IIT JEE , AIEEE Forum - Mathematics , Trignometry

ashish_banga

in a triangle ABC, E and F are the points on AC and AB respectively. the lines BE and CF intersect at P. if the area( BPF) =5 , area (PFAE) = 22 and area ( CPE) = 8 then area (BPC) is

ashish_banga

solve quickly

Greatdreams

Given :

area PBC = ?

area BPF = 5

area PFAE = 22

area CPE = 8

area CFA = 30

so , by the problem

BPC / APB = CPE / APE & PBC / PAC = BPF / PAF

So let PBC = x and APF = y

So,

x / 5+y = 8/22 - y

So x = 8 / 22 - y * ( 5 + y)

and x/30 - y = 5/y

so x = (30 - y) * 5/y

Equating them you have

8(5 + y)/ (22 - y) = 5 (30 - y) / y

or 40 + 8y/22 - y = 150 - 5y / y

cross multiplying,

40 y + 8 y ² = (150 - 5y)(22 - y) = 3300 - 110 y - 150 y + 5y²

or 3 y² + 300y - 3300 = 0

or y² + 100y - 1100 = 0

so y = 10 or - 110

So x = 8 . (5+y)/(22 - y) = 10 = area of BPC .

ashish_banga

how do you get
BPC / APB = CPE / APE & PBC / PAC = BPF / PAF

Ask & Discuss Questions with Community & Experts