Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • (i)Rationalize the denominator in each of the following and hence evaluate by taking sqroot =1.414 sqroot 3 =1.732 and sqroot 5 =2.236 upto three places of decimal : 4 by sqroot 3

(i) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{4}{\sqrt{3}}
(ii) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{6}{\sqrt{6}}
(iii) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}
(iv) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{2}{2+\sqrt{2}}
(v) Rationalize the denominator in each of the following and hence evaluate by taking \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732 and \sqrt{5}= 2\cdot 236, upto three places of decimal : \frac{1}{\sqrt{3}+\sqrt{2}}

Answers (1)

(i) Answer.   2.3093
Solution. Given: \frac{4}{\sqrt{3}}
Rationalising,
\frac{4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}= \frac{4\sqrt{3}}{3}
(Given that \sqrt{3}= 1.732)
= \frac{4\times 1\cdot 732}{3}
= 2.3093
Hence the answer is 2.3093

(ii) Answer.  2.449
Solution.  Given: \frac{6}{\sqrt{6}}
Rationalising,
\frac{6}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}= \frac{6\sqrt{6}}{6}
= \frac{6\times \sqrt{2}\sqrt{3}}{6}
Putting the given values,
\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732
We get :
= \sqrt{2}\cdot \sqrt{3}
= 1\cdot 414\times 1\cdot 732= 2\cdot 449
Hence the answer is 2.449

(iii) Answer.  0.462852
Solution.   Given that \frac{\sqrt{10}-\sqrt{5}}{2}
This can be written as

\frac{\sqrt{2}\times \sqrt{5}-\sqrt{5}}{2}
Now putting the given values,

\sqrt{2}= 1\cdot 414,\sqrt{5}= 2\cdot 236
We get :
\Rightarrow \frac{1\cdot 414\times 2\cdot 236-2\cdot 236}{2}
= 0.462852
  Hence the answer is 0.462852

(iv) Answer.  0.414
Solution.  Given: \frac{\sqrt{2}}{2+\sqrt{2}}
Rationalising,
\frac{\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}
Using   (a – b) (a + b) = a2 – b2
= \frac{\sqrt{2}\left ( 2-\sqrt{2} \right )}{2^{2}-\sqrt{2}^{2}}
= \frac{2\sqrt{2}-2}{4-2}
= \frac{2\left ( \sqrt{2}-1 \right )}{2}
= \sqrt{2}-1
Putting the given value of \sqrt{2}= 1\cdot 414
We get
= 1.414 – 1
= 0.414
 Hence the answer is 0.414

(v)  Answer.  0.318
Solution. Given that \frac{1}{\sqrt{3}+\sqrt{2}}
Rationalising,
\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}
Using   (a – b) (a + b) = a2 – b2
= \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}^{2}-\sqrt{2}^{2}}
= \frac{\sqrt{3}-\sqrt{2}}{3-2}
= \sqrt{3}-\sqrt{2}
Putting the given values, \sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732
We get,
= 1.732 – 1.414
= 0.318
Hence the answer is 0.318

Posted by

infoexpert27

View full answer

Similar Questions

Latest Question