Q.41. In a traingle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y.If x2-c2 = y, where c is the length of the third side of the triangle,then the circumradius of the traingle 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 traingle.
According to the given condition
a+b = x and ab=y
x2-c2=y
(a+b)2 – c2 = ab
a2+b2-c2 = -ab
\frac{a^{2}+b^{2}-c^{2}}{2ab}=-\frac{1}{2} Cos \; C=\frac{1}{2} C=\frac{2\pi }{3}Circumradius = C=\frac{c}{2 sin\; C}=\frac{c}{2 sin\left ( \frac{2\pi }{3} \right)}=\frac{2\pi }{3}