TS EAMCET 2015 Solved Paper

Consider the standard notation followed in triangle geometry.Let $r$ be the radius of the incircle of the triangle.Let $r_1, r_2, r_3$ be the radii of the excircles drawn opposite to the vertices $A, B, C$ respectively.Let $s$ be the semi-perimeter of the triangle where $s=\frac{a+b+c}{2}$Let $a, b, c$ be the lengths of the sides of the triangles opposite to the vertices $A, B, C$ respectively..Let $\Delta$ be the area of the triangle. We know, $\Delta = r \cdot s = r_1(s-a) = r_2(s-b) = r_3(s-c)$$\therefore \frac{1}{r} = \frac{s}{\Delta}$$\frac{1}{r_1} = \frac{s-a}{\Delta}$$\frac{1}{r_2} = \frac{s-b}{\Delta}$$\frac{1}{r_3} = \frac{s-c}{\Delta}$ $\Rightarrow \frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2} = \frac{1}{\Delta^2} ((s-a)^2+(s-b)^2+(s-c)^2+s^2)$$= \frac{1}{\Delta^2} (s^2-2as+a^2+s^2-2bs+b^2+s^2-2cs+c^2+s^2)$ $= \frac{1}{\Delta^2} (4s^2-2s(a+b+c)+ (a^2+b^2+c^2))$$= \frac{1}{\Delta^2} (4s^2-4s^2+ (a^2+b^2+c^2)) ( \because \overline{a}+\overline{b}+\overline{c}=\overline{2}\overline{s})$$\therefore \frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_{3^2}}+\frac{1}{r^2} = \frac{ (a^2+b^2+c^2) }{\Delta^2}$