how to do this elegantly ?

In ΔABC , E and F are pts on AC and AB . Lines BE and CF intersect at P

area of

BPF = 5
PFAE=22
CPE=8

wats area of BPC

options

22
16
10
not uniquely decidable

10 Answers

1
skygirl ·

is it 10 ?

11
Subash ·

can you post your "elegant" method too sky :P

1
skygirl ·

hehhe :D

jus drew and ''felt'' [3]

62
Lokesh Verma ·

In ΔABC , E and F are pts on AC and AB . Lines BE and CF intersect at P

area of BPF = 5 = IV, PFAE=22 = I+III , CPE=8 = II
wats area of BPC = V

Using same base and same height
AC/EC= I/(I+III+IV) = II/(II+V)
I/(27)= 8/(8+V)
EF/FB=III/(I+II+III) = IV/(IV+V)
III/(30)=5/(5+V)

add I + III

you will get the answer :)

1
skygirl ·

how and why did u use symmetry ?

pls explain ...

9
Celestine preetham ·

thanks sir

sky thats based on fact altitude remains same for APE and ABE

62
Lokesh Verma ·

sky .. did u get it?

1
skygirl ·

yeah yeah ... tx :)

21
tapanmast Vora ·

Wow!!! Awesome [1]

62
Lokesh Verma ·

on btw i used the wrong work similarity..

it is equal distance ...

will fix that :)

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