Question

In rectangle $ABCD,$ $AB=6,$ $AD=30,$ and $G$ is the midpoint of $\overline{AD}.$ Segment $AB$ is extended 2 units beyond $B$ to point $E,$ and $F$ is the intersection of $\overline{ED}$ and $\overline{BC}.$ What is the area of quadrilateral $BFDG?$

$\dfrac{133}{2}$

$67$

$\dfrac{135}{2}$

$68$

$\dfrac{137}{2}$

Solution:

The polygon $BFDG$ is a trapezoid with bases $DG$ and $BF$ and height $6.$ Also, since $G$ is the midpoint between $A$ and $D,$ we have $GD= 15.$

We can see that $EBF \sim EAD ,$ so $\dfrac{BF}{AD} = \dfrac{EB}{EA}$ $\dfrac{BF}{30} = \dfrac 28$ $BF = 7.5$

This makes the area of $BFDG$ equal to $\dfrac{6(15+7.5)}2 = \dfrac{135}2.$

Thus, the correct answer is C .

2012 AMC 10B Problem 19

Solution:

Problem 19 in Other Years