Evaluate each of the following using identities:

(i) $(399)^{2}$

(ii) $(0.98)^{2}$

(iii) $991 \times 1009$

(iv) $117 \times 83$

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Solution

$(i) (399)^{2} = (400 - 1)^{2} = (400)^{2} - 2 \times 400 \times 1 + (1)^{2} {∵ (a - b)^{2} = a^{2} - 2 a b + b^{2}} = 160000 - 800 + 1 = 160001 - 800 = 159201 (i i) (0.98)^{2} = (1 - 0.02)^{2} {∵ (a - b)^{2} = a^{2} - 2 a b + b^{2}} = (1)^{2} - 2 \times 1 \times 0.02 + (0.02)^{2} = 1 - 0.04 + 0.0004 = 1.0004 - 0.04 = 1.0004 - 0.0400 = 0.9604 (i i i) 991 \times 1009 = (1000 - 9) (1000 + 9) {∵ (a + b) (a - b) = a^{2} - b^{2}} = (1000)^{2} - (9)^{2} = 1000000 - 81 = 999919 (i v) 117 \times 83 = (100 + 17) (100 - 17) {∵ (a + b) (a - b) = a^{2} - b^{2}} = (100)^{2} - (17)^{2} = 10000 - 289 = 9711$

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