Question

Two sides of a triangle are given by the roots of the equation $x^2-2\sqrt{5}x+4=0$ and the angle between them is $\frac{\pi }{3}$, then the semiperimeter of the triangle is-

• A. $2(\sqrt{5}-\sqrt{2})$
• B. $\sqrt{5}-\sqrt{2}$
• C. $\frac{6}{\sqrt{5}-\sqrt{2}}$
• D. $\frac{3}{\sqrt{5}-\sqrt{2}}$

Answer 1

The correct option is D $\frac{3}{\sqrt{5}-\sqrt{2}}$

$x^2-2\sqrt{5}x+4=0$

Let the sides of length of $a,b$ are roots of given equation and $\angle C={60}^{\circ }$

$a+b=2\sqrt{5}ab=4{(a+b)}^2=a^2+b^2+2ab\Rightarrow a^2+b^2=12\cos C=\frac{a^2+b^2-c^2}{2ab}\cos \frac{\pi }{3}=\frac{12-c^2}{8}\frac{1}{2}=\frac{12-c^2}{8}{c}^2=8\Rightarrow c=2\sqrt{2}s=\frac{a+b+c}{2}=\frac{2\sqrt{5}+2\sqrt{2}}{2}=\sqrt{5}+\sqrt{2}s=\sqrt{5}+\sqrt{2}\times \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{3}{\sqrt{5}-\sqrt{2}}$