Question

In rectangle $ABCD,$ $AB = 5$ and $BC = 3.$ Points $F$ and $G$ are on $\overline{CD}$ so that $DF = 1$ and $GC = 2.$ Lines $AF$ and $BG$ intersect at $E.$ Find the area of $\triangle AEB.$

$10$

$\dfrac{21}{2}$

$12$

$\dfrac{25}{2}$

$15$

Answer: D

Concepts:similarity triangle area

Difficulty rating: 1580

Solution:

Since $FG = 5 - 1 - 2 = 2$ and $\overline{FG} \parallel \overline{AB},$ triangles $FEG$ and $AEB$ are similar with ratio $\dfrac{FG}{AB} = \dfrac{2}{5}.$

Let the distance from $E$ to line $CD$ be $k.$ Then the distance from $E$ to $AB$ is $k + 3,$ and $\frac{k}{k + 3} = \frac{2}{5},$ giving $k = 2.$

The height of $\triangle AEB$ is $k + 3 = 5,$ so its area is $\frac{1}{2}(5)(5) = \frac{25}{2}.$

Thus, the correct answer is D.

2003 AMC 12B Problem 14

Answer: D

Difficulty rating: 1580

Solution:

Problem 14 in Other Years