Question

Using standard identity, find the value of ${99}^2$.

• A. 9901
• B. 9801
• C. 1001
• D. 9701

Answer 1

The correct option is B 9801

Given,

$={99}^2$

$=(100-1{)}^2$

We know, $(a-b{)}^2=a^2-2ab+b^2$

Then,

${99}^2$

$=(100-1{)}^2$

$={100}^2+1^2-2\times 100\times 1$
$=10000+1-200$

$=10001-200$

$=9801$

Therefore, option $B$ is correct.