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If `H` is the orthocentre of triangle `A B C ,R=` circumradius and`P=A H+B H+C H` , then `P=2(R+r)` (b) `maxof P is 3R` `minofP is 3R` (d) `P=2(R-r)`

A

`P = 2 (R + r)`

B

max. of P is 3R

C

min. of P is 3R

D

`P = 2 (R -r)`

Text Solution

Verified by Experts

The correct Answer is:
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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