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6.Using identities, evaluate:
(i) \(71^2\)

(ii) \(99^2\)

(iii) \(102^2\)

(iv) \(998^2\)

(v) \(5.2^2\)

(vi) \(297 \times 303\)

(vii) \(78 \times 82\)

(viii) \(8.9^2\)

(ix) \(1.05 \times 9.5\)


Answer :

(i)To find: \(71^2\)

\(=(70+1)^2\)

\(=(70)^2+2(70)(1)+1^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=4900+140+1\)

\(=5041\)

Thus, we have \(71^2=5041\)

(ii)To find: \(99^2\)

\(=(100-1)^2\)

\(=(100)^2-2(100)(1)+1^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=10000-200+1\)

\(=9801\)

Thus, we have \(99^2=9801\)

(iii)To find: \(102^2\)

\(=(100+2)^2\)

\(=(100)^2+2(100)(2)+2^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=10000+400+4\)

\(=10404\)

Thus, we have \(102^2=10404\)

(iv)To find: \(998^2\)

\(=(1000-2)^2\)

\(=(1000)^2-2(1000)(2)+2^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=1000000-4000+4\)

\(=996004\)

Thus, we have \(998^2=996004\)

(v)To find: \(5.2^2\)

\(=(5+0.2)^2\)

\(=(5)^2+2(5)(0.2+(0.2)^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=25+2+0.04\)

\(=27.04\)

Thus, we have \((5.2)^2=27.04\)

(vi)To find:\(297\times 303\)

\((300-3)(300+3)=(300)^2-(3)^2\quad\)[using \((a-b)(a+b)=a^2-b^2\)]>

\(=90000-9\)

\(=89991\)

Hence we have,\(297\times303=89991\)

(vii)To find:\(78\times 82\)

\((80-2)(80+2)=(80)^2-(2)^2\quad\)[using \((a-b)(a+b)=a^2-b^2\)]>

\(=6400-4\)

\(=6396\)

Hence we have,\(78\times82=6396\)

(viii)To find: \(8.9^2\)

\(=(9-0.1)^2\)

\(=(9)^2-2(9)(0.1)+(0.1)^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=81-1.8+0.01\)

\(=79.21\)

Thus, we have \((8.9)^2=79.21\)

(ix)To find:\(1.05\times 9.5\)

\((1+0.05)(10-0.5)=1(10-0.5)+0.05(10-0.5)\)>

\(=10-0.5+0.05\times10-0.05\times0.5\)

\(=10-0.5+0.5-0.025\)

\(=10.5-0.525\)

\(=9.975\)

Hence we have,\(1.05\times9.5=9.975\)