Question

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 $\triangle ACE?$

$\dfrac{1}{4}$

$\dfrac{9}{25}$

$\dfrac{3}{8}$

$\dfrac{11}{25}$

$\dfrac{7}{16}$

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 $\dfrac{[BDF]}{[ACE]} = 1 - 3 \cdot \dfrac{3}{16} = 1 - \dfrac{9}{16} = \dfrac{7}{16}.$

Thus, the correct answer is E.

2004 AMC 10B Problem 18

Solution:

Problem 18 in Other Years