1.

If H is the orthocentre of triangle ABC, R = circumradius and P = AH + BH + CH, then

Answer»

<P>`P = 2 (R + r)`
MAX. of P is 3R
min. of P is 3R
`P = 2 (R -r)`

SOLUTION :`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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