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 rectangle

Difficulty rating: 1480

Solution:

Here $FG = CD - DF - GC$ $= 5 - 1 - 2$ $= 2.$ Let $h$ be the distance from $E$ down to line $CD.$ Since $\triangle FEG \sim \triangle AEB$ with ratio $\dfrac{FG}{AB}=\dfrac25,$ we have $\dfrac{h}{h+3}=\dfrac25,$ so $h=2.$

The height of $\triangle AEB$ from $E$ to $AB$ is $h+3=5,$ giving area $\dfrac12 \cdot 5 \cdot 5 = \dfrac{25}{2}.$

Thus, the correct answer is D.

2003 AMC 10B Problem 20

Answer: D

Difficulty rating: 1480

Solution:

Problem 20 in Other Years