If $$\frac{\sqrt{121}\times \sqrt{52-14\sqrt{3}}}{\sqrt{28+10\sqrt{3}}}=a+b\sqrt{3},$$​​
then find the value of (a+b).

Correct option is D

Given:

​$$\frac{\sqrt{121}\times \sqrt{52-14\sqrt{3}}}{\sqrt{28+10\sqrt{3}}}=a+b\sqrt{3}$$​​

Solution :

​$$11\times\sqrt{\frac{52-14\sqrt{3}}{28+10\sqrt{3}}}$$​​
​$$=11\times\frac{7-\sqrt{3}}{5+\sqrt{3}}$$​​

Rationalising:
​$$=11\times\frac{(7-\sqrt{3})(5-\sqrt{3})}{25-3}$$​​
​$$=11\times\frac{38-12\sqrt{3}}{22}$$​​

=19$$-6\sqrt{3}$$​​

So,
a = 19,b = -6

a + b = 19 - 6 = 13