Let the centroid of an equilateral triangle ABC be at the origin.

Question

Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line x + y = 3. If R and r be the radius of circumcircle and incircle respectively of $$\Delta$$ABC, then (R + r) is equal to :

Accepted Answer

$$r = \left| {{{0 + 0 - 3} \over {\sqrt 2 }}} \right| = {3 \over {\sqrt 2 }}$$

$$\sin 30^\circ  = {r \over R} = {1 \over 2}$$

R = 2r

So, $$r + R = 3r = 3 \times \left( {{3 \over {\sqrt 2 }}} \right) = {9 \over {\sqrt 2 }}$$

Solutions · 1