Question

Evaluate each of the following using identities :
i) $(399{)}^2$
ii) $(0.98{)}^2$
iii) $991\times 1009$

• A. i) 159876ii) 0.91
  iii) 876590
  
• B. i) 135879ii) 0.87
  iii) 896750
  
• C. i) 159201ii) 0.9604
  iii) 999919
  
• D. i) 138760ii) 0.9
  iii) 999999
  

Answer 1

Answer: (C)

i) 159201

ii) 0.9604

iii) 999919

$\left(i\right)$

$(399{)}^2=(400-1{)}^2$

$=(400{)}^2+(1{)}^2-2\times 400\times 1$ [ $(a-b)=a^2+b^2-2ab$ ]

$=160000+1-800$

$=159201$

$\therefore$ $(399{)}^2=159201$

$\left(ii\right)$

$(0.98{)}^2=(1-0.02{)}^2$

$=(1{)}^2+(0.02{)}^2-2\times 1\times 0.02$ [ $(a-b)=a^2+b^2-2ab$ ]

$=1+0.0004-0.04$

$=0.9604$

$\left(iii\right)$

$991\times 1009=(1000-9)(1000+9)$

$=(1000{)}^2-(9{)}^2$ [ $(a+b)(a-b)=a^2-b^2$ ]

$=1000000-81$

$=999919$