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The medians $AD$, $BE$, and $CF$ of triangle $ABC$ intersect at the centroid $G$. The line through $G$ that is parallel to $BC$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. If the area of triangle $ABC$ is 144, then find the area of triangle $ENG$.

Guest Mar 31, 2018
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See the setup below :

 

Let  A  = (4, 12)

B = (0, 0)

C  = (24, 0)

So......the height of ABC  is 12  and  the base  is 24...so the area  is 12 * 24 / 2  =  144

E  is  ( 14,6)     and D  is  ( 12, 0)

 

Since AD  is a mediah....then GD  is (1/3)  of AD.....so.....the y coordinate of G  will be  (1/3) the y coordinate of A  = (1/3)(12)  =  4

So  MN  has the equation  y  = 4

And since the y coordinate of E  is 6  and E  is the apex  of ENG...the height of ENG  is

[ y coordinate of E - 4]  =  [ 6 - 4 ]   =  2

 

And since MN is parallel to BC....then  trinagle AGN  is similar to triangle ADC

And the height of AGN  = 8   and the  height of ADC  = 12

So each dimension of AGN  is 2/3  of ADC

And the base of ADC  = 12

So....the base of AGN  = GN  = (2/3)base of ADC  = (2/3)(12)  =  8

 

So....the area of ENG  =  (1/2)*GN * height of ENG  = 

 

(1/2) *  8  *   2  =

 

8  units^2

 

cool cool cool

CPhill  Mar 31, 2018

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