Discussion Forum

 In a triangle, the sum of two sides is x  and the product of the same two sides is y. If x²-  c²  = y,  where c is the third side of the triangle, then the ratio of the inradius to the circumradius  of the triangele is 

Answer:

Explanation:

  Plan

    (i)  $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$

(ii)    R= $\frac{abc}{4\triangle},r=\frac{\triangle}{s}$

  where, R,r,$\triangle$  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

     $\Rightarrow$         (a+b)²-c²=y

$\Rightarrow$      a²+b²+2ab-c²=ab

$\Rightarrow$     a²+b²-c²=-ab

$\Rightarrow   $     $\frac{a^{2}+b^{2}-c^{2}}{2ab}=-\frac{1}{2}$

                                  = $\cos 120^{0}$

$\Rightarrow$                   $\angle  C= \frac{2\pi}{3}$

$\therefore$                        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)}$