What word describes polygons that have the same shape and proportional corresponding angles?

In this explainer, we will learn how to use the properties of similar polygons to solve algebraic expressions and equations.

We begin by recapping what it means for two polygons to be similar.

Two polygons are similar if they have the same number of sides, their corresponding angles are congruent, and the lengths of their corresponding sides are proportional.

Consider two similar polygons, 𝐴𝐵𝐶𝐷 and 𝐸𝐹𝐺𝐻, as shown below. The similarity of the two polygons can be written as 𝐴𝐵𝐶𝐷∼𝐸𝐹𝐺𝐻. The ordering of the letters is important and indicates which vertices of the two polygons correspond to one another. In this example, vertex 𝐴 corresponds to vertex 𝐸, vertex 𝐵 corresponds to vertex 𝐹, and so on.

For two similar polygons, the ratio of each pair of corresponding side lengths is the same. This is known as the similarity ratio. For the similar polygons 𝐴𝐵𝐶𝐷 and 𝐸𝐹𝐺𝐻, the similarity ratio is equal to each of the four ratios below: 𝐴𝐵𝐸𝐹=𝐵𝐶𝐹𝐺=𝐶𝐷𝐺𝐻=𝐷𝐴𝐻𝐸.

We complete our similarity statement for these two polygons by listing the pairs of congruent corresponding angles: ∠𝐴=∠𝐸,∠𝐵=∠𝐹,∠𝐶=∠𝐺,∠𝐷=∠𝐻.

When calculating a similarity ratio, the direction in which we work is important: if the similarity ratio of polygons 𝐴𝐵𝐶𝐷 to 𝐸𝐹𝐺𝐻 is 𝑘, then the similarity ratio of polygons 𝐸𝐹𝐺𝐻 to 𝐴𝐵𝐶𝐷 is 1𝑘. We must ensure that we always divide the lengths of the same polygon by the corresponding lengths of the other.

In our first example, we recall how to use proportions to calculate an unknown length in a quadrilateral when we can calculate the similarity ratio and the length of the corresponding side is given numerically.

If 𝐴𝐵𝐶𝐷 is similar to 𝑋𝑌𝑍𝐿, what is the length of 𝐴𝐷?

Answer

From the figure, it appears as if the two polygons have been drawn in the same orientation and so, for example, side 𝐵𝐶 on the larger polygon corresponds to side 𝑌𝑍 on the smaller polygon. This is confirmed by the ordering of the letters in the written statement of similarity: if 𝐴𝐵𝐶𝐷 is similar to 𝑋𝑌𝑍𝐿, then vertex 𝐴 corresponds to vertex 𝑋, vertex 𝐵 corresponds to vertex 𝑌, and so on.

From the figure, we identify that we are given a pair of corresponding lengths, the lengths of sides 𝐵𝐶 and 𝑌𝑍. The side whose length we want to calculate, side 𝐴𝐷, corresponds to side 𝑋𝐿 on the smaller polygon. As the two polygons are similar, the lengths of their corresponding sides are proportional and so we can write down the similarity ratio for these two pairs of corresponding sides: 𝐴𝐷𝑋𝐿=𝐵𝐶𝑌𝑍.

By substituting the lengths of 𝑋𝐿, 𝐵𝐶, and 𝑌𝑍, we form an equation: 𝐴𝐷14=1812.

To solve for 𝐴𝐷, we multiply both sides of the equation by 14 and evaluate: 𝐴𝐷=18×1412=21.

The length of 𝐴𝐷 is 21 cm.

Our first example allowed us to recap the basic principles of how to use the proportionality of corresponding sides in similar polygons to calculate an unknown side length. We will now extend these skills to problems in which at least some of the side lengths are expressed algebraically. We will begin each problem in the same way: using the side lengths we are given, either numerically or algebraically, to set up an equation using the proportionality of pairs of corresponding side lengths. We will then solve this equation to find the unknown value or values.

In our first example of this type, we consider a pair of similar right triangles in which the lengths of two sides of one triangle have been given numerically and the lengths of the corresponding two sides of the other triangle have been expressed algebraically.

Given that triangles 𝐴𝐵𝐶 and 𝐴′𝐵′𝐶′ are similar, work out the value of 𝑥.

