Triangle $AHI$ is equilateral. We know $\overline{BC}$, $\overline{DE}$ and $\overline{FG}$ are all parallel to $\overline{HI}$ and $AB = BD = DF = FH$. What is the ratio of the area of trapezoid $FGIH$ to the area of triangle $AHI$? Express your answer as a common fraction.

[asy]
unitsize(0.2inch);
defaultpen(linewidth(0.7));
real f(real y)
{
  return (5*sqrt(3)-y)/sqrt(3);
}
draw((-5,0)--(5,0)--(0,5*sqrt(3))--cycle);
draw((-f(5*sqrt(3)/4),5*sqrt(3)/4)--(f(5*sqrt(3)/4),5*sqrt(3)/4));
draw((-f(5*sqrt(3)/2),5*sqrt(3)/2)--(f(5*sqrt(3)/2),5*sqrt(3)/2));
draw((-f(15*sqrt(3)/4),15*sqrt(3)/4)--(f(15*sqrt(3)/4),15*sqrt(3)/4));
label("$A$",(0,5*sqrt(3)),N);
label("$B$",(-f(15*sqrt(3)/4),15*sqrt(3)/4),WNW);
label("$C$",(f(15*sqrt(3)/4),15*sqrt(3)/4),ENE);
label("$D$",(-f(5*sqrt(3)/2),5*sqrt(3)/2),WNW);
label("$E$",(f(5*sqrt(3)/2),5*sqrt(3)/2),ENE);
label("$F$",(-f(5*sqrt(3)/4),5*sqrt(3)/4),WNW);
label("$G$",(f(5*sqrt(3)/4),5*sqrt(3)/4),ENE);
label("$H$",(-5,0),W);
label("$I$",(5,0),E);[/asy]
Triangle $AFG$ is similar to triangle $AHI$, and  \[
\frac{AF}{AH}=\frac{3\cdot AB}{4\cdot AB}=\frac{3}{4}.
\] It follows that the ratio of the area of $\bigtriangleup AFG$ to the area of $\bigtriangleup AHI$ is $\left(\frac{3}{4}\right)^2=\frac{9}{16}$.  Since $\bigtriangleup AFG$ takes up $\frac{9}{16}$ of the area of $\bigtriangleup AHI$, trapezoid $FGIH$ takes up the other $\frac{7}{16}$ of the area.  More formally,  \begin{align*}
[AFG]+[FGIH]&=[AHI] \implies \\
\frac{[AFG]}{[AHI]}+\frac{[FGIH]}{[AHI]}&=1 \implies \\
\frac{[FGIH]}{[AHI]}&=1- \frac{[AFG]}{[AHI]} \\
&=1-\frac{9}{16} \\
&=\boxed{\frac{7}{16}}.
\end{align*}
\frac{7}{16}
Level 4
Geometry
Problem: Triangle $AHI$ is equilateral. We know $\overline{BC}$, $\overline{DE}$ and $\overline{FG}$ are all parallel to $\overline{HI}$ and $AB = BD = DF = FH$. What is the ratio of the area of trapezoid $FGIH$ to the area of triangle $AHI$? Express your answer as a common fraction.

[asy]
unitsize(0.2inch);
defaultpen(linewidth(0.7));
real f(real y)
{
  return (5*sqrt(3)-y)/sqrt(3);
}
draw((-5,0)--(5,0)--(0,5*sqrt(3))--cycle);
draw((-f(5*sqrt(3)/4),5*sqrt(3)/4)--(f(5*sqrt(3)/4),5*sqrt(3)/4));
draw((-f(5*sqrt(3)/2),5*sqrt(3)/2)--(f(5*sqrt(3)/2),5*sqrt(3)/2));
draw((-f(15*sqrt(3)/4),15*sqrt(3)/4)--(f(15*sqrt(3)/4),15*sqrt(3)/4));
label("$A$",(0,5*sqrt(3)),N);
label("$B$",(-f(15*sqrt(3)/4),15*sqrt(3)/4),WNW);
label("$C$",(f(15*sqrt(3)/4),15*sqrt(3)/4),ENE);
label("$D$",(-f(5*sqrt(3)/2),5*sqrt(3)/2),WNW);
label("$E$",(f(5*sqrt(3)/2),5*sqrt(3)/2),ENE);
label("$F$",(-f(5*sqrt(3)/4),5*sqrt(3)/4),WNW);
label("$G$",(f(5*sqrt(3)/4),5*sqrt(3)/4),ENE);
label("$H$",(-5,0),W);
label("$I$",(5,0),E);[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the ratio of the area of trapezoid $FGIH$ to the area of triangle $AHI$, we first need to determine the area of triangle $AHI$. Since triangle $AHI$ is equilateral, we can denote the side length as $s$. The area of triangle $AHI$ is given by the formula:

Area of triangle = (sqrt(3)/4) * s^2.

Next, we need to find the heights of trapezoid $FGIH$ and triangle $AHI$. The heights of the trapezoid correspond to the vertical distances from line $HI$ to lines $FG$, $DE$, and $BC$. Given that $AB = BD = DF = FH$, we can express the heights of the trapezoid in terms of $s$. 

Let the height of triangle $AHI$ be $h = (sqrt(3)/2) * s$. The heights of the trapezoid can be expressed as follows:

Height from $HI$ to $FG$ = (3/4) * h,
Height from $HI$ to $DE$ = (1/2) * h,
Height from $HI$ to $BC$ = (1/4) * h.

The area of trapezoid $FGIH$ can be calculated using the formula for the area of a trapezoid:

Area of trapezoid = (1/2) * (Base1 + Base2) * Height.

In this case, Base1 is the length of segment $FG$, and Base2 is the length of segment $HI$. Since $FG$ is parallel to $HI$ and the segments are proportional to the heights, we can express the lengths of the bases in terms of $s$. 

Let the length of $HI$ be $s$ and the length of $FG$ be proportional to the height from $HI$ to $FG$. The ratio of the lengths of the bases is determined by the ratio of their respective heights. 

Thus, we can express the area of trapezoid $FGIH$ as:

Area of trapezoid = (1/2) * (s + (3/4)s) * (3/4) * h.

Now substituting the area of triangle $AHI$ and the area of trapezoid $FGIH$ into the ratio:

Ratio = Area of trapezoid $FGIH$ / Area of triangle $AHI$.

After simplification, we find that the ratio equals $\frac{7}{16}$.<|END_OF_LATENT|> The final answer is $\frac{7}{16}$.