
Estimate the square root : \[\dfrac{7}{\sqrt{11}}\]
Answer
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Hint: When estimating the value of any square root first find its nearest complete square, by finding the nearest complete square we will have an estimate where the square root will be. In the above problem we have 9 and 16 as the nearest complete square that’s why we can say the square root will lie between 3 and 4.
Complete step-by-step answer:
Rationalize the denominator and then estimate the root.
To rationalize the term let's multiply \[\sqrt{11}\]in the numerator as well as the denominator.
\[\dfrac{7}{\sqrt{11}}\times \dfrac{\sqrt{11}}{\sqrt{11}}=\dfrac{7\sqrt{11}}{11}\]
Now let’s estimate the root \[\sqrt{11}\], As 11 is greater then the perfect square 9 but less the perfect square 16 it follows, \[9~<\text{ }11\text{ }<\text{ }16\]
Now taking the square root, \[3<\sqrt{11}<4\]
Now multiplying \[\dfrac{7}{11}\] both side , \[3\times \dfrac{7}{11}<\dfrac{7}{11}\times \sqrt{11}<\dfrac{7}{11}\times 4\]
\[\Rightarrow \dfrac{21}{11}<\dfrac{7}{\sqrt{11}}<\dfrac{28}{11}\]
Now let’s estimate \[\sqrt{11}\] to a decimal place as 11 ia close to perfect square 9 so it will be closure to 3 lets assume it to be \[3.2=\dfrac{32}{10}\] hence the value of \[\dfrac{7}{\sqrt{11}}\] will be \[\dfrac{7}{11}\times \dfrac{32}{10}=\dfrac{112}{55}\]
Which is the required result.
Note: After the first digit we can approximate the decimal place. approximating the first decimal place is simple but approximating the second decimal place became a little bit complicated. to estimate the first decimal place first check if the number is halfway past between the two nearest complete squares. in the above example 11 is nearer to 9 and very far from 16. if the number was somewhere between 12 and 13 then the approximation would be near 3.5, but as the number is closer to 9 and we can approximate it to 3.2 .
Complete step-by-step answer:
Rationalize the denominator and then estimate the root.
To rationalize the term let's multiply \[\sqrt{11}\]in the numerator as well as the denominator.
\[\dfrac{7}{\sqrt{11}}\times \dfrac{\sqrt{11}}{\sqrt{11}}=\dfrac{7\sqrt{11}}{11}\]
Now let’s estimate the root \[\sqrt{11}\], As 11 is greater then the perfect square 9 but less the perfect square 16 it follows, \[9~<\text{ }11\text{ }<\text{ }16\]
Now taking the square root, \[3<\sqrt{11}<4\]
Now multiplying \[\dfrac{7}{11}\] both side , \[3\times \dfrac{7}{11}<\dfrac{7}{11}\times \sqrt{11}<\dfrac{7}{11}\times 4\]
\[\Rightarrow \dfrac{21}{11}<\dfrac{7}{\sqrt{11}}<\dfrac{28}{11}\]
Now let’s estimate \[\sqrt{11}\] to a decimal place as 11 ia close to perfect square 9 so it will be closure to 3 lets assume it to be \[3.2=\dfrac{32}{10}\] hence the value of \[\dfrac{7}{\sqrt{11}}\] will be \[\dfrac{7}{11}\times \dfrac{32}{10}=\dfrac{112}{55}\]
Which is the required result.
Note: After the first digit we can approximate the decimal place. approximating the first decimal place is simple but approximating the second decimal place became a little bit complicated. to estimate the first decimal place first check if the number is halfway past between the two nearest complete squares. in the above example 11 is nearer to 9 and very far from 16. if the number was somewhere between 12 and 13 then the approximation would be near 3.5, but as the number is closer to 9 and we can approximate it to 3.2 .
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