Question

Two sides of a triangle are given by the roots of the quadratic equation $x^2-2\sqrt{3}x+2=0$ and the angle between the sides is $\frac{\pi }{3}$ . The perimeter of the triangle is

• A. $2\sqrt{3}$ units
• B. $\sqrt{6}$ units
• C. $2\sqrt{3}+\sqrt{6}$ units
• D. $2\left(\sqrt{3}+\sqrt{6}\right)$ units

Answer 1

Answer: (C)

Let the two sides $a$ and $b$ are the roots of the quadratic equation $x^2-2\sqrt{3}x+2=0$
.
$\therefore a+b=2\sqrt{3}$ and $ab=2$ . Also, $\angle C=\frac{\pi }{3}$
Using cosine rule,
$cosC=\frac{a^2+b^2-c^2}{2ab}\Rightarrow \frac{1}{2}=\frac{a^2+b^2-c^2}{2ab}$
$\Rightarrow a^2+b^2-c^2=ab\Rightarrow {\left(a+b\right)}^2-c^2=3ab$
$=>{\left(2\sqrt{3}\right)}^2-c^2=3\times 2\Rightarrow c=\sqrt{6}$
$\therefore$ Perimeter $=a+b+c=2\sqrt{3}+\sqrt{6}$ units