Similar Triangles and Polygons - Examples, Exercises and Solutions

To find out if the triangles are similar, we can check if there is an appropriate similarity ratio between their sides.

The similarity ratio is the constant difference between the corresponding sides.

In this case, we can check if:

$\frac{62.5}{50}=\frac{100}{80}=\frac{100}{80}$

$\frac{62.5}{50}=\frac{125}{100}=1\frac{25}{100}=1\frac{1}{4}$

$\frac{100}{80}=\frac{10}{8}=1\frac{2}{4}=1\frac{1}{4}$

 Therefore:$1\frac{1}{4}=1\frac{1}{4}=1\frac{1}{4}$

Therefore, we can say that there is a constant ratio of$1\frac{1}{4}$ between the sides of the triangles and therefore the triangles are similar.

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$

To answer the question, we first need to understand what "similarity ratio" means.

In similar triangles, the ratio between the sides is constant.

In the statement, we do not have data on any of the sides.

However, a similarity ratio of 1 means that the sides are exactly the same size.

That is, the triangles are not only similar but also congruent.

In the drawing, you can clearly see that the triangles are of different sizes and, therefore, clearly the similarity ratio between them is not 1.

The square of the ratio between the sides is equal to the ratio between the areas of the parallelograms:

$3^2:4^2=9:16$

The triangles are similar according to the angle-angle theorem.

Having two pairs of equal angles is sufficient to conclude that the triangles are similar.

Since we are given that line BC is parallel to DE

Angle E equals angle C and angle D equals angle B - corresponding angles between parallel lines are equal.

Now let's observe that angle D is opposite to side AE and angle B is opposite to side AC, meaning:

$\frac{AE}{AC}$

Now let's observe that angle E is opposite to side AD and angle C is opposite to side AB, meaning:

$\frac{AD}{AB}$

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:

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

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

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

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:

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

Using the given data, the side ratios can be written as follows:

$\frac{FD}{AB}=\frac{X}{2X}=\frac{1}{2}$

$\frac{FE}{AC}=\frac{\frac{y}{2}}{y}=\frac{y}{2y}=\frac{1}{2}$

$\frac{DE}{BC}=\frac{2Z}{4Z}=\frac{2}{4}=\frac{1}{2}$

We can therefore deduce that the ratio is compatible with the S.S.S theorem (Side-Side-Side):

$\frac{FD}{AB}=\frac{FE}{AC}=\frac{DE}{BC}=\frac{1}{2}$

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.

If the sum of the angles in a triangle equals 180, then angle F equals 75 and therefore angle C also equals 75.

The triangles are similar according to the angle-angle theorem

Answer D is correct

We will mark the side in square A as X

Therefore the perimeter will be:
$4x=56$

$x=14$

Now we can calculate the area of square A:

$14\times14=196$

As we are given the ratio between the areas:

$\frac{196}{S_2}=(\frac{2}{3})^2=\frac{4}{9}$

That is, the ratio will be:

$\frac{196}{X}=\frac{4}{9}$

The area of the square will be equal to:

$\frac{9\times196}{4}=\frac{1764}{4}=441$

The sides of the triangles are not equal and, therefore, the triangles are not similar.

Triangle a and triangle b are similar according to the S.S.S (side side side) theorem

And the relationship between the sides is identical:

$\frac{GH}{DE}=\frac{HI}{EF}=\frac{GI}{DF}$

$\frac{9}{6}=\frac{3}{1}=\frac{6}{2}=3$

That is, the ratio between them is 1:3.