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If H is the orthocentre of triangle ABC, R = circumradius and `P = AH + BH + CH`, thenA. `P = 2 (R + r)`B. max. of P is 3RC. min. of P is 3RD. `P = 2 (R -r)` |
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Answer» Correct Answer - A::B `AH = 2R cos A, BH = 2R cos B, CH = 2R cos C` `:. P = 2R (cos A + cos B+ cos C)` `= 2R (1+(r)/(R))` `=2 (R + r)` We know that in any triangle, `r le(R)/(2)` `:. P le 2R + R` `rArr P le 3R` |
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