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. $\frac{3y}{2x(x+c)}$
• B. $\frac{3y}{2c(x+c)}$
• C. $\frac{3y}{4x(x+c)}$
• D. $\frac{3y}{4c(x+c)}$

Answer 1

The correct option is B $\frac{3y}{2c(x+c)}$

Let two sides of $\Delta$ be a and b.
Then a+b=x and ab=y
Also give $x^2-c^2=y$, where c is the third side of $\Delta$
$\Rightarrow (a+b^2)-c^2=ab\Rightarrow a^2+b^2-c^2=-ab$
$\Rightarrow \frac{a^2+b^2-c^2}{2ab}=-\frac{1}{2}\Rightarrow cosc=-\frac{1}{2}\Rightarrow c={120}^{\circ }$
$\therefore \frac{r}{R}=\frac{\Delta }{s}\times \frac{4\Delta }{abc}$ where $\Delta$ = area of triangle
$\Rightarrow \frac{r}{R}=\frac{4{\Delta }^2}{\frac{(a+b+c)}{2}abc}=\frac{8\times {\left(\frac{1}{2}absinc\right)}^2}{(a+b+c)abc}$
$\Rightarrow \frac{2a^2{b}^{2}si{n}^2{120}^{\circ }}{(a+b+c)abc}=\frac{2ab\times \frac{3}{4}}{(x+c)}=\frac{3y}{2c(x+c)}$