GKT101: General Knowledge for Teachers – Math

Create similar triangles in order to solve for $x$.

[Figure 6]

[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:

$\begin{aligned} \frac{D G}{A G} &=\frac{G C}{G B}=\frac{D C}{A B} \\ \frac{D C}{G C} &=\frac{A B}{G B} \\ \frac{2}{3.33} &=\frac{x}{1.5+2.5+3.33} \\ \frac{2}{3.33} &=\frac{x}{7.33} \\ x &=4.4 \end{aligned}$