Question

Question 13
Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414,\sqrt{3}=1.732$ and $\sqrt{5}=2.236$ upto three places of decimal.
$\left(i\right)\frac{4}{\sqrt{3}}\left(ii\right)\frac{6}{\sqrt{6}}\left(iii\right)\frac{\sqrt{10}-\sqrt{5}}{2}\left(iv\right)\frac{\sqrt{2}}{2+\sqrt{2}}\left(v\right)\frac{1}{\sqrt{3}+\sqrt{2}}$

Answer 1

(i) Let $E=\frac{4}{\sqrt{3}}$
For rationalising the denominator, multiply numerator and denominator by $\sqrt{3}$. We get,
$E=\frac{4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{4\sqrt{3}}{3}=\frac{4}{3}\times 1.732=\frac{6.3928}{3}=2.309[Put\sqrt{3}=1.732]$

(ii) For rationalising denominator, multiply numerator and denominator by $\sqrt{6}$. We get,
$E=\frac{6}{\sqrt{6}}\times \frac{\sqrt{6}}{6}=\frac{6\sqrt{6}}{6}=\sqrt{2}\times \sqrt{3}=1.414\times 1.732=2.449[Put\sqrt{2}=1.414and\sqrt{3}=1.732]$

(iii) Let $E=\frac{\sqrt{10}-\sqrt{5}}{2}=\frac{\sqrt{5}\sqrt{2}-\sqrt{5}}{2}=\frac{\sqrt{5}(\sqrt{2}-1)}{2}[\because \sqrt{10}=\sqrt{2}\sqrt{5}]$
$=\frac{2.236(1.414-1)}{2}=1.118\times 0.414=0.46285\cong 0.463$

(iv) Let $E=\frac{\sqrt{2}}{2+\sqrt{2}}$
For rationalizing the denominator, multiply numerator and denominator by $2-\sqrt{2}$.
We get,
$=\frac{\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{\sqrt{2}(2-\sqrt{2})}{(2{)}^2-(\sqrt{2}{)}^2}$
[Using identity, $(a-b)(a+b)=a^2-b^2]$
$=\frac{\sqrt{2}\times \sqrt{2}(\sqrt{2}-1)}{2}=\frac{2(\sqrt{2}-1)}{2}=\sqrt{2}-1=1.414=0.414[Put\sqrt{2}=1.414]$

(v)
Let $E=\frac{1}{\sqrt{3}+\sqrt{2}}$
For rationalising the denominator multiply numerator and denominator by $\sqrt{3}-\sqrt{2}$.
We get,
$\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}$
[Using identity,$(a-b)(a+b)=a^2-b^2]$
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}$
$=1.732-1.414=0.318$ [Put $\sqrt{3}=1.732$ and $\sqrt{2}=1.414]$