Geometry

Geometry

As shown in the diagram, \(\dfrac{BD}{DC}=2\) , \(\dfrac{CE}{EA}=3\), and \(\dfrac{AF}{FB}=4\). Find \(\dfrac{[DEF]}{[ABC]}\).

\(\begin{array}{|rcll|} \hline [ABC] &=& \dfrac{4EA\cdot3DC}{2}\sin(C) \\ \mathbf{[ABC]} &\mathbf{=}& \mathbf{6 EA\cdot DC \sin(C)} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline [AFE] &=& \dfrac{1EA\cdot4FB}{2}\sin(A) \quad | \quad \sin(A)=\dfrac{3DC}{5FB}\sin(C) \\ [AFE] &=& \dfrac{1EA\cdot4FB}{2}\cdot \dfrac{3DC}{5FB}\sin(C) \\ \mathbf{[AFE]} &\mathbf{=}& \mathbf{ \dfrac{6}{5} EA\cdot DC \sin(C)} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline [FBD] &=& \dfrac{2DC\cdot 1FB}{2}\sin(B) \quad | \quad \sin(B)=\dfrac{4EA}{5FB}\sin(C) \\ [FBD] &=& \dfrac{2DC\cdot 1FB}{2}\cdot \dfrac{4EA}{5FB}\sin(C) \\ \mathbf{[FBD]} &\mathbf{=}& \mathbf{ \dfrac{4}{5} EA\cdot DC \sin(C)} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline [EDC] &=& \dfrac{1DC\cdot 3EA}{2}\sin(C) \\ \mathbf{[EDC]} &\mathbf{=}& \mathbf{ \dfrac{3}{2} EA\cdot DC \sin(C)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline [AFE]+[FBD]+[EDC]+[DEF] &=& [ABC] \\ [DEF] &=& [ABC] -([AFE]+[FBD]+[EDC]) \quad | \quad : [ABC] \\\\ \dfrac{[DEF]}{[ABC]} &=& \dfrac{[ABC] -([AFE]+[FBD]+[EDC])} {[ABC]} \\\\ \dfrac{[DEF]}{[ABC]} &=&1 -\dfrac{[AFE]+[FBD]+[EDC]} {[ABC]} \\\\ \dfrac{[DEF]}{[ABC]} &=&\small{1 -\dfrac{\dfrac{6}{5} EA\cdot DC \sin(C)+\dfrac{4}{5} EA\cdot DC \sin(C)+\dfrac{3}{2} EA\cdot DC \sin(C)} {6 EA\cdot DC \sin(C)} }\\\\ \dfrac{[DEF]}{[ABC]} &=&1 -\dfrac{\dfrac{35}{10}} {6 } \\\\ \dfrac{[DEF]}{[ABC]} &=&1 - \dfrac{7}{12} \\\\ \mathbf{\dfrac{[DEF]}{[ABC]}} &\mathbf{=}& \mathbf{\dfrac{5}{12}} \\ \hline \end{array}\)

Coolstuff, could you put only one of these questions per post.

It makes it more confusing to hve 2 or more  plus there should only be one question per post anyway. :)

Here's the second one

Angle AEB= Angle DEC

Angle ABE = Angle DCE 

So.....triangle AEB is similar to Triangle DEC

So 

DC / AB = 12/18 = 2/3

So

DE /  AE =  2 / 3   ⇒    AE = (3/2)DE

Which implies that

DE / AE  =  EC / EB

2 / 3  =   EC / EB

EC = (2/3)EB

EB + DE = 19

EC + AE = 21

EB + DE =  19

(2/3)EB  + (3/2)DE = 21

EB + DE  = 19

-EB - (9/4)DE = -63/2

-(5/4)DE = -25/2

DE = (25/2) (4/5) = 10

So

AE = (3/2)DE =  15

And

EB + DE = 19

EB + 10 = 19

EB = 9

So

EC = (2/3)EB

EC = (2/3)(9) = 6

And

[ BEC] = (1/2)(EB) (EC) sin BEC  =  (1/2) (9)(6) sin BEC

[ AED ] = (1/2)(DE) (AE) sinAED = (1/2) (10)( 15) sin AED

And sin BEC = sin AED

So

[BEC ] / [ AED] =   (9)(6) / [ 10 * 15]  =   54 / 150  =   9 /  25