Q.42. In a triangle, the sum of the 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{3}{2} y
D. \frac{y}{\sqrt{3}}
Answer :- B. \frac{c}{\sqrt{3}}
Explanation :-
Let a and b be the lengths of two sides of a triangle.
\therefore \quad According to the given condition,
\mathrm{a}+\mathrm{b}=x and \mathrm{ab}=y
\therefore \quad x^{2}-c^{2}=y \Rightarrow(a+b)^{2}-c^{2}=a b \Rightarrow a^{2}+b^{2}+2 a b-c^{2}=a b \Rightarrow a^{2}+b^{2}-c^{2}=-a b \Rightarrow \frac{a^{2}+b^{2}-c^{2}}{2 a b}=-\frac{1}{2} \Rightarrow \cos \mathrm{C}=\frac{1}{2} \Rightarrow \mathrm{C}=\frac{2 \pi}{3}\Rightarrow circumradius =\frac{c}{2 \sin C}=\frac{c}{2 \sin \left(\frac{2 \pi}{3}\right)}=\frac{c}{\sqrt{3}}