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Lesson Explainer: Similarity of Polygons Mathematics

In this explainer, we will learn how to identify and prove the similarity of polygons, write the order of the corresponding vertices, and use the similarity to solve problems.

We can begin by recalling that polygons are two-dimensional shapes with straight sides. For example, squares, rectangles, triangles, hexagons, and octagons are all polygons. Polygons that have exactly the same shape and size are congruent, whereas similar polygons have the same shape and may have a different size.

We can define similar polygons more formally below.

Definition: Similar Polygons

Two polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion.

Let’s consider the two quadrilaterals 𝐴𝐡𝐢𝐷 and 𝑃𝑄𝑅𝑆 below.

If we are given that π΄π΅πΆπ·βˆΌπ‘ƒπ‘„π‘…π‘† (𝐴𝐡𝐢𝐷 is similar to 𝑃𝑄𝑅𝑆), we have π‘šβˆ π΄=π‘šβˆ π‘ƒ,π‘šβˆ π΅=π‘šβˆ π‘„,π‘šβˆ πΆ=π‘šβˆ π‘…,π‘šβˆ π·=π‘šβˆ π‘†.and

We can also observe the corresponding sides.

These are 𝐴𝐡 and 𝑃𝑄, 𝐡𝐢 and 𝑄𝑅, 𝐢𝐷 and 𝑅𝑆, and 𝐷𝐴 and 𝑆𝑃.

Since corresponding sides are in the same proportion, we can write 𝐴𝐡𝑃𝑄=𝐡𝐢𝑄𝑅=𝐢𝐷𝑅𝑆=𝐷𝐴𝑆𝑃.

The proportional relationship can also be given with all the numerators and denominators swapped in the entire statement; that is, 𝑃𝑄𝐴𝐡=𝑄𝑅𝐡𝐢=𝑅𝑆𝐢𝐷=𝑆𝑃𝐷𝐴.

We should use the similarity statement to identify corresponding vertices, rather than solely using any given diagrams. For example, if we have two triangles such that β–³πΈπΉπΊβˆΌβ–³π‘‹π‘Œπ‘, then π‘šβˆ πΈ=π‘šβˆ π‘‹, π‘šβˆ πΉ=π‘šβˆ π‘Œ, and π‘šβˆ πΊ=π‘šβˆ π‘. We could also note that side 𝐹𝐺 would be corresponding to π‘Œπ‘.

In the first example, we will use corresponding sides and angles to identify whether two polygons are similar.

Example 1: Verifying Whether Two Given Polygons Are Similar

Are the two polygons similar?


We recall that two polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion.

Inspecting the angles in the figure, we have two pairs of congruent angles: π‘šβˆ π΄π΅πΆ=π‘šβˆ πΉπΊπ»(=103),π‘šβˆ π΅πΆπ·=π‘šβˆ πΊπ»πΈ(=84).∘∘

We can calculate π‘šβˆ π·π΄π΅ in quadrilateral 𝐴𝐡𝐢𝐷 using the property that the internal angles in a quadrilateral sum to 360∘. Hence, we have π‘šβˆ π·π΄π΅=360βˆ’(103+84+95)=360βˆ’282=78.∘∘∘∘∘∘∘

We can use the same property of the angle measures in quadrilaterals to calculate π‘šβˆ πΉπΈπ». We have π‘šβˆ πΉπΈπ»=360βˆ’(78+103+84)=360βˆ’265=95.∘∘∘∘∘∘∘

Therefore, we have 4 pairs of corresponding angles that are congruent.

We now check whether we have a proportional relationship between the lengths of corresponding sides; that is, we check whether 𝐴𝐡𝐹𝐺=𝐡𝐢𝐺𝐻=𝐢𝐷𝐻𝐸=𝐷𝐴𝐸𝐹.

We could also write the proportionality as 𝐹𝐺𝐴𝐡=𝐺𝐻𝐡𝐢=𝐻𝐸𝐢𝐷=𝐸𝐹𝐷𝐴.

Substituting the given measurements, we have 𝐴𝐡𝐹𝐺=2014=107,𝐡𝐢𝐺𝐻=1611,𝐢𝐷𝐻𝐸=2014=107,𝐷𝐴𝐸𝐹=18.613=9365.

