2003 AMC 12B Problem 14

Problem 14 of 25IntermediateGeometry

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.

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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.

More practice

Concepts: similarity · triangle area

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.