In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y.

Question:

In a triangle, the sum of lengths of two sides is $x$ and the product of the lengths of the same two sides is $\mathrm{y}$. If $\mathrm{x}^{2}-\mathrm{c}^{2}=\mathrm{y}$, where $\mathrm{c}$ is the length of the third side of the triangle, then the circumradius of the triangle is :

  1. $\frac{y}{\sqrt{3}}$

  2. $\frac{c}{\sqrt{3}}$

  3. $\frac{c}{3}$

  4. $\frac{3}{2} y$


Correct Option: , 2

Solution:

Given $a+b=x$ and $a b=y$

If $x^{2}-c^{2}=y \Rightarrow(a+b)^{2}-c^{2}=a b$

$\Rightarrow a^{2}+b^{2}-c^{2}=-a b$

$\Rightarrow \frac{\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}}{2 \mathrm{ab}}=-\frac{1}{2}$

$\Rightarrow \cos \mathrm{C}=-\frac{1}{2}$

$\Rightarrow \angle \mathrm{C}=\frac{2 \pi}{3}$'

$R=\frac{c}{2 \sin C}=\frac{c}{\sqrt{3}}$

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