Similar Triangles and Polygons Practice Problems

First, let's look at angles C and E, which are equal to 30 degrees.

Angle C is opposite side AB and angle E is opposite side BD.

$\frac{AB}{DB}$

Now let's look at angle B, which is equal to 90 degrees in both triangles.

In triangle ABC the opposite side is AC and in triangle EBD the opposite side is ED.

$\frac{AC}{ED}$

Let's look at angles A and D, which are equal to 60 degrees.

Angle A is the opposite side of CB, angle D is the opposite side of EB

$\frac{CB}{EB}$

Therefore, from this it can be deduced that:

$\frac{AB}{BD}=\frac{AC}{ED}$

And also:

$\frac{CB}{ED}=\frac{AB}{BD}$

From the data in the drawing, it seems that angle M is equal to angle B

Also, angle A is an angle shared by both triangles ABC and AMN

That is, triangles ABC and AMN are similar respectively according to the angle-angle theorem.

According to the letters, the sides that are equal to each other are:

$\frac{AB}{AM}=\frac{BC}{MN}=\frac{AC}{AN}$

Now we can calculate the ratio between the sides of the given triangles:

$MN=3,BC=6$$\frac{6}{3}=2$

From the drawing it appears that angle E equals angle A

Since angle D equals 90 degrees, its adjacent angle also equals 90 degrees.

In other words, angle D1 equals angle D2 and both equal 90 degrees.

Since we have two pairs of equal angles, the triangles are similar.

Also angle B equals angle C

Now let's write the similar triangles according to their corresponding angle letters:

$ABC=ECD$

Let's write the ratio of sides according to the corresponding letters of the similar triangles:

$\frac{AB}{EC}=\frac{AD}{ED}=\frac{BD}{CD}$

Given that the data shows that there are two pairs with equal angles:

$B=E=40$

$C=F=60$

The triangles are similar according to the angle-angle theorem, therefore triangle ABC is similar to triangle DEF.

Let's look at the order of letters of the triangles that match each other and see the ratio of the sides.

We will write accordingly:

Triangle ABC is similar to triangle DFE

The order of similarity ratio will be:

$\frac{AB}{DF}=\frac{BC}{FE}=\frac{AC}{DE}$

Now let's insert the existing data we have in the diagram:

$\frac{8y}{9y}=\frac{7m}{FE}$

Let's reduce y and we get:

$\frac{8}{9}FE=7m$

$FE=\frac{9}{8}\times7m$

$FE=7\frac{7}{8}m$