Although we have two pairs of side lengths in the same proportion, we do not have all four pairs of sides in the same proportion. Hence, we can give the answer: no, the two polygons are not similar.

Similar polygons can also be considered as a dilation of each other. If the scale factor is 1, then the polygons are congruent. We can use the scale factor of this dilation to work out the measure of unknown sides. This scale factor may also be referred to as the ratio of enlargement. This may be particularly useful when the ratio of sides is clearer, or more intuitive.

Look at the figure below, where β–³π½πΎπΏβˆΌβ–³π‘€π‘π‘‚.

To find the length of 𝐽𝐿, we could observe that the lengths of 𝑀𝑁𝑂 must be double the lengths of 𝐽𝐾𝐿. This is because we can write that 𝑀𝑁𝐽𝐾=84=2.

The scale factor from △𝐽𝐾𝐿 to △𝑀𝑁𝑂 is 2. The scale factor in the opposite direction is found by dividing by 2. But, we must express scale factors in terms of a multiplier, and dividing by 2 is equivalent to multiplying by 12.

Therefore, to find the length of 𝐽𝐿, we multiply the corresponding length in 𝑀𝑁𝑂 by 12. This gives us 𝐽𝐿=11Γ—

This is an equivalent approach to finding the length of 𝐽𝐿 by writing a single equation involving the 2 pairs of sides: 𝐽𝐿𝑀𝑂=𝐽𝐾𝑀𝑁𝐽𝐿11=48𝐽𝐿11=12𝐽𝐿=11Γ—

We will now check whether another pair of polygons are similar.

Example 2: Finding the Scale Factor between Two Similar Polygons

Are these two polygons similar? If yes, find the scale factor from π‘‹π‘Œπ‘πΏ to 𝐴𝐡𝐢𝐷.


Two polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion.

We can see from the diagram that there are pairs of angles that are marked as having equal measures: π‘šβˆ π‘‹=π‘šβˆ π΄,π‘šβˆ πΏ=π‘šβˆ π·,π‘šβˆ π‘=π‘šβˆ πΆ.

We recall that the angles in a quadrilateral sum to 360∘; hence, we can write the measures of βˆ π‘Œ and ∠𝐡 as π‘šβˆ π‘Œ=360βˆ’(π‘šβˆ π‘+π‘šβˆ πΏ+π‘šβˆ π‘‹),π‘šβˆ π΅=360βˆ’(π‘šβˆ πΆ+π‘šβˆ π·+π‘šβˆ π΄).∘∘

Using the congruent angles above, we can also write π‘šβˆ π‘Œ as π‘šβˆ π‘Œ=360βˆ’(π‘šβˆ πΆ+π‘šβˆ π·+π‘šβˆ π΄).∘

Thus, π‘šβˆ π‘Œ=π‘šβˆ π΅.

Therefore, we have found that there are four congruent angles. This alone is insufficient to prove similarity, so, we must also determine whether the corresponding sides of the polygon are in proportion.

The sides that are corresponding are 𝑍𝐿 and 𝐢𝐷, 𝐿𝑋 and 𝐷𝐴, π‘‹π‘Œ and 𝐴𝐡, and π‘Œπ‘ and 𝐡𝐢. The sides are in proportion if 𝑍𝐿𝐢𝐷=𝐿𝑋𝐷𝐴=π‘‹π‘Œπ΄π΅=π‘Œπ‘π΅πΆ.

Considering each ratio in turn, we have 𝑍𝐿𝐢𝐷=3.22.56=54,𝐿𝑋𝐷𝐴=3.42.72=54,π‘‹π‘Œπ΄π΅=4.83.84=54,π‘Œπ‘π΅πΆ=3.22.56=54.

As each of these proportions can be simplified to 54, the proportions are equal. Since the angles are also congruent, we have proved that the polygons are similar.

Although it is not required here to write a similarity statement between the polygons, we do need to take into account the letter ordering if doing so. If we write polygon π‘‹π‘Œπ‘πΏ with the letters in that order, then we must place the congruent vertices in the similar polygon in the same order. Hence, π‘‹π‘Œπ‘πΏβˆΌπ΄π΅πΆπ·. Alternatively, π‘πΏπ‘‹π‘ŒβˆΌπΆπ·π΄π΅ would be another valid statement.

