Lesson Explainer: Congruent Polygons | Nagwa Lesson Explainer: Congruent Polygons | Nagwa

Lesson Explainer: Congruent Polygons Mathematics • First Year of Preparatory School

Join Nagwa Classes

Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

In this explainer, we will learn how to identify congruent polygons and use their properties to find a missing side length or angle.

Recall that polygons are two-dimensional shapes with straight sides.

Each point where two sides of a polygon meet is called a vertex (the plural is β€œvertices”).

Recalling also that congruent angles are angles that have the same measure and congruent sides are sides that have the same length, we can give a definition of congruent polygons as follows.

Definition: Congruent Polygons

Two polygons are congruent if there is a correspondence between their vertices such that all corresponding interior angles and all corresponding sides are congruent.

Conversely, if two polygons are congruent, then there is a correspondence between their vertices such that all corresponding interior angles and all corresponding sides are congruent.

In other words, congruent polygons are polygons whose vertices and sides coincide exactly. Another way to think of this concept is that congruent polygons are polygons with the same shape and size, although they can be rotations, translations, or mirror images of each other.

In order to prove that two polygons are congruent, we need to show that

  1. all corresponding sides are congruent, which means they have the same length;
  2. all corresponding interior angles are congruent, which means they have the same measure.

On the other hand, if we are told that two polygons are congruent, this immediately implies that conditions (i) and (ii) must hold.

Note that the polygons with the smallest number of sides (three) are triangles. There are special rules for proving the congruency of triangles, and these are covered in another lesson. Here, we start with an example about a type of four-sided polygon: the square.

Example 1: Congruence of Squares

Are two squares congruent if the side length of one square is equal to the side length of the other?

Answer

All squares have four vertices, so we can always form a correspondence between the vertices of one and the vertices of another. To prove that two squares with the same side length are congruent, we need to show that all corresponding interior angles are congruent and all corresponding sides are congruent.

We know that all squares have four equal interior angles of 90∘ (i.e., four right angles).

This means that for every interior angle in one square, any corresponding interior angle in the other square must measure the same. Therefore, all corresponding interior angles are congruent.

Now, checking the sides, we are told that the two squares have the same side length, which we can label as 𝑙. Since both squares have four sides of length 𝑙, then for every side in one square, any corresponding side in the other square must have the same length.

Therefore, all corresponding sides are congruent.

Since we have shown that all corresponding interior angles and all corresponding sides are congruent, this implies that the two squares themselves are congruent.

We conclude that the answer to the question is yes, two squares are congruent if the side length of one is equal to the side length of the other.

Before moving on to more complicated problems, we introduce a helpful piece of mathematical notation.

For objects 𝑆 and 𝑇, we write 𝑆≅𝑇 to mean that 𝑆 and 𝑇 are congruent.

For polygons 𝐴𝐡𝐢 and π‘‹π‘Œπ‘, the notation π΄π΅πΆβ‰…π‘‹π‘Œπ‘ implies that the interior angle at vertex 𝐴 is congruent to the one at vertex 𝑋, the interior angle at vertex 𝐡 is congruent to the one at vertex π‘Œ, and the interior angle at vertex 𝐢 is congruent to the one at vertex 𝑍. Furthermore, side 𝐴𝐡 is congruent to side π‘‹π‘Œ, side 𝐡𝐢 is congruent to side π‘Œπ‘, and side 𝐢𝐴 is congruent to side 𝑍𝑋.

The same labeling convention applies to all congruent polygons, irrespective of their number of sides. For example, for two four-sided polygons, we would use the notation π΄π΅πΆπ·β‰…π‘Šπ‘‹π‘Œπ‘.

By using this notation, we can express detailed information about the properties of congruent polygons in a very concise way. In particular, the order of the vertex letters tells us which interior angle is congruent to which and also which side is congruent to which. Our next example shows how to apply this knowledge.

Example 2: Understanding the Notation for Congruence

The symbol β‰… means that the two objects are congruent. Which statement is true?

  1. △𝐴𝐡𝐢≅△𝐷𝐴𝐢
  2. △𝐡𝐢𝐴≅△𝐷𝐴𝐢
  3. △𝐴𝐡𝐢≅△𝐢𝐴𝐷
  4. △𝐴𝐢𝐡≅△𝐷𝐴𝐢

Answer

The diagram shows a four-sided polygon (or quadrilateral) split into two triangles that share the side 𝐴𝐢. From the wording of the question, we know that one of the four answer options is correct, so we may assume the two triangles are congruent.

