2003 AMC 10B Problem 20

Attempt Problem 20 of the 2003 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10B solutions, or check the answer key.

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20.

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

Small hint:

Compute

FG = CD - DF - GC

Big hint:

\triangle FEG \sim \triangle AEB;

use the ratio

\tfrac{FG}{AB}

to locate the height of

E

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.