In a triangle,the sum of lengths of two sides | vedclass

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(B) Let the sides of the triangle be $a, b,$ and $c$. Given $a + b = x$ and $ab = y$.Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$:$(a

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

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(B) Let the sides of the triangle be $a, b,$ and $c$. Given $a + b = x$ and $ab = y$.Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$:$(a + b)^2 - c^2 = ab$$a^2 + b^2 + 2ab - c^2 = ab$$a^2 + b^2 - c^2 = -ab$Dividing by $2ab$,we get:$\frac{a^2 + b^2 - c^2}{2ab} = -\frac{1}{2}$By the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$,so $\cos C = -\frac{1}{2}$.Thus,$C = 120^\circ$ or $\frac{2\pi}{3}$ radians.Then $\sin C = \sin(120^\circ) = \frac{\sqrt{3}}{2}$.Using the Sine Rule,$\frac{c}{\sin C} = 2R$,where $R$ is the circumradius:$R = \frac{c}{2 \sin C} = \frac{c}{2(\frac{\sqrt{3}}{2})} = \frac{c}{\sqrt{3}}$.

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

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