Answer

We are told that these two triangles are similar and so we begin by identifying pairs of corresponding sides. From the ordering of the letters in the similarity statement, we know that side 𝐴𝐶 corresponds to side 𝐴′𝐶′ and side 𝐵𝐶 corresponds to side 𝐵′𝐶′. We have been given numerical values for the lengths of these two sides in triangle 𝐴′𝐵′𝐶′ and algebraic expressions for the lengths of the corresponding sides in triangle 𝐴𝐵𝐶. As the two triangles are similar, corresponding side lengths are proportional and so, using the similarity ratio, we have 𝐵𝐶𝐵′𝐶′=𝐴𝐶𝐴′𝐶′.

Substituting the values and expressions for these four side lengths gives an equation in 𝑥: 2𝑥+16=𝑥+34.

We solve this equation to determine the value of 𝑥. To eliminate the denominators, we can cross multiply: 4(2𝑥+1)=6(𝑥+3).

Distributing each set of parentheses gives 8𝑥+4=6𝑥+18.

Finally, we solve for 𝑥 by collecting like terms: 8𝑥=6𝑥+142𝑥=14𝑥=7.

It is sensible to check our answers whenever possible. In the previous example, we could use the value of 𝑥 we found to calculate the lengths of sides 𝐴𝐶 and 𝐵𝐶. Substituting 𝑥=7 into the expressions for each of these side lengths gives 𝐴𝐶=𝑥+3=7+3=10,𝐵𝐶=2𝑥+1=(2×7)+1=15.

We can then use these lengths to check that corresponding pairs of sides are indeed proportional: 𝐴𝐶𝐴′𝐶′=104=52,𝐵𝐶𝐵′𝐶′=156=52.

The ratio is the same for both pairs of corresponding sides and so this confirms that our value of 𝑥 is correct.

In each of the two examples we have considered so far, the two similar polygons have been drawn in the same orientation. However, this will not always be the case and great care needs to be taken when approaching any problem to ensure that we check which sides of the two polygons correspond to one another before we begin any calculations. Let us now consider an example in which the two polygons are drawn in different orientations.

Given that the two polygons are similar, find the value of 𝑥.

Answer

We note that the two polygons are clearly drawn in different orientations and so we must first identify which vertices correspond to one another. From the figure, we see that the angle at 𝑊 is congruent to the angle at 𝑆 as both are marked with single arcs. We also observe that the angle at 𝐽 is congruent to the angle at 𝑅 as both are marked with double arcs. We can, therefore, state that 𝑊𝐽𝐶𝑍𝑉∼𝑆𝑅𝑃𝑄𝑇, with the ordering of the letters representing which vertices correspond to one another.

The unknown we want to calculate, 𝑥, forms part of the expressions we have been given for the lengths of sides 𝑊𝐽 and 𝐽𝐶. These sides correspond to sides 𝑆𝑅 and 𝑅𝑃 of the second polygon, the lengths of which are both given. We can, therefore, form an equation using the similarity ratio for these two polygons: 𝑊𝐽𝑆𝑅=𝐽𝐶𝑅𝑃2𝑥+624=7𝑥−728.

We solve this equation to determine the value of 𝑥. First, the left-hand side can be simplified by factoring the numerator and then canceling the common factor of 2 in the numerator and denominator. The right-hand side can be simplified by factoring the numerator and then canceling the common factor of 7 in the numerator and denominator: 2(𝑥+3)24=7(𝑥−1)28𝑥+312=𝑥−14.

As 4 is a factor of 12, we can eliminate both denominators by multiplying both sides of the equation by 12 and then canceling common factors: 12(𝑥+3)12=12(𝑥−1)4𝑥+3=3(𝑥−1).

Distributing the parentheses and collecting like terms gives 𝑥+3=3𝑥−3𝑥+6=3𝑥6=2𝑥3=𝑥.

Using the similarity ratio for these two similar polygons, we find that 𝑥=3.

Some problems may involve more than one unknown variable. In such cases we will need to form and solve more than one equation, but the process will always be the same: we determine the similarity ratio and set up equations using each pair of corresponding side lengths. In our next example, we will use the proportionality of side lengths in a pair of similar polygons to determine two unknowns.

