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 :$\frac{1}{\sqrt{3}+\sqrt{2}}$

• A. $0.310$
• B. $0.320$
• C. $0.318$
• D. $0.308$

Answer 1

Answer: (C)

$0.318$

Given, $\frac{1}{\sqrt{3}+\sqrt{2}}$
Rationalizing the denominator, we get

$\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$

$=\frac{1.732-1.414}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}$

$=\frac{0.318}{3-2}$

$=0.318$