Question
In an equilateral triangle sum of the distances of orthocenter from all the vertices is equal to -
(Where 'R' is the circumradius of triangle)
(Where 'R' is the circumradius of triangle)
Answer: Option C
:
C
Let the triangle be ABC and the orthocenter be O.
We know that OA = 2RcosA
It is given that triangle is equilateral. So cosA = cos (60°) = 12
And OA = R
Similarly, OB = OC = R
And the sum of them = OA + OB + OC = 3R
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:
C
Let the triangle be ABC and the orthocenter be O.
We know that OA = 2RcosA
It is given that triangle is equilateral. So cosA = cos (60°) = 12
And OA = R
Similarly, OB = OC = R
And the sum of them = OA + OB + OC = 3R
Was this answer helpful ?
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