GKT101: General Knowledge for Teachers – Math

Once you know that two triangles are similar, you can use the fact that their corresponding sides are proportional and their corresponding angles are congruent to solve problems.

Consider the triangles in the image below.

a. Prove that the triangles are similar.

The triangles are similar by $A A \sim$ because they have at least two pairs of congruent angles, i.e.,

$\angle A B C=\angle D E F=90^{\circ}, \angle B C A=\angle E F D=30^{\circ}, \angle C A B=\angle F D E=60^{\circ}$

b. Use the Pythagorean Theorem to find $D E$.

According to the Pythagorean Theorem, in $\triangle D E F$ :

$\begin{aligned}E F^{2}+D E^{2} &=F D^{2} \\(3 \sqrt{3})^{2}+D E^{2} &=6^{2} \\(9 \cdot 3)+D E^{2} &=6^{2} \\27+D E^{2} &=36 \\D E^{2} &=36-27 \\D E^{2} &=9 \\D E &=3\end{aligned}$

c. Use the fact that the triangles are similar to find the missing sides of $\triangle A B C$.

$\triangle A B C \sim \triangle D E F$, which means that the corresponding sides are proportional. Therefore,

$\begin{aligned}&\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F} \\&\frac{A C}{D F}=\frac{18}{6}=3, \text { so the scale factor is } 3\end{aligned}$

• $\frac{A B}{D E}=3 \rightarrow \frac{A B}{3}=3 \rightarrow A B=9$
• $\frac{B C}{E F}=3 \rightarrow \frac{B C}{3 \sqrt{3}}=3 \rightarrow B C=9 \sqrt{3}$

Triangles of this configuration are called $30-60-90$ triangles because of their angle measures.

d. Explain why all 30-60-90 triangles are similar.

All 30-60-90 triangles are similar by $A A \sim$ because they all have at least two pairs of congruent angles.