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

#1**+1 **

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

CPhill
Mar 31, 2018