Similar Triangle Applications

Create similar triangles in order to solve for \(x\).


[Figure 6]

Extend \(\overline{A D}\) and \(\overline{B C}\) to create point \(G\).


[Figure 7]

\(\triangle D G C \sim \triangle E G F \sim \triangle A G B\) by \(A A \sim\) because angles \(\angle D C G, \angle E F G, \angle A B G\) are all right angles and are therefore congruent and all triangles share \(\angle G\). This means that their corresponding sides are proportional. First, solve for \(G C\) by looking at \(\triangle D G C\) and \(\triangle E G F . \triangle D G C \sim \triangle E G F\) which means that corresponding sides are proportional.

Therefore:

\(

\begin{aligned}

\frac{D G}{E G} &=\frac{G C}{G F}=\frac{D C}{E F} \\

\frac{D C}{G C} &=\frac{E F}{G F} \\

\frac{2}{G C} &=\frac{3.5}{2.5+G C} \\

5+2 G C &=3.5 G C \\

5 &=1.5 G C \\

G C & \approx 3.33

\end{aligned}

\)

Next, solve for \(x\) by looking at \(\triangle D G C\) and \(\triangle A G B . \triangle D G C \sim \triangle A G B\) which means that corresponding sides are proportional.

Therefore:

\(

\begin{aligned}

\frac{D G}{A G} &=\frac{G C}{G B}=\frac{D C}{A B} \\

\frac{D C}{G C} &=\frac{A B}{G B} \\

\frac{2}{3.33} &=\frac{x}{1.5+2.5+3.33} \\

\frac{2}{3.33} &=\frac{x}{7.33} \\

x &=4.4

\end{aligned}

\)