b rectangles counted separately (not doubled as a shortcut), and the square of side b — mirroring exactly the four regions of the diagram rather than the already-simplified 2ab term.
Why the Two Rectangles Aren't Counted as One
It's worth noticing that the diagram keeps the two a-by-b rectangles as separate pieces rather than immediately combining them into "one 2ab piece." That choice isn't arbitrary — it mirrors exactly how the square was cut, along two perpendicular lines that create four genuinely distinct regions, not three. Only after all four areas are added together does 2ab emerge, as ab + ab collapsing into a single term because they happen to be like terms. Skipping straight to 2ab without first seeing two separate ab regions would still give the right final identity, but it would hide the actual geometric reason a coefficient of 2 shows up in the middle term at all — it isn't an arbitrary multiplier, it's a count of how many identical rectangles the cut produced.
Cutting a Square Down Instead of Building One Up
The minus-sign identity works by taking area away rather than adding it. Start with a square of side a, and cut two strips of width b out of it along two adjacent sides. What's left in the corner is a smaller square of side (a − b) — call it Part I — and the three strips removed are Part II, Part III, and Part IV.
Part I = a² − (area II + III + IV)
(a−b)² = a² − (a−b)b − b(a−b) − b²
= a² − ab + b² − ab + b² − b² = a² − 2ab + b²
∴ (a−b)² ≡ a² − 2ab + b²This works from the opposite direction of the first proof: instead of building (a+b)² up from four added pieces, (a−b)² is found by starting with the full a² square and subtracting exactly the pieces that don't belong to the smaller inner square. Notice too that Parts II and III are subtracted once each — not combined into a single "2ab" subtraction from the start — for the same reason the two rectangles in the (a+b)² proof were kept separate: the diagram genuinely has two distinct strips removed, of dimensions (a−b)×b and b×(a−b), and it's only after both are subtracted individually that they combine into the familiar −2ab term.
Checking (a − b)² Two Different Sizes
| a, b | L.H.S. = (a−b)² | R.H.S. = a² − 2ab + b² |
|---|---|---|
| a=3, b=1 | 2² = 4 sq.u. | 9 − 3 − 3 + 1 = 4 sq.u. |
| a=5, b=2 | 3² = 9 sq.u. | 25 − 10 − 10 + 4 = 9 sq.u. |
Both rows subtract the middle term twice separately (−ab and −ab again) rather than jumping straight to −2ab, keeping the arithmetic lined up exactly with the four labelled regions of the diagram above.
Removing a Corner Instead of a Border
The third identity starts differently again: take a square of side a, and remove a smaller square of side b from one of its corners entirely, rather than cutting strips along the edges.
a² − b² = area of Part I + area of Part II = a(a−b) + b(a−b)
= (a−b)(a+b)
∴ a² − b² ≡ (a − b)(a + b)Sliding Part II around to sit alongside Part I turns the L-shaped remainder into one plain rectangle, (a − b) wide and (a + b) tall — which is exactly why a² − b² factors into (a−b)(a+b) rather than needing any further explanation. This rearrangement step is the one genuinely new idea across all three proofs in this lesson: the first two identities only ever added or subtracted areas in place, but this third one physically moves a piece to a new position, relying on the fact that cutting and sliding a shape never changes its area, only where it sits.
Confirming the Factoring Both Ways
| a, b | a² − b² | (a−b)(a+b), as two rectangle areas |
|---|---|---|
| a=3, b=2 | 9 − 4 = 5 sq.u. | 3(1) + 2(1) = 3 + 2 = 5 sq.u. |
| a=2, b=1 | 4 − 1 = 3 sq.u. | 2(1) + 1(1) = 2 + 1 = 3 sq.u. |
Both checks compute the rearranged rectangle's area as two separate pieces — a×(a−b) and b×(a−b) — rather than multiplying the full (a+b) by (a−b) directly, which keeps the verification tied visibly back to the two labelled regions in the diagram rather than collapsing straight to the already-known algebraic shortcut.
What the Pictures Add That the Algebra Alone Doesn't
All three identities were already proven algebraically in Exercise 11.4, by direct multiplication of the brackets. Nothing here changes which identities are true — the geometric versions are a different kind of evidence for the same three facts, built from areas that can be measured and rearranged by hand rather than symbols that can only be manipulated on paper. This closes out the chapter's arc: from naming the pieces of an expression, through every pattern of multiplying them together, to identities that shortcut the most common of those multiplications, and finally to seeing why those shortcuts were never just algebraic tricks — they were true about area and length all along.