Given that 𝐴𝐵𝐶𝐷∼𝐸𝐹𝐺𝐻, find the values of 𝑥 and 𝑦.

Answer

We note first that the two polygons have been drawn in different orientations. Using the ordering of the letters in the statement of similarity, we can determine which vertices correspond to one another: 𝐴 corresponds to 𝐸, 𝐵 to 𝐹, and so on. We may find it helpful to redraw the second polygon in the same orientation as the first, although this is not essential.

We have two unknowns to calculate, 𝑥 and 𝑦. To begin, we can calculate the similarity ratio using the lengths of the corresponding sides 𝐵𝐶 and 𝐹𝐺, both of which have been given numerically: 𝐵𝐶𝐹𝐺=105=2.

A similarity ratio of 2 means that the sides of polygon 𝐴𝐵𝐶𝐷 are each twice as long as the lengths of the corresponding sides of polygon 𝐸𝐹𝐺𝐻. To determine the value of 𝑥, we consider the similarity ratio between sides 𝐶𝐷 and 𝐺𝐻. We now know that 𝐶𝐷𝐺𝐻=2 or, using the fact that the lengths of polygon 𝐴𝐵𝐶𝐷 are twice as long as the corresponding lengths of polygon 𝐸𝐹𝐺𝐻, we can take a slightly less formal approach and immediately state that 𝐶𝐷=2𝐺𝐻.

Substituting 𝐶𝐷=𝑥 and 𝐺𝐻=8, we obtain 𝑥=2×8=16.

To determine the value of 𝑦, we consider the similarity ratio for sides 𝐸𝐻 and 𝐴𝐷. Using a logical approach, we know that the length of 𝐸𝐻 is half the length of 𝐴𝐷, as 𝐸𝐻 is a side of the smaller polygon. Substituting the expression for the length of 𝐸𝐻 and the value for the length of 𝐴𝐷 gives 𝐸𝐻=12𝐴𝐷2𝑦−14=12×8.

We solve this equation to determine the value of 𝑦: 2𝑦−14=42𝑦=18𝑦=9.

Hence, the solution is 𝑥=16, 𝑦=9.

All of the problems we have considered so far have been related to calculating an unknown when it is used to express the length of a side. In our next example, we consider instead how we can use the properties of similar polygons to calculate the value of an unknown when it forms part of the expression for the measure of an angle. This will require a different property of similar polygons: rather than the proportionality of corresponding side lengths, we will use the congruence of corresponding angles.

Given that 𝐴𝐵𝐶𝐷 is similar to 𝑄𝑆𝑅𝑃, find the values of 𝑥 and 𝑦.

Answer

We recall first that corresponding angles in similar polygons are congruent. We therefore need to determine which angles in the two quadrilaterals are corresponding. Upon closer inspection of the figure, it is apparent that the two quadrilaterals have not been drawn in the same orientation, as the angles at the top of each are not of equal measure.

Given the order of the letters in the similarity statement, we deduce that

• vertex 𝐴 corresponds to vertex 𝑄,
• vertex 𝐵 corresponds to vertex 𝑆,
• vertex 𝐶 corresponds to vertex 𝑅,
• vertex 𝐷 corresponds to vertex 𝑃.

We may find it helpful to use colors to indicate corresponding angles in the two polygons, as shown below.

We can now form some equations by equating the expressions or values of the measures of corresponding angles. Considering angles 𝐷 and 𝑃, we have 3𝑥+65=98.

To solve for 𝑥, we subtract 65 from each side of the equation and then divide by 3: 3𝑥=33𝑥=11.

Next, equating the value for angle 𝐶 with the expression for the measure of angle 𝑅 gives 𝑦+35=84.

Subtracting 35 from each side of the equation gives 𝑦=49.

Thus, using the congruence of corresponding angles in similar polygons, we find that 𝑥=11 and 𝑦=49.

We can check our answer to the previous problem by calculating the measures of each angle and confirming that the angle sum in each quadrilateral is indeed 360∘. Using 𝑥=11, the measure of angle 𝐷 is (3𝑥+65)=((3×11)+65)=98.∘∘∘

The measure of angle 𝐵 is the same as the measure of angle 𝑆, which is 97∘. Summing the four angles in 𝐴𝐵𝐶𝐷 gives 81+97+84+98=360,∘∘∘∘∘ which confirms that our value of 𝑥 is correct.

