Equivalent expressions from areas and patterns

Why does $(a+b)^2$ always equal $a^2+2ab+b^2$ , no matter what numbers $a$ and $b$ are? You could expand it term by term and trust the algebra — or you could draw a single square, split it into four pieces, and see the identity sitting there as areas. That is the idea behind equivalent expressions from areas and patterns: the same total area, counted in two different ways, must give two expressions that are equal. This is the geometric companion to the algebraic identities you already expand symbolically in the chapter We Distribute, Yet Things Multiply. Instead of memorising $(a+b)(a-b)=a^2-b^2$ as a rule, you build a picture where cutting and rearranging a square makes the identity obvious. The same "count it two ways" trick also explains number patterns, such as why the sum of the first few odd numbers always lands on a perfect square. Once you can read a rectangle as a sum of smaller areas, expanding brackets stops being a memorised procedure and becomes something you can draw.

The core insight

When one area (or one counted pattern) is computed two different ways, the two resulting expressions must be equivalent — this is what an area diagram or dot pattern actually proves.

The core idea

Splitting a rectangle proves the distributive law. Take a rectangle of height $a$ and width $(b+c)$ . Its area is $a(b+c)$ by the usual length-times-width rule. Now slice the width into two pieces, $b$ and $c$ . The same rectangle is now two smaller rectangles side by side, with areas $ab$ and $ac$ . Both descriptions measure the identical physical region, so:

a(b+c) = ab + ac

where: $a$ is the common height, and $b$ , $c$ are the two pieces the width is split into. Nothing algebraic happened here — it is one shape, counted twice.

Splitting a square proves $(a+b)^2$ . Draw a square of side $(a+b)$ and split each side into a piece $a$ and a piece $b$ . This carves the square into four regions: a small square of side $a$ , a small square of side $b$ , and two matching rectangles of size $a \times b$ . Summing the four regions gives the identity directly — see the figure below.

Cut-and-rearrange proves $(a+b)(a-b)$ . Start with a square of side $a$ (area $a^2$ ) and remove a smaller square of side $b$ from one corner. What remains is an L-shaped region of area $a^2-b^2$ . That same L-shape can be cut once and reassembled, with no gaps or overlaps, into a plain rectangle of dimensions $(a+b)$ by $(a-b)$ . Since cutting and rearranging never changes total area, $a^2-b^2$ and $(a+b)(a-b)$ must be the same number.

The same logic works for counting patterns, not just areas. Arrange dots into nested L-shaped layers around a corner: $1$ dot, then a strip of $3$ , then a strip of $5$ , then a strip of $7$ , and so on. After $n$ layers the dots always form a perfect $n \times n$ square, so the running total after $n$ odd numbers is always $n^2$ — a fact about numbers proved by watching a shape grow, exactly as the area diagrams prove algebraic identities.

Figure

A square of side a plus b split into four regions showing a squared plus two ab rectangles plus b squared

Derivation AProving (a+b)(a-b) equals a squared minus b squared by cutting and rearranging

The goal is to show, using only areas and a single physical cut — no algebraic expansion at all — that the two expressions $a^2-b^2$ and $(a+b)(a-b)$ describe the exact same region, for a positive number $a$ larger than a positive number $b$ , so the identity must hold.

Step 1. Draw a square of side $a$ ; its area is $a^2$ by the definition of a square's area. Mark a smaller square of side $b$ sitting in one corner of it, and remove that smaller square. The leftover, L-shaped piece has area $a^2 - b^2$ , since we started with $a^2$ and took away $b^2$ .

Step 2. Cut the L-shape into two rectangular strips: a horizontal strip of dimensions $a \times (a-b)$ and a vertical strip of dimensions $b \times (a-b)$ . Check these are exactly the L-shape: the horizontal strip is the part of the square below the removed corner, and the vertical strip is the part beside it, and together with the removed corner they tile the whole original square with no gaps.

Step 3. Slide the vertical strip so it sits at the end of the horizontal strip, joined along their matching side of length $(a-b)$ . Because sliding a rigid piece never changes its area, the two strips together still have total area $a^2-b^2$ .

