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# Find the value of three places of decimal of the following. It is given that $\sqrt{2}=1.414$, $\sqrt{3}=1.732$, $\sqrt{5}=2.236$, and $\sqrt{10}=3.162$$\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$A) 7.936B) 4.975C) 6.364D) None of the above

Last updated date: 16th Jul 2024
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Hint: In this question, we will first take out the common factor $\sqrt{5}$ from the numerator of the term $\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$. After that, we will do the rationalization. For rationalization, we will multiply the term $\sqrt{2}$ in both denominator and numerator of the term $\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$ to solve this question easily. After that, we will put the values which are given in the question and then solve it, and then we will get the answer.

Complete step by step solution:
Let us solve this question.
In this question, we have given that the value of $\sqrt{2}$ is 1.414, the value of $\sqrt{3}$ is 1.732, the value of $\sqrt{5}$ is 2.236, and the value of $\sqrt{10}$ is 3.162
From these given values, we have to find the value of the term $\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$ up to three decimal places.
As we know that 10 can be written as 5 multiplied by 2 and 15 can be written as 3 multiplied by 5.
So, we can write the term $\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$ as
$\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}=\dfrac{\sqrt{2\times 5}+\sqrt{3\times 5}}{\sqrt{2}}$
The above equation can also be written as
$\Rightarrow \dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}=\dfrac{\sqrt{2}\times \sqrt{5}+\sqrt{3}\times \sqrt{5}}{\sqrt{2}}$
Now, we can see that there is a common factor of $\sqrt{5}$ in the numerator of the right side of the above equation.
So, after taking out that common factor, we can write the above equation as
$\Rightarrow \dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}=\dfrac{\sqrt{5}\left( \sqrt{2}+\sqrt{3} \right)}{\sqrt{2}}$
Now, here we will do the rationalization. For that, we will multiply $\sqrt{2}$ in both the numerator and denominator of the right side of the equation.
$\Rightarrow \dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}=\dfrac{\sqrt{2}\times \sqrt{5}\left( \sqrt{2}+\sqrt{3} \right)}{\sqrt{2}\times \sqrt{2}}$
We can write the above equation as
$\Rightarrow \dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}=\dfrac{\sqrt{10}\left( \sqrt{2}+\sqrt{3} \right)}{2}$
We know that $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$, so $\sqrt{2}+\sqrt{3}=1.414+1.732=3.146$
The above equation can also be written as
$\Rightarrow \dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}=\dfrac{\sqrt{10}\left( 3.146 \right)}{2}=\sqrt{10}\times 1.573=3.162\times 1.573=4.973816$
Hence, the value of $\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$ will be 4.973816
The value of $\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$ up to three places of decimal will be 4.974
This value is approx 4.975

Hence, the option is B.

Note:
We should know how to do the rationalization. We can solve this question by different methods.
We can write the term $\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$ after rationalizing, we get
$\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}=\dfrac{\sqrt{2}\times \left( \sqrt{10}+\sqrt{15} \right)}{\sqrt{2}\times \sqrt{2}}=\dfrac{\sqrt{2}\times \sqrt{10}+\sqrt{2}\times \sqrt{15}}{2}$
We can write the above term as
$\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}=\dfrac{\sqrt{2}\times \sqrt{10}+\sqrt{2}\times \sqrt{3}\times \sqrt{5}}{2}=\dfrac{1.414\times 3.162+1.414\times 1.732\times 2.236}{2}=\dfrac{9.947}{2}=4.9746$
So, the value of the term $\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$ after rounding up to three places of decimal will be 4.975