Answer:
Option B
Explanation:
Plan
(i) cosC=a2+b2−c22ab
(ii) R= abc4△,r=△s
where, R,r,△ denote the circumradius , inradius and area of traingle , respectively
Let the sides of triangle be a,b, and c
Given x=a+b+c
y=ab
x2-c2=y
⇒ (a+b)2-c2=y
⇒ a2+b2+2ab-c2=ab
⇒ a2+b2-c2=-ab
⇒ a2+b2−c22ab=−12
= cos1200
⇒ ∠C=2π3
∴ R= \frac{abc}{4\triangle},r=\frac{\triangle}{s}
\Rightarrow \frac{r}{R}=\frac{4\triangle^{2}}{s(abc)}
=\frac{4\left[\frac{1}{2}ab\sin\left(\frac{2\pi}{3}\right)\right]^{2}}{\frac{x+y}{2}.y.c}
\therefore \frac{r}{R}=\frac{3y}{2c(x+c)}