Label the squares and rectangles in the figure so that it represents the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

Step-by-step solution

Idea: The whole square has area (a + b + c)². The same area is the sum of the nine pieces. A piece in column p and row q is a rectangle of width p and height q, so its area is p × q.

1. Top row (height a): widths a, b, c give areas a², ab, ac.½ mark
2. Middle row (height b): areas ab, b², bc.½ mark
3. Bottom row (height c): areas ac, bc, c².½ mark
4. So there are three squares (a², b², c², along the diagonal) and six rectangles in matching pairs: two ab, two bc, two ca. See the labelled diagram below.½ mark
5. Adding all nine areas: a² + b² + c² + 2ab + 2bc + 2ca.
   This equals the area of the big square, (a + b + c)², which shows the identity.1 mark

Check: Take a = 3, b = 2, c = 1: whole square = 6² = 36. Pieces: 9 + 4 + 1 + 2(6) + 2(2) + 2(3) = 14 + 12 + 4 + 6 = 36 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Labelling a rectangle with the sum of its sides (for example a + b) instead of the product ab. Area is length × breadth.
• Counting only one ab, one bc and one ca. Each rectangle appears twice, once on each side of the diagonal.
• Mixing up which side is which: the piece in the c column and b row is bc, not b² or c².

How this can come in the exam

Try one yourself

Using the same kind of model, find the area of a square of side (5 + 3 + 2) cm by adding its nine pieces.

More questions like this

• Let us use this identity to find the square of a number, say 119:
• Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
• Factor using suitable identities:
• Expand the following using the identity
  (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:
• Is this an identity?
  (a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c².

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