Question

If $\sqrt{3}:(1+\sqrt{2})::\sqrt{6}:x$, then $x$ is equal to

• A. $\sqrt{3}+1$
• B. $\sqrt{2}-1$
• C. $\sqrt{2}-2$
• D. $\sqrt{2}+2$

Answer 1

Answer: (D)

$\sqrt{2}+2$

Given, $\sqrt{3}:(1+\sqrt{2})::\sqrt{6}:x$

$\Rightarrow \frac{\sqrt{3}}{1+\sqrt{2}}=\frac{\sqrt{6}}{x}$

On cross multiplying, we get

$\Rightarrow x=\frac{\sqrt{6}.(1+\sqrt{2})}{\sqrt{3}}=\frac{\sqrt{2}.\sqrt{3}(1+\sqrt{2})}{\sqrt{3}}$

$=\sqrt{2}(1+\sqrt{2})=\sqrt{2}+\sqrt{2}.\sqrt{2}$

$=\sqrt{2}+2$