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  • Rationalise the denominator of the following : 2 by 3 sqroot 3

(i) Rationalise the denominator of the following : \frac{2}{3\sqrt{3}}
(ii)Rationalise the denominator of the following : \frac{\sqrt{40}}{\sqrt{3}}
(iii) Rationalise the denominator of the following : \frac{3+\sqrt{2}}{4\sqrt{2}}
(iv)Rationalise the denominator of the following :\frac{16}{\sqrt{41}-5}
(v) Rationalise the denominator of the following : \frac{2+\sqrt{3}}{2-\sqrt{3}}
(vi) Rationalise the denominator of the following  : \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}
(vii) Rationalise the denominator of the following : \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}

(viii) Rationalise the denominator of the following : \frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}

(ix) Rationalise the denominator of the following : \frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}


 

Answers (1)

(i) Answer.  \frac{2\sqrt{3}}{9}  
Solution.         We have, \frac{2}{3\sqrt{3}}
Rationalising the denominator, we get:
\frac{2}{3\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}
= \frac{2\sqrt{3}}{3\sqrt{3}\sqrt{3}}
= \frac{2\sqrt{3}}{9}
Hence the answer is \frac{2\sqrt{3}}{9}
 

(ii) Answer.  \frac{2\sqrt{30}}{3}  
Solution. We have ,\frac{\sqrt{40}}{\sqrt{3}}
We know that, 40 = (2) (2) (10)
\frac{\sqrt{40}}{\sqrt{3}}= \frac{\sqrt{2\cdot 2\cdot 10}}{\sqrt{3}}= \frac{2\sqrt{10}}{\sqrt{3}}
Rationalising the denominator, we get:
= \frac{2\sqrt{10}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}
= \frac{2\sqrt{10}\sqrt{3}}{\sqrt{3}\sqrt{3}}
= \frac{2\sqrt{30}}{3}
Hence the answer is: \frac{2\sqrt{30}}{3}

(iii) Answer.  \frac{3\sqrt{2}+2}{8}
Solution.  We have \frac{3+\sqrt{2}}{4\sqrt{2}}
Rationalising the denominator, we get:
\frac{3+\sqrt{2}}{4\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}
= \frac{\left ( 3+\sqrt{2} \right )\sqrt{2}}{4\sqrt{2}\sqrt{2}}
= \frac{3\sqrt{2}+2}{8}
Hence the answer is \frac{3\sqrt{2}+2}{8}

(iv) Answer. \sqrt{41}+5
Solution. We have \frac{16}{\sqrt{41}-5}
Rationalising the denominator, we get:
\frac{16}{\sqrt{41}-5}\times \frac{\sqrt{41}+5}{\sqrt{41}+5}
= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41}-5 \right )\sqrt{41}+5}
Using  the identity (a – b) (a + b) = a2 – b2
We get:
= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41} \right )^{2}-\left ( 5 \right )^{2}}
= \frac{16\left ( \sqrt{41}+5 \right )}{41-25}
= \frac{16\left ( \sqrt{41}+5 \right )}{16}
= \sqrt{41}+5
Hence the answer is \sqrt{41}+5

(v) Answer.    7+4\sqrt{3}
Solution. We have, \frac{2+\sqrt{3}}{2-\sqrt{3}}
Rationalising the denominator, we get:
\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2-\sqrt{3}}
= \frac{\left ( 2+\sqrt{3} \right )^{2}}{\left ( 2-\sqrt{3} \right )\left ( 2+\sqrt{3} \right )}

Using   (a – b) (a + b) = a2 – b2
and      (a + b)2 = a2 + b2 + 2ab
= \frac{2^{2}+\left ( \sqrt{3} \right )^{2}+2\cdot 2\cdot \sqrt{3}}{2^{2}-\left ( \sqrt{3} \right )^{2}}
= \frac{4+3+4\sqrt{3}}{4-3}
= 7+4\sqrt{3}
Hence the answer is 7+4\sqrt{3}

(vi)Answer.  3\sqrt{2}-2\sqrt{3}
Solution.   We have, \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}
Rationalising the denominator, we get:
= \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}
= \frac{\sqrt{6}\left ( \sqrt{3}-\sqrt{2} \right )}{\left ( \sqrt{2}+\sqrt{3} \right )\left ( \sqrt{3}-\sqrt{2} \right )}
Using  the identity (a – b) (a + b) = a2 – b2
We get:
= \frac{\sqrt{18}-\sqrt{12}}{\left ( \sqrt{3} \right )^{2}\left ( \sqrt{2} \right )^{2}}
= \frac{\sqrt{3\cdot 3\cdot 2}-\sqrt{2\cdot 2\cdot 3}}{3-2}
= \frac{3\sqrt{2}-2\sqrt{3}}{1}
= 3\sqrt{2}-2\sqrt{3}
Hence the answer 3\sqrt{2}-2\sqrt{3}

(vii) Answer. 5+2\sqrt{6}
Solution. We have, \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}

Rationalising the denominator, we get:
\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}
= \frac{\left ( \sqrt{3} +\sqrt{2}\right )^{2}}{\left ( \sqrt{3} -\sqrt{2} \right )\left ( \sqrt{3} +\sqrt{2} \right )}

Using   (a – b) (a + b) = a2 – b2
and      (a + b)2 = a2 + b2 + 2ab
= \frac{\left ( \sqrt{3} \right )^{2}+\left ( \sqrt{2} \right )^{2}+2\sqrt{3}\sqrt{2}}{\left ( \sqrt{3} \right )^{2}-\left ( \sqrt{2} \right )^{2}}
= \frac{3+2+2\sqrt{6}}{3-2}
= 5+2\sqrt{6}
Hence the answer is 5+2\sqrt{6}

(viii)

Answer: 9+2 \sqrt{15}

Solution:

We have \frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}

Rationalize

=\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{3 \sqrt{5}(\sqrt{5}+\sqrt{3})+\sqrt{3}(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}

=\frac{15+3 \sqrt{15}+\sqrt{15}+3}{5-3}=\frac{18+4 \sqrt{15}}{2}

=9+2 \sqrt{15}

(ix) Answer: \frac{9+4 \sqrt{6}}{15}

Solution:

We have \frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}

=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{16 \times 3}+\sqrt{9 \times 2}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}}

Rationalize

=\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}} \times \frac{(4 \sqrt{3}-3 \sqrt{2})}{(4 \sqrt{3}-3 \sqrt{2})}

=\frac{4 \sqrt{3}(4 \sqrt{3}-3 \sqrt{2})+5 \sqrt{2}(4 \sqrt{3}-3 \sqrt{2})}{(4 \sqrt{3})^{2}-(3 \sqrt{2})^{2}}

=\frac{48-12 \sqrt{6}+20 \sqrt{6}-30}{30}

=\frac{18+8 \sqrt{6}}{30}

=\frac{9+4 \sqrt{6}}{15}

 

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