Let side a,b and c of `DeltaABC` be related by the relation a : b : c = 3 : 5 : 4. Altitudes AD,BE and CF are dropped on BC, CA and AB, respectively. If `P_(1)D+P_(2)E+P_(3)F=42`, then the value of a + b + c is

Options

A. 1200
B. 120
C. 12
D. none of these

Answer

A

Explanation

`P_(1)D=|BP_(1)-BD|=|(a)/(2)-c cos B|`
`=|(a)/(2)-(c(a^(2)+b^(2)-b^(2)))/(2ac)|`
`=|(b^(2)-c^(2))/(2a)|`
`=|((b-c)(b+c))/(2a)|`
Similarly `P_(2)E=|(c^(2)-a^(2))/(2b)|=|((c-a)(c+a))/(2b)|`
and `P_(3)F=|((a-b)(a+b))/(2c)|`
Given `a:b:c=3:5:4`
`therefore a=3k, b=5k, c=4k`
`P_(1)D=(3)/(2)k,P_(2)E=(7k)/(10),P_(3)F=2k`
`therefore (3)/(2)k + 2k+(7k)/(10)=(42k)/(10)=42`
`therefore k = 10`
`therefore a+b+c=120k=1200`