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Points A(0,0), B(9,6) and C(6,12) are vertices of triangle ABC. Point D is on segment AB such that 3(AD) = DB, point E is on segment BC such that 3(BE) = EC and point F is on segment CA such that 3(CF) = FA. What is the ratio of the area of triangle DEF to the area of triangle ABC? Express your answer as a common fraction.

 Nov 30, 2020
 #1
avatar+114592 
+1

Since 3(AD) = DB, then D  must  be (1/4)  the distance from  A to B

We  can  figure its  coordinates as

[ 0 +  (1/4)(9-0) , 0 + (1/4)(6-0) ] =  ( 9/4, 6/4)  = (9/4, 3/2)

 

Likewise, we can find E  as

[ 9 + (1/4)(6-9) , 6 + (1/4)(12 - 6) ] =   [ 9 - 3/4 ,  6 + 6/4 ]  =  [ 33/4 , 6 + 3/2 ] = [33/4 , 15/2 ] 

 

Finally, we note  that F is  3/4  of the distance  from A to C

To find its coordinates we have

[ 0 + (3/4)(6-0) , 0  + (3/4) (12-0) ]  =  [ 18/4, 36/4 ]  =  [ 9/2, 9]

 

To find the area of ABC we  can use this formula  where

Let (0,0)  =  (x1, y1)    (9,6)  = (x2, y2)    (6,12)  = (x3, y3)

Area = (1/2)  [ x1 (y2 - y3)  + x2 (y3 - y1) + x3(y1 - y2) ]

(1/2) [ 0 ( 6 - 12)  + 9(12 -0) + 6(0 - 6) ]  =  (1/2 [ 9 *12 + 6(-6) ] = 

(1/2) [ 108 - 36 ]  =

(1/2) 72 =

36

 

Likewise

Let  (9/4, 3/2) =  (x1,y1)     (33/4, 15/2 )  =    (x2, y2)        (9/2, 9)  = (x3, y3)

Area = (1/2)  [ (9/4) ( (15/2) - 9)  + (33/4) (9 - 3/2)  + (9/2)(3/2 - 15/2) ]  =   63/4

 

Area of DEF  to  Area of ABC  =

 

(63/4)  / 36  =   

 

63  / 144  = 

 

7 / 16

 

cool cool cool

 Nov 30, 2020
 #2
avatar+895 
+2

Points A(0,0), B(9,6) and C(6,12) are vertices of triangle ABC. Point D is on segment AB such that 3(AD) = DB, point E is on segment BC such that 3(BE) = EC, and point F is on segment CA such that 3(CF) = FA. What is the ratio of the area of triangle DEF to the area of triangle ABC? Express your answer as a common fraction.

 

Area of ΔABC = 36 u2             area of ΔDEF = 15.75 u2

 

So, the ratio...         [DEF] / [ABC] = 7/16 smiley

 Nov 30, 2020
edited by jugoslav  Nov 30, 2020

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