Page 108 - Start Up Mathematics_8 (Non CCE)

Observations: The remaining portion of I (purple colour) RUSK is a square of side 6 cm.
(a – b = 10 – 4 = 6 cm)

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Area of RUSK = (a – b) = (6) = 36 cm 2
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\ LHS = (a – b) = (10 – 4) = 6 = 36
Area (RUSK) = Area (RING) – {Area UINO + Area KSOG}
= Area (RING) – {(Area SUIT + Area of OSTN) + Area KSOG} ...(1)
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Area of RING = (a) = (10) = 100 cm 2
Area of SUIT = (a – b) ¥ b = 6 ¥ 4 = 24 cm 2
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Area of OSTN = (b) = (4) = 16 cm 2
Area of KSOG = (a – b) ¥ b = 6 ¥ 4 = 24 cm 2
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RHS of (1) = a – {(a – b) ¥ b + b + (a – b) ¥ b}
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= a – {ab – b + b + ab – b }
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= a – (2ab – b ) = a – 2ab + b 2
Area of RUSK = 100 – (24 + 16 + 24) = 100 – 64 = 36 cm 2
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\ RHS = a – 2ab + b = 36 cm 2
\ LHS = RHS
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Conclusion: Hence the identity (a – b) = a – 2ab + b is verified.
Special Product
(x + a)(x + b) = x(x + b) + a(x + b) (Distributive property of multiplication over addition)
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= x + xb + ax + ab
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= x + bx + ax + ab (Commutative property xb = bx)
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= x + ax + bx + ab ( bx + ax = ax + bx)
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= x + (a + b)x + ab (Taking x common from ax + bx)

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So, you have the identity: \ (x + a)(x + b) = x + (a + b)x + ab
This identity can be used for the following three identities:

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(i) (x + a)(x – b) = x + (a – b)x – ab
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(ii) (x – a)(x + b) = x + (b – a)x – ab
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(iii) (x – a)(x – b) = x – (a + b)x + ab
All these identities are very helpful in simplifying and evaluating algebraic expressions.
Example 33: Find the products: (a) (x + 4)(x + 5) (b) (4x – 3)(4x + 5)

Solution: (a) (x + 4)(x + 5)
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= (x) + (4 + 5)x + 4 ¥ 5 {Using (x + a)(x + b) = x + (a + b)x + ab}
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= x + 9x + 20
100