Question

Compare the given numbers by comparing their approximate values: $2\sqrt{3},\sqrt{10}$

• A. $2\sqrt{3}<\sqrt{10}$
• B. $2\sqrt{3}>\sqrt{10}$
• C. It cannot be determined as there is no common term.

Answer 1

The correct option is B $2\sqrt{3}>\sqrt{10}$

As $2\sqrt{3}$ and $\sqrt{10}$ do not have like terms, find the approximate value of both the numbers using the perfect square method.

Let's find the approximate values up to one decimal place.

For $2\sqrt{3},$ first we need to find approximate value of $\sqrt{3}.$

The largest perfect square smaller than $300$ is ${17}^2.$
And, the smallest perfect square larger than $300$ is ${18}^2.$

${17}^2<300<{18}^2$
(dividing by $100$ throughout the inequality)
$\Rightarrow \frac{{17}^2}{100}<3.00<\frac{{18}^2}{100}$
(taking square root throughout the inequality)
$\Rightarrow \sqrt{\frac{{17}^2}{{10}^2}}<\sqrt{3}<\sqrt{\frac{{18}^2}{{10}^2}}$
$\Rightarrow 1.7<\sqrt{3}<1.8\Rightarrow 3.4<2\sqrt{3}<3.6\therefore 2\sqrt{3}=3.\underset{–}{\mathrm{4...}}$

Similarly, the approximate value of $\sqrt{10}$ is $3.\underset{–}{\mathrm{1...}}$

Clearly, $3.\underset{–}{\mathrm{4...}}>3.\underset{–}{\mathrm{1...}}$

$\therefore 2\sqrt{3}>\sqrt{10}$

Alternatively, the given surds can be squared and compared.
$(2\sqrt{3}{)}^2=12(\sqrt{10}{)}^2=10$
$\because 12>10\Rightarrow (2\sqrt{3}{)}^2>(\sqrt{10}{)}^2$

If $a\&b$ are positive numbers and $a^2>b^2\Rightarrow a>b$

$\therefore 2\sqrt{3}>\sqrt{10}$