Question

In square $ABCE,$ $AF=2FE$ and $CD=2DE.$ What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE?$

$\ \dfrac{1}{6}$

$\ \dfrac{2}{9}$

$\ \dfrac{5}{18}$

$\ \dfrac{1}{3}$

$\ \dfrac{7}{20}$

Because the answer is a ratio, choose the side length of the square to be $3$ . Then $AF=2$ , $FE=1$ , $CD=2$ , and $DE=1$ .

The square has area $9$ . The areas of triangles $ABF$ and $BCD$ are each $\dfrac{1}{2}\cdot3\cdot2=3$ .

Triangle $DEF$ has area $\dfrac{1}{2}\cdot1\cdot1=\dfrac{1}{2}$ .

So $[BFD]=9-3-3-\dfrac{1}{2}=\dfrac{5}{2}$ , and the desired ratio is $\dfrac{5/2}{9}=\dfrac{5}{18}$ .

Thus, C is the correct answer.