Question

The tenth place value of $\sqrt{8}$ .

Answer 1

We know that tenth place value of a number is the digit just after decimal point of the number.

Let's find the square root of $8$ using the perfect square method.

The largest perfect square smaller than $8$ is $4.$
And, the smallest perfect square larger than $8$ is $9.$

$4<8<9$
(taking square root throughout the inequality)
$\Rightarrow \sqrt{4}<\sqrt{8}<\sqrt{9}$
$\Rightarrow 2<\sqrt{8}<3$
$\therefore \sqrt{8}=2.\underset{–}{}$

To find the tenth place digit of $\sqrt{8},$ we start with $800$


The largest perfect square smaller than $800$ is $784\text{i.e.}{28}^2.$
And, the smallest perfect square larger than $800$ is $841\text{i.e.}{29}^2.$

${28}^2<800<{29}^9$
$\Rightarrow 784<800<841$
(dividing by $100$ throughout the inequality)
$\Rightarrow \frac{784}{100}<8.00<\frac{841}{100}$
(taking square root throughout the inequality)
$\Rightarrow \sqrt{\frac{{28}^2}{{10}^2}}<\sqrt{8}<\sqrt{\frac{{29}^2}{{10}^2}}$
$\Rightarrow 2.8<\sqrt{8}<2.9$
$\therefore \sqrt{8}=2.\underset{–}{8}$

Hence, the tenths place digit of $\sqrt{8}$ is $8.$