Question

A triangle EFG is inscribed in a unit square ABCD with E on AB, F on DA, G on CD such that AE = DF = CD = 1/3. The area of the triangle EFG is

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

Answer 1

The correct option is A $\frac{5}{18}$

$ABCD$ is a unit square so that ar $ABCD$ $=1$

& side of square $=\sqrt{1}=1$

Now, $AE=DF=GC=\frac{1}{3}$

$\therefore AF=1-\frac{1}{3}=2/3$

$DG=1-1/3=2/3$

Now in $△DFG$,

$FG=\sqrt{{\left(DF\right)}^2+{\left(DG\right)}^2}=\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(\frac{2}{3}\right)}^2}=\frac{\sqrt{5}}{3}$

Also, $FE=\frac{\sqrt{5}}{3}$

$\therefore ar△EFG=\frac{1}{2}\times GF\times FE=\frac{1}{2}\times \frac{\sqrt{5}}{3}\times \frac{\sqrt{5}}{3}=\frac{5}{18}$