Step 4. Read off the shape of the rearranged piece. Its total length is now $a + b$ (the horizontal strip's length $a$ plus the vertical strip's width $b$ , laid end to end), and its height throughout is $(a-b)$ . So the rearranged shape is a single rectangle of dimensions $(a+b) \times (a-b)$ , and its area is $(a+b)(a-b)$ .

Step 5. Since Step 1 gave area $a^2-b^2$ and Step 4 gave the same physical region with area $(a+b)(a-b)$ , and cutting-and-sliding never changes area, the two expressions must be equal: $a^2-b^2=(a+b)(a-b)$ .

Worked example

Use an area diagram to justify $(x+3)(x+5)=x^2+8x+15$ for a positive value of $x$ , and check the result against the odd-number square pattern.

Step 1. Draw a rectangle with one side of length $(x+3)$ and the adjacent side of length $(x+5)$ . Split the first side into pieces $x$ and $3$ , and split the second side into pieces $x$ and $5$ . This divides the rectangle into four smaller regions, exactly like the $(a+b)^2$ picture but with unequal splits on each side.

Step 2. Read off each region's area: top-left is $x \times x = x^2$ , top-right is $x \times 5 = 5x$ , bottom-left is $3 \times x = 3x$ , and bottom-right is $3 \times 5 = 15$ .

Step 3. Add the four regions to get the total area of the big rectangle a second way: $x^2 + 5x + 3x + 15$ . Combining the like terms $5x$ and $3x$ gives $x^2+8x+15$ .

Step 4. This matches $(x+3)(x+5)$ found by algebraic expansion, confirming the identity — the diagram and the algebra describe the same rectangle, so they must agree. Answer: $(x+3)(x+5)=x^2+8x+15$ , verified by both the area diagram and direct expansion.

Competitive stretch

Using the nested-L-shape dot pattern from the core idea, check the identity for $n=4$ : starting from $1$ dot and adding L-strips of $3$ , $5$ , then $7$ dots builds squares of side $1$ , $2$ , $3$ , $4$ with areas $1$ , $4$ , $9$ , $16$ . Running totals confirm $1=1^2$ , $1+3=4=2^2$ , $1+3+5=9=3^2$ , and $1+3+5+7=16=4^2$ — the sum of the first four odd numbers is $16$ , matching $4^2$ exactly.

Common traps

Miscounting the regions of $(a+b)^2$ , forgetting it splits into four regions, with two equal $ab$ rectangles, not three regions total.
Treating the diagram as a mere illustration rather than a genuine proof — the total area computed two different ways must be equal, so the picture is the argument, not decoration.
Mismatching split lengths across the two sides, pairing the wrong pieces from each side and reading off the wrong region dimensions.
Forcing an area diagram onto three-or-more-term expressions, where a simple rectangle split no longer applies; this method works cleanly for two-term-by-two-term products.

Ask the tutor: if you want to see the area diagram drawn out step by step for any specific pair of brackets you are expanding, ask the AI tutor to sketch it with you.

Why does an area diagram count as an actual proof and not just a nice picture?

Because both expressions describe measurements of the same physical region. The total area of a shape does not depend on how you choose to slice it up and add the pieces — it is a fixed number. So if one way of counting gives $a(b+c)$ and another way of counting the identical rectangle gives $ab+ac$ , those two expressions have to be equal for every value of $a$ , $b$ , and $c$ , not just as an approximation. That is exactly what an algebraic identity claims, so the diagram genuinely establishes it.

Does this method work for expressions like (a+b+c) squared with three terms?

Not with a simple rectangle split, no — a two-piece-by-two-piece rectangle naturally gives four regions, matching a two-term-by-two-term product. For three terms you would need a square split into nine regions (a three-by-three grid), which is a valid extension of the same idea but is beyond what this topic covers here; stick to two-term splits for now and expand three-term products algebraically instead.

How is the odd-number dot pattern related to area diagrams at all?

Both rely on the same "count the same total two ways" idea, just with dots instead of continuous area. Each L-shaped strip of dots wraps around an existing square and turns it into the next-size square, so the number of dots added at each step (the next odd number) always equals the extra area needed to grow the square by one unit on each side. Counting the dots strip by strip and counting the final square's area both give the same total, which is why the sum of odd numbers lands exactly on a perfect square.

Equivalent expressions from areas and patterns

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads