Question

Find the approximate value of the following numbers to nearest tenths.

• A. 7.1
• B. 5.6
• C. 13.4

Answer 1

$\sqrt{31}$
$5^2=25and{6}^2=36$

Now, 25 < 31 < 36.

So, √31 has to lie between 5 and 6.
As 31 is closer to 36, multiply the numbers close to the square root of 36 i.e., 6.

∴ By trial-and-error method you get,

$5.5\times 5.5=30.255.6\times 5.6=31.25$
$\text{So,}\sqrt{31}=5.6$

$\sqrt{180}$
${13}^2=169and{14}^2=196$

Now, 169 < 180 < 196.

So, $\sqrt{180}$ has to lie between 13 and 14.
180 is closer to 169.

∴ By trial-and-error method you get,

$13.1\times 13.1=171.6113.2\times 13.2=174.2413.3\times 13.3=176.8913.4\times 13.4=179.5613.5\times 13.5=182.25$
So, $\sqrt{180}=13.4$

$\sqrt{51}$
$7^2=49and{8}^2=64$

Now, 49 < 51 < 64.

So, $\sqrt{49}<\sqrt{51}<\sqrt{64}\Rightarrow 7<\sqrt{51}<8$
51 is closer to 49.

∴ By trial-and-error method you get,

$7.1\times 7.1=50.41617.2\times 7.2=51.84$
So, $\sqrt{51}=7.1$