Question

In rectangle $ABCD,$ points $F$ and $G$ lie on $\overline{AB}$ so that $AF = FG = GB$ and $E$ is the midpoint of $\overline{DC}.$ Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J.$ The area of rectangle $ABCD$ is $70.$ Find the area of triangle $EHJ.$

$\dfrac{5}{2}$

$\dfrac{35}{12}$

$3$

$\dfrac{7}{2}$

$\dfrac{35}{8}$

Solution:

Triangle $EFG$ has base $FG = \dfrac{1}{3}AB$ and height equal to the rectangle's height, so its area is $\dfrac{1}{6}(70) = \dfrac{35}{3}.$

Because $EC \parallel AF,$ triangles $AFH$ and $CEH$ are similar with ratio $\dfrac{EC}{AF} = \dfrac{3}{2},$ so $\dfrac{EH}{EF} = \dfrac{3}{5}.$ Likewise $\dfrac{EJ}{EG} = \dfrac{3}{7}.$

Then $\dfrac{[EHJ]}{[EFG]} = \dfrac{EH}{EF}\cdot\dfrac{EJ}{EG} = \dfrac{3}{5}\cdot\dfrac{3}{7} = \dfrac{9}{35},$ giving $[EHJ] = \dfrac{9}{35}\cdot\dfrac{35}{3} = 3.$

Thus, the correct answer is C.

2001 AMC 12 Problem 22

Solution:

Problem 22 in Other Years