In triangle $\triangle ABC$, $D$ lies between $A$ and $C$ and $AC=3AD$, $E$ lies between $B$ and $C$ and $BC=4EC$. $B,G,F,D$ in that order, are on a straight line and $BD=5GF=5FD$. Suppose the area of $\triangle ABC$ is $900$, find the area of the triangle $\triangle EFG$.
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Tags: geometry
byrmc
03.06.2014 22:37
area of $\triangle BDC=600$
let area of $\triangle EFG=3S$ then area of $\triangle BDC=20S$
$600=20S$
$S=30$
so area of $\triangle EFG=90$
YESMAths
04.06.2014 07:40
Since $AD/DC=1/2$ we get that $ar(\triangle{ABD})/ar(\triangle{DBC})=1/2$
So $ar(\triangle{ABD})= 300$
Now similarly we have that $\frac{1}{3}ar(\triangle{BGE})=ar(\triangle{EFG})=ar(\triangle{EFD})$
So, $ar(\triangle{DCE})+5ar(\triangle{EFG})=600$
Since $ar(\triangle{DCE})=\frac{1}{6}ar(\triangle{ABC})=150$(using the trigonometric identity for the area of a triangle)e have that $ar(\triangle{EFG})=\frac{600-150}{5}=90.$