Word Problems on Square Roots

Solution:

In the given number, mark off the digits in pairs starting from the unit’s digit.

Each pair and the remaining one digit is called a period.

Now, 1² = 1. On subtracting, we get 0 as reminder.

Now, bring down the next period i.e., 47.

So, we take 22 as divisor is 1 * 2 = 2 and trial dividend is 47.

So, we take 22 as divisor and put 2 as quotient.

The reminder is 72.

Bring down the next period i.e., 69.

Now, the trial divisor is 121 * 2 = 242 and the trial dividend is 7269.

So, we take 3 as quotient and 2423 as divisor.

The reminder is then zero.

Hence, $\sqrt(1471369)$ = 1213.

This is shown by the below Figure.

Solution:

Given expression = $\sqrt{248\sqrt{51\sqrt{169}}}$

= $\sqrt{248\sqrt{51\sqrt{13^2}}}$

= $\sqrt{248\sqrt{51+13}}$

= $\sqrt{248\sqrt{64}}$

= $\sqrt{248\sqrt{8^2}}$

= $\sqrt{248+8}$

= $\sqrt{256}$

= $\sqrt{16^2}$

= 16.

Therefore, the answer for the given equation is 16.

Solution:

Let $\sqrt{86.49}$ +$\sqrt{5+(x)^2 }$ =12.3.

Then, 9.3 + $\sqrt{5+(x)^2 }$ =12.3.

$\sqrt{5+(x)^2 }$ = 12.3 - 9.3.

$\sqrt{5+(x)^2 }$ = 3.

Square the values on both side,

5 + $x^2$ = 9.

$x^2$ = 9 - 5.

$x^2$ = 4.

x = $\sqrt{4}$

x = 2.

Solution:

Express the given number as the product of prime factors.

Now, take the product of these prime factors choosing one out of every pair of same primes.

This product gives the square root of the given number.

Thus, resolving 6084 into prime factors, We get:

6084=2² * 3² * 13²

$\sqrt(6084)$ = ( 2 * 3 * 13 ) = 78.

This is shown by the below Figure.

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