In a triangle, the sum of lengths of two sides is $\mathrm{x}$ and the product of the lengths of the same…
In a triangle, the sum of lengths of two sides is $\mathrm{x}$ and the product of the lengths of the same two sides is $y$. if $x^{2}-c^{2}=y,$ where $c$ is the length of the third side of the triangle, then the circumradius of the triangle is
$\frac{3}{2} y$
$\frac{c}{\sqrt{3}}$
$\frac{c}{3}$
$\frac{y}{\sqrt{3}}$
Solution
Let two sides of triangle are $a$ and $b$.
$
\begin{aligned}
& a+b=x \\
& a b=y \\
& x^{2}-c^{2}=y \Rightarrow(a+b)^{2}-c^{2}=a b \\
\Rightarrow &(a+b-c)(a+b+c)=a b \\
\Rightarrow & 2(s-c)(2 s)=a b \\
\Rightarrow & 4 s(s-c)=a b \\
\Rightarrow & \frac{s(s-c)}{a b}=\frac{1}{4} \\
\Rightarrow & \cos ^{2} \frac{c}{2}=\frac{1}{4} \\
\Rightarrow & \cos c=-\frac{1}{2} \Rightarrow c=120^{\circ}
\end{aligned}
$
$\therefore$ Area of triangle is,
$\Delta=\frac{1}{2} a b\left(\sin 120^{\circ}\right)=\frac{\sqrt{3}}{4} a b$
$\because \quad R=\frac{a b c}{4 \Delta}$
$\because \quad \mathrm{R}=\frac{a b c}{\sqrt{3} a b}=\frac{c}{\sqrt{3}}$