Recall that if two polygons are congruent, then there is a correspondence between their vertices such that all the corresponding interior angles and sides are congruent. By comparing the two triangles, we need to work out which interior angle is congruent to which. This will then enable us to use mathematical notation to describe the congruence relationship between the triangles, so that we can pick the correct answer option.

The three different side lengths in each triangle are marked with either a single dash, a double dash, or with no dashes (the shared side). In △𝐴𝐡𝐢, the interior angle at vertex 𝐴 (written ∠𝐴) is between the shared side and the side with a single dash. Similarly, ∠𝐡 is between the sides with single and double dashes and ∠𝐢 is between the side with a double dash and the shared side.

In △𝐴𝐢𝐷, tracking the interior angles in the same order by sides, we see that ∠𝐢 is between the shared side and the side with a single dash, ∠𝐷 is between the sides with single and double dashes, and ∠𝐴 is between the side with a double dash and the shared side.

These correspondences tell us that βˆ π΄β–³π΄π΅πΆβ‰…βˆ πΆβ–³π΄πΆπ·,βˆ π΅β–³π΄π΅πΆβ‰…βˆ π·β–³π΄πΆπ·,βˆ πΆβ–³π΄π΅πΆβ‰…βˆ π΄β–³π΄πΆπ·,ofofofofofof as shown in the diagram below.

We can express this congruence relationship by the notation △𝐴𝐡𝐢≅△𝐢𝐷𝐴, but this is not one of the four available answer options.

Consequently, it is important to remember that there is more than one way to describe the same congruence relationship, depending on the vertex we start at and the direction of travel around the polygon. For instance, instead of starting at vertex 𝐴 in △𝐴𝐡𝐢, we could have started at 𝐡 or 𝐢, so the following three statements are equivalent: △𝐴𝐡𝐢≅△𝐢𝐷𝐴,△𝐡𝐢𝐴≅△𝐷𝐴𝐢,△𝐢𝐴𝐡≅△𝐴𝐢𝐷.

Additionally, if we travel around the polygon in the opposite direction, we get three more equivalent statements: △𝐢𝐡𝐴≅△𝐴𝐷𝐢,△𝐡𝐴𝐢≅△𝐷𝐢𝐴,△𝐴𝐢𝐡≅△𝐢𝐴𝐷.

We have now listed all possible congruence relationships between the two triangles. The only one from this list that appears as an answer option is △𝐡𝐢𝐴≅△𝐷𝐴𝐢, so statement B is correct.

In the above example, we knew that the given triangles had a congruence relationship, but in many questions, we will be asked to check whether or not two polygons are congruent.

Example 3: Identifying Congruent Polygons

Are the polygons shown congruent?

Answer

Recall that two polygons are congruent if there is a correspondence between their vertices such that all the corresponding interior angles and sides are congruent. Therefore, if we can show that these conditions are satisfied, then the polygons must be congruent.

From the diagram, the polygons 𝐢𝐷𝐸𝐹 and 𝑀𝑁𝑂𝑃 are both parallelograms, so in theory, we can form a correspondence between their vertices. Starting with vertex 𝐢 of parallelogram 𝐢𝐷𝐸𝐹, it has an interior angle of 76∘. Comparing with parallelogram 𝑀𝑁𝑂𝑃, we see that the only possible corresponding vertices are 𝑀 or 𝑂, so βˆ πΆβ‰…βˆ π‘€βˆ π‘‚.or

Repeating this step for vertices 𝐷, 𝐸, and 𝐹 of 𝐢𝐷𝐸𝐹, we deduce that βˆ π·β‰…βˆ π‘βˆ π‘ƒ,βˆ πΈβ‰…βˆ π‘€βˆ π‘‚,βˆ πΉβ‰…βˆ π‘βˆ π‘ƒ.ororor

Next, we compare side lengths. Starting with side 𝐢𝐷 of parallelogram 𝐢𝐷𝐸𝐹, we see that its length is marked with a single dash. Comparing with parallelogram 𝑀𝑁𝑂𝑃, we see that the only possible corresponding sides are 𝑀𝑁 or 𝑂𝑃, so 𝐢𝐷≅𝑀𝑁𝑂𝑃.or

