Question

A $\Delta EFG$ is inscribed in a unit square $ABCD$ with $E$ on $AB,F$ on $DA,G$ on $CD$ such that $AE=DF=CG=\frac{1}{3}$ The area of the $\Delta EFG$ is :

• A. $\frac{5}{18}$
• B. $\frac{1}{3}$
• C. $\frac{5}{9}$
• D. $\frac{4}{9}$

Answer 1

Answer: (A)

$\frac{5}{18}$

Given $ABCD$ square of unit one and $AE=DF=CG=\frac{1}{3}$ unit

Then $AF=EB=DG=\frac{2}{3}$ unit

So area of triangle $FDG=\frac{1}{2}\times \frac{1}{3}\times \frac{2}{3}=\frac{2}{18}$

And area of triangle $FAE=\frac{1}{2}\times \frac{1}{3}\times \frac{2}{3}=\frac{2}{18}$

Then area of quadrilateral $EADG=\frac{1}{2}(\frac{1}{3}+\frac{2}{3})\times 1=\frac{1}{2}$

So area of triangle $EFG=$ area of quadrilateral $EADG-$area of triangle $FDG - $area of triangle $FAE$

Area of triangle EFG=$\frac{1}{2}-\frac{2}{18}-\frac{2}{18}=\frac{5}{18}$