Question

The roots of $x^2-2\sqrt{3}x+2=0$ represent two sides of a triangle. lf the angle between them is $\frac{\pi }{3}$ , then the perimeter of the triangle is

• A. $2\sqrt{3}+6$
• B. $2\sqrt{3}+\sqrt{6}$
• C. $3\sqrt{2}+6$
• D. $3\sqrt{2}+\sqrt{6}$

Answer 1

Answer: (B)

$2\sqrt{3}+\sqrt{6}$

$\cos \pi /3=\frac{x_1^2+x_2^2-c^2}{2x_1{x}_2}$

$\frac{1}{2}=\frac{{(x_1+x_2)}^2-2x_1{x}_2-c^2}{2x_1{x}_2}$

From the quadratic equation given in the question

$x_1+x_2=2\sqrt{3}$

$x_1{x}_2=2$

now substitute these values in the above cos formula, we get
$\frac{1}{2}=\frac{{\left(2\sqrt{3}\right)}^2-2\times 2-c^2}{2\times 2}$
$2=12-4-c^2$
$c^2=6$
$c=\sqrt{6}$
$Perimeter=x_{1+}{x}_2+c$
$=2\sqrt{3}+\sqrt{6}$