In ` A B C ,ifa=10a n dbcotB+c cot C=2(r+R)` then the maximum area of ` A B C` will be (a) 50 (b) `sqrt(50) ` (c) 25 (d) 5

Options

A. 50
B. `sqrt50`
C. 25
D. 5

Answer

C

Explanation

`b cot B + c cot C = 2 (r + R)`
`rARr 2R sin B .(cosB)/(sinB) + 2R sin C.(cos C)/(sinC) = 2(r + R)`
`rArr cos B + cos C = 1 + (r)/(R)`
`rArr cos B + cos C = 1 + 4 sin.(A)/(2) sin.(B)/(2) sin.(C)/(2)`
`rArr cos B + cos C = cos A + cos B + cos C`
`:. cos A = 0`
`rArr A = (pi)/(2)`
`rArr a^(2) = b^(2) + c^(2)`
`rArr b^(2) + c^(2) = 100`
Using A.M. `ge` G.M. we get
`(b^(2) + c^(2))/(2) ge sqrt(b^(2) c^(2))`
`rArr bc le 50`
Hence, area of `Delta ABC = (1)/(2) bc le 25`