Question

Estimate the value of square root of 110

Answer 1

Hint: We solve this problem by using the long division method of the square root.
The steps involved in finding the square root using the long division method are shown below:
(1) We divide the number into pairs of 2 digits starting from right side and we take four decimal numbers after the decimal point to get the two pairs.
(2) We take the square less than and near to first pair on the left and take the square root of that near number as the first digit of the required value. Let us assume it as \['x'\]
(3) We subtract the first pair and the near square to that pair and take the second pair as a set of number.
(4) If the new set of number is \['n'\] then we double the first digit we got in the step (2) to make the problem as
\[2x\underset{\scriptscriptstyle-}{k}\times \underset{\scriptscriptstyle-}{k}\le n\]
Here \[2x\underset{\scriptscriptstyle-}{k}\] is a number but not \[2x\times k\]
Here, the value of \['k'\] found by train and error method will be the second digit of the required square root.
(5) We repeat the step (4) until all the pairs are completed so that we use the decimal point in the answer when we cross the decimal point in the given number.

Complete step by step answer:
We are asked to find the square root of 110
Let us use the long division method for solving this problem
We know that the steps involved in finding the square root using the long division method are shown below:
(1) We divide the number into pairs of 2 digits starting from right side and we take four decimal numbers after the decimal point to get the two pairs.
(2) We take the square less than and near to first pair on the left and take the square root of that near number as the first digit of the required value. Let us assume it as \['x'\]
(3) We subtract the first pair and the near square to that pair and take the second pair as a set of numbers.
(4) If the new set of number is \['n'\] then we double the first digit we got in the step (2) to make the problem as
\[2x\underset{\scriptscriptstyle-}{k}\times \underset{\scriptscriptstyle-}{k}\le n\]
Here \[2x\underset{\scriptscriptstyle-}{k}\] is a number but not \[2x\times k\]
Here, the value of \['k'\] found by train and error method will be the second digit of the required square root.
(5) We repeat step (4) until all the pairs are completed so that we use the decimal point in the answer when we cross the decimal point in the given number.
Now, let us take the number 110 up to four decimal points so that we get the pairs as
\[\Rightarrow \left| \!{\overline {\,
 \overline{01}\overline{10}.\overline{00}\overline{00} \,}} \right. \]
Here, we can see that 1 is the number near to first pair which is a perfect square.
By taking the square root of 1 then we get
\[\Rightarrow \begin{matrix}
   1 \\
   1\left| \!{\overline {\,
 \begin{align}
  & \overline{01}\overline{10}.\overline{00}\overline{00} \\
 & -\left( 1 \right) \\
 & =0010.0000 \\
\end{align} \,}} \right. \\
\end{matrix}\]
Now, by subtracting the first pair 01 with 1 and taking the second pair then we get
\[\Rightarrow \begin{matrix}
   1 \\
   \left| \!{\overline {\,
 \overline{10}.\overline{00}\overline{00} \,}} \right. \\
\end{matrix}\]
Now, let us double the number 1 to form the equation as
\[2\underset{\scriptscriptstyle-}{k}\times \underset{\scriptscriptstyle-}{k}\le 10\]
Here, we know that \['k'\] is a digit.
By using the trial and error method we get the digit \[k\] as
\[20\times 0=0\]

By using the condition we get
\[\Rightarrow \begin{matrix}
   10 \\
   20\left| \!{\overline {\,
 \begin{align}
  & 10.0000 \\
 & -0 \\
 & =10.0000 \\
\end{align} \,}} \right. \\
\end{matrix}\]
Now, let us double the number 10 to form and we can see that we are crossing the decimal point then the equation as
\[20.\underset{\scriptscriptstyle-}{k}\times 0.\underset{\scriptscriptstyle-}{k}\le 10.00\]
Here, we know that \['k'\] is a digit.
By using the trial and error method we get the digit \[k\] as
\[20.4\times 0.4=8.16\]
By using the condition we get
\[\Rightarrow \begin{matrix}
   10.4 \\
   20.4\left| \!{\overline {\,
 \begin{align}
  & 10.0000 \\
 & -\left( 8.16 \right) \\
 & =1.8400 \\
\end{align} \,}} \right. \\
\end{matrix}\]
Now, let us double the number 10.4 to form and we can see that we are crossing the decimal point then the equation as
\[20.8\underset{\scriptscriptstyle-}{k}\times 0.0\underset{\scriptscriptstyle-}{k}\le 1.8400\]
Here, we know that \['k'\] is a digit.
By using the trial and error method we get the digit \[k\] as
\[20.88\times 0.08=1.6064\]
By using the condition we get
\[\Rightarrow \begin{matrix}
   10.48 \\
   20.88\left| \!{\overline {\,
 \begin{align}
  & 1.8400 \\
 & -\left( 1.6064 \right) \\
 & =0.2336 \\
\end{align} \,}} \right. \\
\end{matrix}\]
Here, we can see that the pairs are completed but we didn’t get the remainder as 0
We know that the number 110 is not a perfect square so that this division will continue forever.

But we can stop after two decimal points and give the answer.
Therefore, we can conclude that the square root of 110 is 10.48 that is
\[\therefore \sqrt{110}=10.48\]


Note:
 Students may do mistakes in taking the decimal point at the quotient after crossing the decimal point in the number.
Here we have the equation after crossing the decimal point as
\[20.\underset{\scriptscriptstyle-}{k}\times 0.\underset{\scriptscriptstyle-}{k}\le 10.00\]
Here we can see that the second term is \[0.\underset{\scriptscriptstyle-}{k}\]
But students may take the equation as
\[20.\underset{\scriptscriptstyle-}{k}\times \underset{\scriptscriptstyle-}{k}\le 10.00\]
This gives the wrong answer.
If you have any difficulty then find the square root of 1100000 and we can divide the number with 100 because
\[\begin{align}
  & \Rightarrow \sqrt{110}=\sqrt{\dfrac{1100000}{10000}} \\
 & \Rightarrow \sqrt{110}=\dfrac{\sqrt{1100000}}{100} \\
\end{align}\]
In this method we may not get any confusion.