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

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

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

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(B) Let $a$ and $b$ be the lengths of two sides of a triangle.According to the given condition,$a+b=x$ and $ab=y$.Given $x^2-c^2=y$,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,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.Therefore,$\cos C = -\frac{1}{2}$.Since $C$ is an angle in a triangle,$C = 120^\circ$ or $\frac{2\pi}{3}$ radians.The circumradius $R$ is given by $R = \frac{c}{2\sin C}$.$R = \frac{c}{2\sin(120^\circ)} = \frac{c}{2(\frac{\sqrt{3}}{2})} = \frac{c}{\sqrt{3}}$.

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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