Question

The value of ${\left(1.02\right)}^2+{\left(0.98\right)}^2$, corrected to three decimal places is:

• A. $2.001$
• B. $2.004$
• C. $2.006$
• D. $1.995$

Answer 1

The correct option is A $2.001$

Given, $(1.02{)}^2+(0.98{)}^2$.

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

Then,

$(1.02{)}^2+(0.98{)}^2$

$=(1.02{)}^2+(0.98{)}^2+(2\times 1.02\times 0.98)-(2\times 1.02\times 0.98$)

$=(1.02+0.98{)}^2-1.9992$

$=(2{)}^2-1.9992$

$=4-1.9992$

$=2.0008$.

When corrected to three decimal digit,

$2.0008$ $=2.001$.

Therefore, option $A$ is correct.