Question

In a triangle the sum of two sides is $x$ and the product of the same two 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. $\frac{3 y}{2 x(x+c)}$
• B. $\frac{3 y}{2 c(x+c)}$
• C. $\frac{3 y}{4 x(x+c)}$
• D. $\frac{3 y}{4 c(x+c)}$

Answer 1

Answer: (B)

$x=a+b$
$y=a b$
$x^{2}-c^{2}=y$
$\Rightarrow \frac{a^{2}+b^{2}-c^{2}}{2 a b}=-\frac{1}{2}=\cos \left(120^{\circ}\right)$
$\Rightarrow \angle C=\frac{2 \pi}{3}$
$\Rightarrow R=\frac{a b c}{4 \Delta}, r=\frac{\Delta}{s}$
$\Rightarrow \frac{r}{R}=\frac{4 \Delta^{2}}{s(a b c)}=\frac{4\left[\frac{1}{2} a b \sin \left(\frac{2 \pi}{3}\right)\right]^{2}}{\frac{x+c}{2} \cdot y \cdot c}$
$\Rightarrow \frac{r}{R}=\frac{3 y}{2 c(x+c)}$