Let origin be the centroid of an equilateral triangle $\mathrm{ABC}$ and one of its sides be along the straight line $x+y=3$. If $\mathrm{R}$ and $\mathrm{r}$ are its circum radius and inradius respectively, then $\mathrm{R}+\mathrm{r}=$

• A

  $2 \sqrt{2}$
• Correct answer
  B

  $\frac{9}{\sqrt{2}}$
• C

  $7 \sqrt{2}$
• D

  $\frac{3}{\sqrt{2}}$

Solution

Given that one of the sides of the equilateral triangle be along the straight line $x+y=3$ And we know in an equilateral triangle each side have an angle $60^{\circ}$. $\therefore \angle \mathrm{OBM}=\frac{60^{\circ}}{2}=30^{\circ}$ From the figure, $r=\mathrm{OM}=$ perpendicular distance $O(0,0) \& x+y=3$ $\Rightarrow r=\left|\frac{(0+0-3)}{\sqrt{1+1}}\right|=\frac{3}{\sqrt{2}}$ ...(i)
Now, $r+\mathrm{R}=\frac{3}{\sqrt{2}}+\frac{6}{\sqrt{2}}=\frac{9}{\sqrt{2}}$

Hence, the correct answer is option: 'B'.