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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,\(1.05\times9.5=9.975\)