Let the origin be the centroid of an equilate | vedclass

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(B) In an equilateral triangle,the centroid coincides with the circumcenter and incenter. Let $O$ be the origin $(0,0)$,which is the centroid.The inra

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$\frac{9}{\sqrt{2}}$

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(B) In an equilateral triangle,the centroid coincides with the circumcenter and incenter. Let $O$ be the origin $(0,0)$,which is the centroid.The inradius $r$ is the perpendicular distance from the centroid $O(0,0)$ to the side $x+y-3=0$.$r = \left| \frac{0+0-3}{\sqrt{1^2+1^2}} ight| = \frac{3}{\sqrt{2}}$.In an equilateral triangle,the circumradius $R$ is twice the inradius $r$,so $R = 2r$.$R = 2 	imes \frac{3}{\sqrt{2}} = \frac{6}{\sqrt{2}}$.Thus,$R+r = \frac{6}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{9}{\sqrt{2}}$.

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

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