Question

If two sides of a triangle are the roots of $x^2-7x+8=0$ and the angle between these sides is $\frac{\pi }{3},$ then the product of inradius and circumradius of the triangle is

• A. $\frac{8}{7}$
• B. $\frac{5}{3}$
• C. $\frac{5\sqrt{2}}{3}$
• D. $8$

Answer 1

The correct option is B $\frac{5}{3}$

Let $a,b,c$ be the sides of the triangle.
$a$ and $b$ are roots of $x^2-7x+8=0$
$\Rightarrow a+b=7$ and $ab=8$
$C=\frac{\pi }{3}$
$\Rightarrow \cos C=\frac{1}{2}=\frac{a^2+b^2-c^2}{2ab}$
$\Rightarrow ab=a^2+b^2-c^2$
$\Rightarrow c^2=(a+b{)}^2-3ab=49-24=25$
$\Rightarrow c=5$

$\therefore rR=\frac{\Delta }{s}\times \frac{abc}{4\Delta }$
$=\frac{abc}{2(a+b+c)}=\frac{5}{3}$