Repeating this process for the other sides of 𝐢𝐷𝐸𝐹, we get 𝐷𝐸≅𝑁𝑂𝑃𝑀,𝐸𝐹≅𝑀𝑁𝑂𝑃,𝐹𝐢≅𝑁𝑂𝑃𝑀.ororor

This means we have a choice of correspondences, but to prove that the two parallelograms are congruent, it is sufficient to find one set of correspondences that works. Choosing βˆ πΆβ‰…βˆ π‘€ implies that βˆ π·β‰…βˆ π‘, βˆ πΈβ‰…βˆ π‘‚, and βˆ πΉβ‰…βˆ π‘ƒ. Therefore, our answer is yes, the two polygons are congruent, with 𝐢𝐷𝐸𝐹≅𝑀𝑁𝑂𝑃. We can see this more clearly if we rotate the polygon 𝐢𝐷𝐸𝐹, as shown in the diagram below.

Note that if we had chosen βˆ πΆβ‰…βˆ π‘‚ instead, it would follow that βˆ π·β‰…βˆ π‘ƒ, βˆ πΈβ‰…βˆ π‘€, and βˆ πΉβ‰…βˆ π‘. Again, the two polygons would be congruent, but this time with 𝐢𝐷𝐸𝐹≅𝑂𝑃𝑀𝑁. This can be seen by rotating the polygon 𝐢𝐷𝐸𝐹 as below.

Once we have identified two polygons as being congruent, we can sometimes use their properties to find a missing side length or angle in geometric problems. Let’s look at an example of this type.

Example 4: Finding the Measure of an Angle Bounded between Two Congruent Quadrilaterals

Given that π‘‹π‘ŒπΎπ‘€β‰…π΄π΅πΆπ‘€, find the measure of βˆ πΎπ‘€πΆ.

Answer

Recall that congruent polygons are the same shape and size, but they can be rotations, translations, or mirror images of each other. We are told that π‘‹π‘ŒπΎπ‘€β‰…π΄π΅πΆπ‘€, and from the diagram, we see that polygon π‘‹π‘ŒπΎπ‘€ is actually a reflection of polygon 𝐴𝐡𝐢𝑀 in the straight line that passes through 𝑀, perpendicular to the line segment 𝐴𝑋.

As βˆ πΎπ‘€π‘‹=53∘ with βˆ πΆπ‘€π΄β‰…βˆ πΎπ‘€π‘‹, we deduce that βˆ πΆπ‘€π΄=53∘. Therefore, we know two of the three angles at 𝑀 above the line segment 𝐴𝑋. The missing angle is βˆ πΎπ‘€πΆ, which we have been asked to work out. Recalling the fact that angles on a straight line sum to 180∘, we have the equation βˆ πΆπ‘€π΄+βˆ πΎπ‘€π‘‹+βˆ πΎπ‘€πΆ=180.∘

Subtracting βˆ πΆπ‘€π΄ and βˆ πΎπ‘€π‘‹ from both sides gives βˆ πΎπ‘€πΆ=180βˆ’βˆ πΆπ‘€π΄βˆ’βˆ πΎπ‘€π‘‹,∘ and substituting the values βˆ πΆπ‘€π΄=53∘ and βˆ πΎπ‘€π‘‹=53∘, we get βˆ πΎπ‘€πΆ=180βˆ’53βˆ’53=74.∘∘∘∘

Thus, we have calculated that βˆ πΎπ‘€πΆ=74∘.

In our final example, we apply the properties of congruent polygons in a geometric context.

Example 5: Using the Properties of Congruence to Solve a Geometry Problem

The perimeter of the polygon 𝐴𝐡𝐢𝐷𝐸 is 176 cm and 𝐴𝐡𝐢𝐷𝐸≅𝐹𝑀𝐿𝐷𝐸. Given that πΈβˆˆβƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΄πΉ and 𝐷𝐸=48cm, find the perimeter of the figure 𝐴𝐡𝐢𝐷𝐿𝑀𝐹.

Answer

Recall that congruent polygons are the same shape and size, but they can be rotations, translations, or mirror images of each other. The question states that 𝐴𝐡𝐢𝐷𝐸≅𝐹𝑀𝐿𝐷𝐸, and from the diagram, we see that polygon 𝐴𝐡𝐢𝐷𝐸 is a reflection of polygon 𝐹𝑀𝐿𝐷𝐸 in the straight line containing the line segment 𝐷𝐸.

