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

Answer 1

Answer: (B)

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