Question
In a triangle the sum of two sides is $$x$$ and the product of the same sides is $$y.$$ If $${x^2} - {c^2} = y,$$ where $$c$$ is the third side of the triangle, then the ratio of the in radius to the circum-radius of the triangle is
A.
$$\frac{{3y}}{{2x\left( {x + c} \right)}}$$
B.
$$\frac{{3y}}{{2c\left( {x + c} \right)}}$$
C.
$$\frac{{3y}}{{4x\left( {x + c} \right)}}$$
D.
$$\frac{{3y}}{{4c\left( {x + c} \right)}}$$
Answer :
$$\frac{{3y}}{{2c\left( {x + c} \right)}}$$
Solution :
Let two sides of $$\Delta $$ be $$a$$ and $$b.$$
Then $$a + b = x$$ and $$ab = y$$
Also given $${x^2} - {c^2} = y,$$ where $$c$$ is the third side of $$\Delta .$$
$$\eqalign{
& \Rightarrow \,\,{\left( {a + b} \right)^2} - {c^2} = ab \cr
& \Rightarrow \,\,{a^2} + {b^2} - {c^2} = - ab \cr
& \Rightarrow \,\,\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} = - \frac{1}{2} \cr
& \Rightarrow \,\,\cos c = - \frac{1}{2} \cr
& \Rightarrow \,\,c = {120^ \circ } \cr} $$
$$\therefore \,\,\frac{r}{R} = \frac{\Delta }{s} \times \frac{{4\Delta }}{{abc}}$$ where $$\Delta =\,\,$$ area of triangle
$$\eqalign{
& \Rightarrow \,\,\frac{r}{R} = \frac{{4{\Delta ^2}}}{{\frac{{\left( {a + b + c} \right)}}{2}abc}} \cr
& = \frac{{8 \times {{\left( {\frac{1}{2}ab\sin c} \right)}^2}}}{{\left( {a + b + c} \right)abc}} \cr
& \Rightarrow \,\,\frac{{2{a^2}{b^2}{{\sin }^2}{{120}^ \circ }}}{{\left( {a + b + c} \right)abc}} \cr
& = \frac{{2ab \times \frac{3}{4}}}{{\left( {x + c} \right)c}} \cr
& = \frac{{3y}}{{2c\left( {x + c} \right)}} \cr} $$