The Ratio of Similarity - Examples, Exercises and Solutions

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.

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:

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

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

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

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):

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

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:

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

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.

We calculate the perimeter of the smaller triangle (top):

$3.5+1.5+4=9$

Due to their similarity, the ratio between the sides of the triangles is equal to the ratio between the perimeters of the triangles.

We will identify the perimeter of the large triangle using $x$:

$\frac{x}{9}=\frac{14}{3.5}$

$3.5x=14\times9$

$3.5x=126$

$x=36$

We square it. 7 squared is equal to 49.

The perimeter of the left triangle: 13+12+5=25+5=30

Therefore, the perimeter of the right triangle divided by 30 is equal to 5.2 divided by 13:

$\frac{x}{30}=\frac{5.2}{13}$

$13x=156$

$x=12$

Let's call the small triangle A and the large triangle B, let's write the ratio:

$\frac{A}{B}=\frac{3}{4}$

Square it:

$\frac{S_A}{S_B}=(\frac{3}{4})^2$

$\frac{S_A}{S_B}=\frac{9}{16}$

Therefore, the ratio is 9:16

The ratio of similarity is 1:4

The length of the corresponding side in the small triangle is:

$\frac{42}{4}=6$

We multiply the ratio by 2

$9:8=18:16$

Raised to the power of 2:

$9^2:8^2=81:64$

To begin with we can determine the perimeter of the second triangle by using the equation below.

$\frac{P_2}{P_1}=\sqrt{\frac{S_2}{S_1}}$

We insert the existing data

$\frac{P_2}{38}=\sqrt{\frac{81}{361}}$

$\frac{P_2}{38}=\frac{\sqrt{81}}{\sqrt{361}}=\frac{9}{19}$

Lastly we multiply by 38 to obtain the following answer:

$P_2=\frac{9}{19}\times38=18$

Let's look at triangle FCE and calculate side FC using the Pythagorean theorem:

$EC^2+FC^2=EF^2$

Let's substitute the known values into the formula:

$8^2+FC^2=10^2$

$64+FC^2=100$

$FC^2=100-64$

$FC^2=36$

Let's take the square root:

$FC=6$

Since we know that the triangles overlap:

$\frac{AD}{FC}=\frac{DE}{CE}=\frac{AE}{FE}$

Let's substitute the known values into the formula:

$\frac{AD}{6}=\frac{16}{8}$

$AD=2\times6=12$

Let's calculate side CD:

$16+8=24$

Since in a rectangle each pair of opposite sides are equal, we can calculate the perimeter of rectangle ABCD

$12+24+12+24=24+48=72$

First, we must remember the formula for the area of a parallelogram:$\text{Lado }x\text{ Altura}$.

In this case, we will try to find the height CH and the side BC.

We start from the side

First, let's observe the small triangle EBG,

As it is a right triangle, we can use the Pythagorean theorem (

$A^2+B^2=C^2$)

$BG^2+4^2=5^2$

$BG^2+16=25$

$BG^2=9$

$BG=3$

Now, let's start looking for GC.

First, remember that the deltoid has two pairs of equal adjacent sides, therefore:$FC=EC=9$

Now we can also do Pythagoras in the triangle GCE.

$GC^2+4^2=9^2$

$GC^2+16=81$

$GC^2=65$

$GC=\sqrt{65}$

Now we can calculate the side BC:

$BC=BG+GT=3+\sqrt{65}\approx11$

Now, let's observe the triangle BGE and DHC

Angle BGE = 90°
Angle CHD = 90°
Angle CDH=EBG because these are opposite parallel angles.

Therefore, there is a ratio of similarity between the two triangles, so:

$\frac{HD}{BG}=\frac{HC}{GE}$

$\frac{HD}{BG}=\frac{7.5}{3}=2.5$

$\frac{HC}{EG}=\frac{HC}{4}=2.5$

$HC=10$

Now that there is a height and a side, all that remains is to calculate.

$10\times11\approx110$