Answer
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Hint: In order to find the square root of 248, we must have prior knowledge of some perfect squares and their square roots. Thus, we shall look for the nearest perfect square less than and greater than the given number, 248. After that we shall use inequalities and find the square root of the two nearest perfect squares. This would help us find the exact location of the square root of 248.
Complete step by step solution:
The perfect square number just less than 248 is 225 and the perfect square number just greater than 248 is 256.
$\Rightarrow 225<248<256$
Now, we shall find the square root of this entire inequality as a result of which we will calculate the square root of each of the three numbers.
$\Rightarrow \sqrt{225}<\sqrt{248}<\sqrt{256}$
We know that $15\times 15=225$.
$\Rightarrow 225={{\left( 15 \right)}^{2}}$
Also, $16\times 16=256$
$\Rightarrow 256={{\left( 16 \right)}^{2}}$
Substituting these values, we get
$\Rightarrow \sqrt{{{\left( 15 \right)}^{2}}}<\sqrt{248}<\sqrt{{{\left( 16 \right)}^{2}}}$
$\Rightarrow 15<\sqrt{248}<16$
Hence, $\sqrt{248}$ is greater than 15 but less than 16.
So, the correct answer is “Option C”.
Note: Another method of solving this problem is by factoring the number. We would first factorize 248 and enlist all those factors. Then, we would start grouping the factors into pairs. For every pair, only one factor would have been able to come out of the square root sign. However, since 248 is not a perfect square, there would have been some prime number under the square root. Further, we would find the square root of the prime number upto two decimal places and multiply it with terms outside the square root and mark it on the number line to get its exact location.
Complete step by step solution:
The perfect square number just less than 248 is 225 and the perfect square number just greater than 248 is 256.
$\Rightarrow 225<248<256$
Now, we shall find the square root of this entire inequality as a result of which we will calculate the square root of each of the three numbers.
$\Rightarrow \sqrt{225}<\sqrt{248}<\sqrt{256}$
We know that $15\times 15=225$.
$\Rightarrow 225={{\left( 15 \right)}^{2}}$
Also, $16\times 16=256$
$\Rightarrow 256={{\left( 16 \right)}^{2}}$
Substituting these values, we get
$\Rightarrow \sqrt{{{\left( 15 \right)}^{2}}}<\sqrt{248}<\sqrt{{{\left( 16 \right)}^{2}}}$
$\Rightarrow 15<\sqrt{248}<16$
Hence, $\sqrt{248}$ is greater than 15 but less than 16.
So, the correct answer is “Option C”.
Note: Another method of solving this problem is by factoring the number. We would first factorize 248 and enlist all those factors. Then, we would start grouping the factors into pairs. For every pair, only one factor would have been able to come out of the square root sign. However, since 248 is not a perfect square, there would have been some prime number under the square root. Further, we would find the square root of the prime number upto two decimal places and multiply it with terms outside the square root and mark it on the number line to get its exact location.
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