​$$\text{If } a^2 + b^2 = 111, \ a \times b = 27, \ \text{and} \ a > b, \ \text{find the value of} \ \frac{a - b}{a + b}.$$​

Correct option is B

Given:
​$$a^2 + b^2 = 111$$​​
a $$\times$$​ b = 27
a > b
Formula Used:
​$$(a + b)^2 = a^2 + b^2 + 2ab \\ (a - b)^2 =a^2 + b^2 - 2ab$$​​
Solution:
$$(a + b)^2 = a^2 + b^2 + 2ab\\(a + b)^2 = 111 + 54 = 165\\a + b = \sqrt{165}\\(a - b)^2 = a^2 + b^2 - 2ab\\(a - b)^2 = 111 - 54 = 57\\a - b = \sqrt{57}\\\frac{a - b}{a + b} = \frac{\sqrt{57}}{\sqrt{165}}\\\frac{a - b}{a + b} = \sqrt{\frac{57}{165}} \\$$​​