2004 AMC 10B Problem 18

Problem 18 of 25IntermediateGeometry

In right triangle

\triangle ACE,

we have

AC = 12,

CE = 16,

and

EA = 20.

Points

B,

D,

and

F

are located on

AC,

CE,

and

EA,

respectively, so that

AB = 3,

CD = 4,

and

EF = 5.

What is the ratio of the area of

\triangle BDF

to that of

\triangle ACE?

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Solution

The area of

\triangle ACE

is

\tfrac12(12)(16) = 96.

Each corner triangle

\triangle ABF,

\triangle BCD,

and

\triangle DEF

has a base and an altitude that are

\tfrac34

and

\tfrac14

of a corresponding base and altitude of

\triangle ACE.

So each has area

\tfrac14 \cdot \tfrac34 = \tfrac{3}{16}

of

\triangle ACE.

Hence

\begin{aligned} \dfrac{[BDF]}{[ACE]} &= 1 - 3 \cdot \dfrac{3}{16} \\ &= 1 - \dfrac{9}{16} = \dfrac{7}{16}. \end{aligned}

Thus, the correct answer is E.

More practice

Concepts: area ratio · area decomposition · triangle area

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