JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 12)
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 :
$$7\sqrt 2 $$
$${9 \over {\sqrt 2 }}$$
$$2\sqrt 2 $$
$$3\sqrt 2 $$
Explanation
_18th_March_Evening_Shift_en_12_1.png)
$$r = \left| {{{0 + 0 - 3} \over {\sqrt 2 }}} \right| = {3 \over {\sqrt 2 }}$$
$$\sin 30^\circ = {r \over R} = {1 \over 2}$$
R = 2r
So, $$r + R = 3r = 3 \times \left( {{3 \over {\sqrt 2 }}} \right) = {9 \over {\sqrt 2 }}$$
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