In 𝑄𝑆𝑅𝑃, the measure of angle 𝑅 is (𝑦+35)=(49+35)=84.∘∘∘

The measure of angle 𝑄 is the same as the measure of angle 𝐴, which is 81∘. Summing the four angles in 𝑄𝑆𝑅𝑃 gives 81+97+84+98=360,∘∘∘∘∘ which confirms that our value of 𝑦 is also correct.

In our final problem, we will consider how we can use the proportionality of side lengths in similar polygons to calculate the perimeter of a triangle in which some of the side lengths are expressed algebraically.

Given that △𝐴𝐵𝐶∼△𝐴′𝐵′𝐶′, determine the perimeter of △𝐴′𝐵′𝐶′.

Answer

To determine the perimeter of triangle 𝐴′𝐵′𝐶′, we first need to calculate the lengths of each of its three sides. We are given the length of side 𝐴′𝐶′, but we do not know either of the others. We do know, however, that the two triangles in the figure are similar and so we can express the similarity ratio between the two triangles using corresponding pairs of sides: 𝐴𝐶𝐴′𝐶′=𝐴𝐵𝐴′𝐵′=𝐵𝐶𝐵′𝐶′.

Using the expressions given for the lengths of sides 𝐴𝐶 and 𝐴′𝐵′ and the numeric values for the lengths of sides 𝐴′𝐶′ and 𝐴𝐵, we can form the equation 𝑥+32=5𝑥.

We solve this equation to determine 𝑥. Multiplying both sides of the equation by 2𝑥 will eliminate both denominators simultaneously: 2𝑥(𝑥+3)2=5(2𝑥)𝑥𝑥(𝑥+3)=10.

Distributing the parentheses and subtracting 10 from each side of the equation gives 𝑥+3𝑥=10𝑥+3𝑥−10=0.

This is a quadratic equation in 𝑥, which can be solved by factoring: (𝑥+5)(𝑥−2)=0𝑥+5=0𝑥−2=0𝑥=−5𝑥=2.oror

As 𝑥 represents the length of a side, its value must be positive, and so the correct value is 𝑥=2.

We now know the lengths of two sides of triangle 𝐴′𝐵′𝐶′ and need to calculate the third.

Using the similarity ratio for sides 𝐴𝐵, 𝐴′𝐵′, 𝐵𝐶, and 𝐵′𝐶′, we have 𝐴𝐵𝐴′𝐵′=𝐵𝐶𝐵′𝐶′.

Substituting 𝐴𝐵=5, 𝐴′𝐵′=2, and 𝐵𝐶=8 gives 52=8𝐵′𝐶′.

We solve it by first multiplying both sides of the equation by 𝐵′𝐶′: 52𝐵′𝐶′=8.

Next, we divide both sides of the equation by 52 and simplify: 𝐵′𝐶′=8÷52=8×25=3.2.

Finally, we calculate the perimeter of triangle 𝐴′𝐵′𝐶′ by summing its three side lengths: perimeterof△𝐴′𝐵′𝐶′=𝐴′𝐵′+𝐵′𝐶′+𝐴′𝐶′=2+3.2+2=7.2.

The perimeter of triangle 𝐴′𝐵′𝐶′ is 7.2 units.

Let us finish by recapping some of the key points from this explainer.

• Two polygons are similar if they have the same number of sides, their corresponding angles are congruent, and the lengths of their corresponding sides are proportional.
• The similarity ratio for a pair of similar polygons is the ratio found by dividing any side length in one polygon by the length of the corresponding side in the other, and it is the same for all pairs of corresponding sides.
• The proportionality of side lengths of similar polygons can be used to find unknown variables when the side lengths have been expressed algebraically. This will require us to form and solve equations using the expressions and values given for each side length.
• As corresponding angles in similar polygons are congruent, unknown values used to express angle measures can be calculated by forming and solving equations in which we first equate the expressions for corresponding angles.
• These skills can be applied to solve problems involving the geometry of similar polygons, such as calculating their perimeters.