The Ratio of Similarity - Examples, Exercises and Solutions

What is the ratio between the sides of the triangles ΔABC and ΔMNA?

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$

$\frac{BC}{MN}=2$

Is the similarity ratio between the three triangles equal to one?

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.

No

According to which theorem are the triangles similar?

What is their ratio of similarity?

According to the given data, we will write the side ratios as follows:

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

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

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

From this, it follows that the ratio is according to S.S.S (Side-Side-Side):

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

S.S.S., $\frac{1}{2}$

$ΔACB∼ΔBED$

Choose the correct answer.

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:

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

Answers a + b are correct.

In the image there are a pair of similar triangles and a triangle that is not similar to the others.

Determine which are similar and calculate their similarity ratio.

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.

$a$ and $b$, similarity ratio of $3$

The triangles above are similar.

Calculate the perimeter of the larger triangle.

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 $x$:

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

$3.5x=14\times9$

$3.5x=126$

$x=36$

36

The similarity ratio between two similar triangles is 7, so that the area ratio is $_{——}$

We square it. 7 squared is equal to 49.

49

The triangle above are similar.

What is the perimeter of the blue triangle?

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:

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

$13x=156$

$x=12$

12

Here are two similar triangles. The ratio of the lengths of the sides of the triangle is 3:4, what is the ratio of the areas of the triangles?

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

9:16

If the ratio of the areas of similar triangles is 1:16, and the length of the side of the larger triangle is 42 cm, what is the length of the corresponding side in the smaller triangle?

The ratio of similarity is 1:4

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

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

10.5

The triangle ABC is similar to the triangle DEF.

The ratio between the lengths of their sides is 9:8.

What is the ratio between the areas of the triangles?

We multiply the ratio by 2

$9:8=18:16$

Raised to the power of 2:

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

81:64

In similar triangles, the area of the triangles is 361 cm² and 81 cm². If it is known that the perimeter of the first triangle is 38, what is the perimeter of the second triangle?

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:

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

18

ABCD is a parallelogram
BFCE is a deltoid

What is the area of the parallelogram ABCD?

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.

$10\times11\approx110$

$\approx110$

BC is parallel to DE.

Fill in the gap:

$\frac{AD}{}=\frac{AE}{AC}$

AB

Triangle DFE is similar to triangle ABC.

Calculate the length of FE.

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