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Geometrical Representation of (a + b) 2

1. Draw a square ABCD of side (a + b) units and divide it into four parts, a + b
i.e., U, V, X, and W as shown alongside. A a b B

2. Further, area of square ABCD = Area of square U + Area of rectangle V +
Area of rectangle X + Area of square W. U V a
a 2 ab a + b
Since, area of square ABCD = AB × BC = (a + b)(a + b) = (a + b) 2 ...(i)
a
2
2
And, area of square U = a , area of square W = b , b X ab W b
b
2
area of rectangle V = ab, and area of rectangle X = ab. D C
\ Area of square U + area of rectangle V + area of rectangle X + area of square W
2
2
2
= a + ab + ab + b = a + 2ab + b 2 ...(ii)
2
2
Thus, from (i) and (ii) (a + b) = a + 2ab + b 2
Identity 2: (a – b) = a – 2ab + b 2
2
2
i.e., Square of difference of two terms = Square of first term – 2 × (first term) × (second term)
+ Square of second term
2
Proof: We have, (a – b) = (a – b)(a – b) = a(a – b) – b(a – b)
[Distributivity of multiplication over subtraction]
2
2
2
= a – ab – ab + b = a – 2ab + b 2
Geometrical Representation of (a – b) 2
1. Draw a square ABCD of side a units and divide it into four parts, i.e., a
U, V, X, and W as shown alongside. A (a – b) b B

2. Area of square U = Area of square ABCD – [Area of rectangle V U V
+ Area of rectangle X + Area of rectangle W] (a – b) 2 b(a – b) (a – b)
a
2
2
2
2
2
2
fi (a – b) = a – [b(a – b) + b(a – b) + b ] = a – ba + b – ba + b – b 2
X
2
2
Thus, (a – b) = a – 2ab + b 2 b b(a – b) W b
b
2
2
Identity 3: (a + b)(a – b) = a – b 2 D (a – b) C
i.e., Product of sum of two terms and difference of two terms = square of first term – square of
second term.
2
2
Proof: We have, (a + b)(a – b) = a(a – b) + b(a – b) = a – ab + ba – b = a – b 2
2
Geometrical Representation of (a + b)(a – b)
1. Draw a rectangle ABCD of length (a + b) units and breadth a units and a
divide it into four parts U, V, W and X as shown alongside. A (a – b) E b B
2. Area of rectangle ABCD = Area of ADGE + Area of EBHF
+ Area of FHCG. U V
a
Now, Area of ADGE = Area of rectangle ABCD – Area of EBHF a(a – b) ab (a + b)
– Area of FHCG.
F
2
2
2
2
fi (a + b)(a – b) = a(a + b) – ab – b = a + ab – ab – b = a – b 2 I b H
X W
2
Thus, (a + b)(a – b) = a – b 2 b(a – b) b 2
D G C
Note: The identities 1, 2 and 3 are known as standard identities.
207 Algebraic Expressions