In order to find the scale factor from π‘‹π‘Œπ‘πΏ to 𝐴𝐡𝐢𝐷, we can take any pair of corresponding sides and divide the length of the side in 𝐴𝐡𝐢𝐷 by the corresponding side length in π‘‹π‘Œπ‘πΏ.

We have 𝐢𝐷𝑍𝐿=2.563.2=0.8.

Alternatively, we have already calculated that the proportion of the corresponding sides is 54. We must note, however, that this proportion was calculated by dividing a side in π‘‹π‘Œπ‘πΏ by a corresponding side in 𝐴𝐡𝐢𝐷. This would give us the scale factor from 𝐴𝐡𝐢𝐷 to π‘‹π‘Œπ‘πΏ, and we need the scale factor in the reverse direction, from π‘‹π‘Œπ‘πΏ to 𝐴𝐡𝐢𝐷. To do this, we instead multiply by the reciprocal of 54, which is 45. This is equivalent to the decimal 0.8.

We can give the answer: yes, the polygons are similar, and the scale factor from π‘‹π‘Œπ‘πΏ to 𝐴𝐡𝐢𝐷 is 0.8.

It is worth noting some facts about similarity in regular polygons. Recall that a regular polygon is a polygon where all angles are congruent and all sides are congruent. Regular polygons include equilateral triangles, squares, regular pentagons, regular hexagons, and so on.

Considering two squares of different side lengths, since within each square all angles are congruent, then each of these angles are also congruent to their corresponding angles in the other square. Furthermore, the proportion of corresponding side lengths will be the same for each side length. Therefore, we can say that any regular 𝑛-sided polygon is similar to another regular 𝑛-sided polygon, where the values of 𝑛 are the same. That is, all equilateral triangles are similar, all squares are similar, and so on.

Of course, since in similar polygons corresponding angles must be congruent and corresponding sides must be in the same proportion, we cannot say that squares are similar to rectangles (as the corresponding side lengths will not be in the same proportion), nor can we say that rhombuses are similar to squares (as corresponding angles are not congruent).

In the next example, we will see how we can use the information that polygons are similar to determine an unknown side length.

Example 3: Finding the Length of a Side in a Quadrilateral given the Corresponding Sides in a Similar Quadrilateral and Their Lengths

Given that 𝐴𝐡𝐢𝐷∼𝐸𝐹𝐺𝐻, determine the length of 𝐺𝐻.


We are given that the two polygons 𝐴𝐡𝐢𝐷 and 𝐸𝐹𝐺𝐻 are similar. This means that their corresponding angles are congruent, and their corresponding sides are in proportion. We can use the proportionality of the sides to help us find the unknown side length, 𝐺𝐻.

In the figure, we are given the lengths of the two sides, 𝐡𝐢 and 𝐹𝐺. We can identify from the similarity statement that these two sides are corresponding, as 𝐴𝐡𝐢𝐷∼𝐸𝐹𝐺𝐻. We can then use the side 𝐢𝐷 to help us work out the corresponding length of 𝐺𝐻. We can write that 𝐹𝐺𝐡𝐢=𝐺𝐻𝐢𝐷.

Substituting the given lengths, we have 9632=𝐺𝐻353=𝐺𝐻353Γ—35=𝐺𝐻𝐺𝐻

We have determined that the length of 𝐺𝐻 is 105 in.

It is a common error to confuse similarity and congruence. Congruent polygons have equal corresponding pairs of angles and equal corresponding sides. Two errors commonly seen when dealing with similarity are either mistakenly writing that corresponding sides are equal, rather than in proportion, or writing that corresponding sides are in proportion and corresponding angles are in proportion. If we have two similar polygons, for example, triangles, the angle measures in both triangles must still sum to 180∘, regardless of the difference in their sizes.

In the next example, we will see how we use a similarity relationship to determine an unknown side length and an unknown angle measure. We will consider how to find the side length both by using the proportionality relationship between two corresponding pairs of sides and by calculating the scale factor.