We have been asked to find the perimeter of the figure 𝐴𝐡𝐢𝐷𝐿𝑀𝐹, which is the new shape formed from the two original polygons by excluding the shared side 𝐷𝐸, as shown below.

In the question, the notation ⃖⃗𝐴𝐹 tells us that 𝐴𝐹 is a line segment, so the horizontal sides 𝐴𝐸 of polygon 𝐴𝐡𝐢𝐷𝐸 and 𝐸𝐹 of polygon 𝐹𝑀𝐿𝐷𝐸 join to make the single horizontal side 𝐴𝐹 of figure 𝐴𝐡𝐢𝐷𝐿𝑀𝐹.

To calculate the perimeter of 𝐴𝐡𝐢𝐷𝐿𝑀𝐹, we need to add together the perimeters of 𝐴𝐡𝐢𝐷𝐸 and 𝐹𝑀𝐿𝐷𝐸, but in both cases, we must exclude the length of the side 𝐷𝐸 (we write this length as just 𝐷𝐸). Then, writing 𝑝(𝐴𝐡𝐢𝐷𝐿𝑀𝐹) for the perimeter of 𝐴𝐡𝐢𝐷𝐿𝑀𝐹 and so on, we have 𝑝(𝐴𝐡𝐢𝐷𝐿𝑀𝐹)=(𝑝(𝐴𝐡𝐢𝐷𝐸)βˆ’π·πΈ)+(𝑝(𝐹𝑀𝐿𝐷𝐸)βˆ’π·πΈ)=𝑝(𝐴𝐡𝐢𝐷𝐸)+𝑝(𝐹𝑀𝐿𝐷𝐸)βˆ’2×𝐷𝐸.

From the question, the perimeter of the polygon 𝐴𝐡𝐢𝐷𝐸 is 176 cm, with 𝐴𝐡𝐢𝐷𝐸≅𝐹𝑀𝐿𝐷𝐸. This implies that the perimeter of the polygon 𝐹𝑀𝐿𝐷𝐸 is also 176 cm, so substituting these values into the above equation, we get 𝑝(𝐴𝐡𝐢𝐷𝐿𝑀𝐹)=176+176βˆ’2×𝐷𝐸=352βˆ’2×𝐷𝐸.

Finally, we know that 𝐷𝐸=48cm, so substituting this value gives 𝑝(𝐴𝐡𝐢𝐷𝐿𝑀𝐹)=352βˆ’2Γ—48=352βˆ’96=256.

All the lengths were given in centimeters, so the perimeter of the figure 𝐴𝐡𝐢𝐷𝐿𝑀𝐹 is 256 cm.

Let us finish by recapping some key concepts from this explainer.

Key Points

  • Two polygons are congruent if there is a correspondence between their vertices such that all corresponding interior angles and all corresponding sides are congruent. Conversely, if two polygons are congruent, then there is a correspondence between their vertices such that all corresponding interior angles and all corresponding sides are congruent.
  • Congruent polygons are polygons with the same shape and size, but they can be rotations, translations, or mirror images of each other.
  • For objects 𝑆 and 𝑇, we write 𝑆≅𝑇 to mean that 𝑆 and 𝑇 are congruent. For polygons 𝐴𝐡𝐢 and π‘‹π‘Œπ‘, the notation π΄π΅πΆβ‰…π‘‹π‘Œπ‘ implies that the interior angle at vertex 𝐴 is congruent to the one at vertex 𝑋, the interior angle at vertex 𝐡 is congruent to the one at vertex π‘Œ, and the interior angle at vertex 𝐢 is congruent to the one at vertex𝑍. Furthermore, side 𝐴𝐡 is congruent to side π‘‹π‘Œ, side 𝐡𝐢 is congruent to side π‘Œπ‘, and side 𝐢𝐴 is congruent to side 𝑍𝑋.
  • The same labeling convention applies to all congruent polygons, irrespective of their number of sides. For example, for two four-sided polygons, we would use the notation π΄π΅πΆπ·β‰…π‘Šπ‘‹π‘Œπ‘.
  • We can use the properties of congruent polygons to work out a missing side length or angle in geometric problems.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy