Question

The square root of sum of the digits in the square of $121$ is

• A. $4$
• B. $3$
• C. $6$
• D. $9$

Answer 1

The correct option is A $4$

${\left(121\right)}^2$

$={(100+21)}^2$

$={100}^2+{21}^2+2\left(100\right)\left(21\right)$ $[\because (a+b{)}^2=a^2+2ab+b^2]$

$=14641$

Sum of digits $=1+4+6+4+1=16$

Square root$=\sqrt{16}=4$