Example 4: Finding the Side Length and Angle Measure in Similar Quadrilaterals

Given that π΄π΅πΆπ·βˆΌπ‘π‘Œπ‘‹πΏ, find π‘šβˆ π‘‹πΏπ‘ and the length of 𝐢𝐷.


We are given the information that the two polygons are similar. Their corresponding angles are congruent and their corresponding sides are in proportion.

To find π‘šβˆ π‘‹πΏπ‘, we note that we do not have enough information about the angles in polygon π‘π‘Œπ‘‹πΏ to work out π‘šβˆ π‘‹πΏπ‘. However, because π΄π΅πΆπ·βˆΌπ‘π‘Œπ‘‹πΏ, we know that the given angle measure of ∠𝐡𝐢𝐷 is corresponding to βˆ π‘Œπ‘‹πΏ. It must also be 85∘.

Using the property that the sum of the internal angle measures in a quadrilateral is 360∘, we have π‘šβˆ π‘‹πΏπ‘=360βˆ’(105+109+85)=360βˆ’299=61.∘∘∘∘∘∘∘

Next, the length of 𝐢𝐷 can be determined by using the corresponding side, 𝑋𝐿, in π‘π‘Œπ‘‹πΏ. The proportion of these sides will be the same proportion as that between all other pairs of corresponding sides in the polygons. We are given the lengths of another pair of corresponding sides, 𝐴𝐡 and π‘π‘Œ.

Therefore, π΄π΅π‘π‘Œ=𝐷𝐢𝐿𝑋.

Substituting the lengths, we have 75150=𝐷𝐢246.2246.2Γ—75150=𝐷𝐢246.2Γ—12=𝐷𝐢𝐷𝐢

Alternatively, we could have calculated the length of 𝐢𝐷 by finding the scale factor from π‘π‘Œπ‘‹πΏ to 𝐴𝐡𝐢𝐷. In order to find the scale factor, we use a known pair of side lengths. Hence, scalefactor=π΄π΅π‘π‘Œ=75150=12.

Using this scale factor, the length of 𝑋𝐿 must be multiplied by 12 to give the length of 𝐢𝐷. This is given by 𝐢𝐷=12Γ—

Thus, using either method to find the side length, we have determined that π‘šβˆ π‘‹πΏπ‘=61∘ and the length of 𝐢𝐷 is 123.1 cm.

We will now see how we can solve a problem involving similar polygons and a perimeter.

Example 5: Finding the Side Lengths of a Polygon given Its Perimeter and the Side Lengths of a Similar Polygon

A polygon has sides of lengths 2 cm, 4 cm, 3 cm, 8 cm, and 4 cm. A second similar polygon has a perimeter of 31.5 cm. What are the lengths of its sides?


We recall that similar polygons have corresponding angles that are congruent and corresponding sides in proportion.

We are given that the side lengths of one polygon, a pentagon, are 2 cm, 4 cm, 3 cm, 8 cm, and 4 cm. We are required to determine the side lengths of a similar polygon using only the information about its perimeter, which is the distance around the edge of the polygon. As the sides of similar polygons are in proportion, then the perimeter, which is also a measure of length, will be in the same proportion.

We calculate the perimeter of the first polygon as follows: perimetercm=2+4+3+8+4=21.

The scale factor from the first polygon to the second polygon can be found by scalefactorperimeterofsecondpolygonperimeteroffirstpolygon==31.521=32.

In order to find the sides in the second polygon, we multiply each corresponding side length in the first polygon by the scale factor of 32. Hence, we have 2Γ—32=3,4Γ—32=6,3Γ—32=4.5,8Γ—32=12,4Γ—32=6.

The sides in the second polygon can be given as 3 cm, 6 cm, 4.5 cm, 12 cm, and 6 cm.

We can now summarize the key points below.

Key Points

  • Two polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion.
  • We can use the similarity statement to identify corresponding sides and angles, and we must ensure that the letter ordering is correct when writing a similarity relationship between polygons.
  • We can calculate an unknown side by writing the proportional relationship between the side and its corresponding side, along with the proportion between another pair of corresponding sides, or by first calculating the dilation scale factor.
  • The scale factor between the perimeters of two similar polygons is the same as that between corresponding side lengths.
  • All regular polygons are similar to other regular polygons with the same number of sides.

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