In a triangle,the sum of two sides is x and t | vedclass

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(B) Let the two sides be $a$ and $b$. Given $a+b=x$ and $ab=y$.Given $x^2-c^2=y$,substituting $x=a+b$ and $y=ab$,we get $(a+b)^2-c^2=ab$.$a^2+b^2+2ab-

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$\frac{3 y}{2 c(x+c)}$

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(B) Let the two sides be $a$ and $b$. Given $a+b=x$ and $ab=y$.Given $x^2-c^2=y$,substituting $x=a+b$ and $y=ab$,we get $(a+b)^2-c^2=ab$.$a^2+b^2+2ab-c^2=ab \implies a^2+b^2-c^2=-ab$.Using the cosine rule,$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{-ab}{2ab} = -\frac{1}{2}$.Thus,$C = 120^\circ$ or $\frac{2\pi}{3}$.The ratio of in-radius $r$ to circum-radius $R$ is given by $\frac{r}{R} = 4 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2}$.Alternatively,$\frac{r}{R} = \frac{\Delta/s}{abc/4\Delta} = \frac{4\Delta^2}{sabc} = \frac{4 \cdot (\frac{1}{2}ab \sin C)^2}{(\frac{a+b+c}{2})abc} = \frac{a^2b^2 \sin^2(120^\circ)}{(\frac{x+c}{2})abc} = \frac{y^2 \cdot (\frac{\sqrt{3}}{2})^2}{(\frac{x+c}{2})abc} = \frac{3y^2}{2(x+c)abc}$.Since $ab=y$,this simplifies to $\frac{3y}{2c(x+c)}$.Wait,re-evaluating: $\frac{r}{R} = \frac{4\Delta^2}{sabc} = \frac{a^2b^2 \sin^2 C}{sabc} = \frac{y^2 \cdot (3/4)}{(\frac{x+c}{2})abc} = \frac{3y^2}{2(x+c)abc} = \frac{3y^2}{2(x+c)yc} = \frac{3y}{2c(x+c)}$.

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

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