(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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(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 $x+y-3=0$.$r = \left| \frac{0+0-3}{\sqrt{1^2+1^2}} \right| = \frac{3}{\sqrt{2}}$.In an equilateral triangle,the circumradius $R$ is twice the inradius $r$,so $R = 2r$.$R = 2 \times \frac{3}{\sqrt{2}} = \frac{6}{\sqrt{2}}$.Thus,$R+r = \frac{6}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{9}{\sqrt{2}}$.

Class 11 Mathematics Straight Line Distance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line

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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 $R+r=$

AP EAMCET 2023 All AP EAMCET

A
$2 \sqrt{2}$
B
$\frac{9}{\sqrt{2}}$
C
$7 \sqrt{2}$
D
$\frac{3}{\sqrt{2}}$

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Distance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line

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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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