Question

ABCD is a parallelogram, G is the point on AB such that $AG=2GB$. E is a point on DC such that $CE=2DE$ and F is a point of BC such that $BF=2FC$.Then area of $△EFG$ =$\frac{5}{18}$ of area of ABCD.

• A. True
• B. False

Answer 1

The correct option is A True

REF. Image.

Let $FB=2CE$ , here $CF=x\&FB=2x$

$\because △CEI\cong △CBH$

$\Rightarrow CI=K\&IH=2K$

Now in $△EGFar\left(△EFG\right)=ar\left(△ESF\right)+ar\left(△SGF\right)$

$=\frac{1}{2}SF\times K+\frac{1}{2}SF\times 2K$

$ar\left(△EFG\right)=\frac{3}{2}SF\times K...\left(1\right)$

Now $ar\left(△EGBC\right)=ar\left(SGBF\right)+ar\left(ESFC\right)$

$=\frac{1}{2}(SF+GB)\times 2K+\frac{1}{2}(SF+EC)\times K$

$\frac{1}{2}\left(ar\right(△ABCD\left)\right)=\frac{3}{2}K\times SE+\frac{2}{3}AB\times K$

$ar\left(△ABCD\right)=3K\times SF+\frac{4}{3}AB\times K$

$ar\left(△ABCD\right)=3K\times SF+\frac{4}{9}\left(ar\right(ABCD\left)\right)$

$\Rightarrow K\times SF=\frac{5}{27}\left(ar\right(ABCD\left)\right)...\left(2\right)$

From (1) & (2), we get

$ar\left(△EFG\right)=\frac{3}{2}\times \frac{5}{27}\left(ar\right(ABCD\left)\right)$

$ar\left(△EFG\right)=\frac{5}{18}\left(ar\right(ABCD\left)\right)$

Hence Proved