Question

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 :$\frac{\sqrt{2}}{2+\sqrt{2}}$

• A. $0.441$
• B. $0.444$
• C. $0.414$
• D. $0.404$

Answer 1

Answer: (C)

$0.414$

Given, $\frac{\sqrt{2}}{2+\sqrt{2}}$

Rationalizing the denominator, we get

$\frac{\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}$

$=\frac{\sqrt{2}(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}$

$=\frac{2\sqrt{2}-2}{2^2-{\left(\sqrt{2}\right)}^2}$

$=\frac{2\times 1.414-2}{4-2}$

$=\frac{2.828-2}{2}$

$=0.414$