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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 $\mathrm{R}+\mathrm{r}=$
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The correct answer is:
$\frac{9}{\sqrt{2}}$
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, $r+\mathrm{R}=\frac{3}{\sqrt{2}}+\frac{6}{\sqrt{2}}=\frac{9}{\sqrt{2}}$
$\therefore \angle \mathrm{OBM}=\frac{60^{\circ}}{2}=30^{\circ}$
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, $r+\mathrm{R}=\frac{3}{\sqrt{2}}+\frac{6}{\sqrt{2}}=\frac{9}{\sqrt{2}}$
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