Question

Using identities, evaluate ${99}^2$

• A. $9135$
• B. $1694$
• C. $9471$
• D. $9801$

Answer 1

The correct option is D $9801$

${99}^2$
$=(100-1{)}^2$
Using,
$(a-b{)}^2=a^2-2ab+b^2$ we get,
$=(100{)}^2-2(100\left)\right(1)+(1{)}^2$
$=10000-200+1$
$=9801$