Question

Approximate $\sqrt{6}$ to two decimal places.

Answer 1

Step 1: Find the whole number of the decimal form of $\sqrt{6}$ by perfect square method.

The largest perfect square less than 6 is 4.
The smallest perfect square greater than 6 is 9.

$4$ < $6$ < $9$
$\sqrt{4}$ < $\sqrt{6}$ < $\sqrt{9}$
2 < $\sqrt{6}$ < 3

$\therefore$ The whole number of decimal form of $\sqrt{6}$ lies between 2 and 3.

$\sqrt{6}$ = 2._____

Step 2: Then, find the first decimal place of the decimal form of $\sqrt{6}$ by perfect square method. So, consider 600.

The largest perfect square less than 600 is 576.
The smallest perfect square greater than 6 is 625.

576 < 600 < 625
${24}^2$ < 600 < ${25}^2$ ($\because {24}^2=576;{25}^2=625$)

Now place the decimal point
5.76 < 6.00 < 6.25
${2.4}^2$ < 6 < ${2.5}^2$
$\sqrt{{2.4}^2}$ < $\sqrt{6}$ < $\sqrt{{2.5}^2}$
$2.4$ < $\sqrt{6}$ <2.5

$\therefore$ The first decimal place of $\sqrt{6}$ is 4.
$\sqrt{6}$ = 2.4

Step 3: Then, find the second decimal place of the decimal form of $\sqrt{6}$ by perfect square method. So, consider 60,000.

The largest perfect square less than 60,000 is 59,536.
The smallest perfect square greater than 6 is 60,025.

59,536 < 60,000 < 60,025
${244}^2$ < 60,000 < ${245}^2$
($\because {244}^2=59,536;{245}^2=60,025$)

Now place the decimal point
5.9536 < 6.0000 < 6.0025
${2.44}^2$ < 6.0000 < ${2.45}^2$
$\sqrt{{2.44}^2}$ < $\sqrt{6}$ < $\sqrt{{2.45}^2}$
$2.44$ < $\sqrt{6}$ <2.45

$\therefore$ The second decimal place of $\sqrt{6}$ is 4.
So we can write $\sqrt{6}$ as 2.44 .

$\therefore$ The approximate value of $\sqrt{6}$ is 2.44.