Squares and Roots III

$\frac { \sqrt { 2 } }{ 2 } \times \frac { \sqrt { 3 } }{ 3 } =\sqrt { \frac { a }{ b } }$

$\frac { \sqrt { 2\times 3 } }{ 2\times 3 } =\sqrt { \frac { a }{ b } }$

$\frac { \sqrt { 6 } }{ 6 } =\sqrt { \frac { a }{ b } }$

$\frac { 6 }{ 36 } =\frac { 1 }{ 6 } =\frac { a }{ b }$

$a=1$

$b=6$

$a+b=7$

$\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{3} = \sqrt{\frac{a}{b}}$

$\frac{\sqrt{2} \times \sqrt{3}}{2 \times 3} = \sqrt{\frac{a}{b}}$

$\frac{\sqrt{{2} \times {3}}}{2 \times 3} = \sqrt{\frac{a}{b}}$

$\frac{\sqrt{6}}{6} = \sqrt{\frac{a}{b}}$

$\frac{\sqrt{6}}{\sqrt{6} \times \sqrt{6}} = \sqrt{\frac{a}{b}}$

$\frac{1}{\sqrt{6}} = \sqrt{\frac{a}{b}}$

$\sqrt{\frac{1}{6}} = \sqrt{\frac{a}{b}}$

Squaring both sides:

$\frac{a}{b} = \frac{1}{6}$

Thus, $a = 1$ and $b = 6$

So, the answer is: $a + b = 1 + 6 = \boxed{7}$

product of dose two values give rise to the value equalent to sqrt(1/6)....whose addition give rise to 7

I mean 6^1/2/6=(a/b)^1/2, sq on both sides we get, 1/6=a/b, =>a=1,b=6 :. a+b=1+6=7.