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 $y$. 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:

• A. $\frac{c}{3}$
• B. $\frac{c}{\sqrt{3}}$
• C. $\frac{y}{\sqrt{3}}$
• D. $\frac{3}{2}y$

Answer 1

The correct option is B $\frac{c}{\sqrt{3}}$

Let the sides of the $△$ be $a,b,c$
$\Rightarrow a+b=x$ and $ab=y$
$\because x^2-c^2=y$
$\Rightarrow (a+b{)}^2-c^2=ab$
$\Rightarrow a^2+b^2+ab-c^2=0$
$\Rightarrow a^2+b^2-c^2=-ab$
$\Rightarrow \frac{a^2+b^2-c^2}{ab}=-1$
$\because \cos \left(C\right)=\frac{a^2+b^2-c^2}{2ab}$
$\Rightarrow \cos \left(C\right)=-\frac{1}{2}$
$\Rightarrow \angle C=\frac{2\pi }{3}$
$\Rightarrow$ Circumradius $R=\frac{c}{2\sin C}=\frac{c}{\sqrt{3}}$