Q22 / 60 · JEE Advanced 2014 (Paper 2) · Mathematics

In a triangle the sum of two sides is x and the product of the same two sides is y. If $x^{2} - c^{2} = y,$ (where c is the third side of the triangle) then the ratio of the inradius to the circumradius of the triangle is

A
$\frac{3 y}{2 x (x + c)}$
Correct answer
B
$\frac{3 y}{2 c (x + c)}$
C
$\frac{3 y}{4 x (x + c)}$
D
$\frac{3 y}{4 c (x + c)}$

Solution

Let

x = a + b

y = a b

x^{2} - c^{2} = y

\Rightarrow \frac{a^{2} + b^{2} - c^{2}}{2 a b} = - \frac{1}{2} = \cos (120^{o})

\Rightarrow ∠ C = \frac{2 π}{3}

\Rightarrow R = \frac{a b c}{4 ∆}, r = \frac{∆}{s}

\Rightarrow \frac{r}{R} = \frac{4 ∆^{2}}{s (a b c)} = \frac{4 {[\frac{1}{2} a b \sin (\frac{2 π}{3})]}^{2}}{\frac{x + c}{2} y . c}

\frac{r}{R} = \frac{3 y}{2 c (x + c)}

Hence, the correct answer is option: 'B'.

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