Lesson Explainer: Areas of Similar Polygons Mathematics

In this explainer, we will learn how to calculate areas of similar polygons given two corresponding side lengths or the scale factor between them and the area of one of the polygons.

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

Definition: Similar Polygons

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

For example, we could consider the following rectangles.

Rectangle 𝐴𝐡𝐢𝐷 is similar to rectangle 𝑃𝑄𝑅𝑆. Both rectangles have 4 sides, all the corresponding angles are congruent, and we can write that 𝐴𝐡𝑃𝑄=𝐡𝐢𝑄𝑅=𝐢𝐷𝑅𝑆=𝐷𝐴𝑆𝑃.

The scale factor from rectangle 𝐴𝐡𝐢𝐷 to rectangle 𝑃𝑄𝑅𝑆 can be found by dividing any of the side lengths of 𝑃𝑄𝑅𝑆 by the corresponding side length in 𝐴𝐡𝐢𝐷. For instance, scalefactor=𝑃𝑄𝐴𝐡=48=12.

If a scale factor in one direction is given as π‘˜, the scale factor in the opposite direction is 1π‘˜.

We can find the length ratio between two similar polygons by writing the ratio of the length of one side in a polygon, with its corresponding side in the other polygon. In the rectangles above, we could write the length ratio of 𝐴𝐡𝐢𝐷:𝑃𝑄𝑅𝑆 as 8∢4=2∢1. Substituting the widths of these rectangles would give us the equivalent ratio, 3∢1.5=2∢1.

We will now explore how we can use the length ratio of similar polygons to find the area ratio.

We can consider the following triangle, 𝐸𝐷𝐹, with base 𝑏 and height β„Ž. A similar triangle, 𝐸′𝐷′𝐹′, is created by a dilation of scale factor π‘˜. Therefore, the dimensions of 𝐸′𝐷′𝐹′ will be π‘˜π‘ and π‘˜β„Ž.

We might pose the following questions: How does the area of △𝐸′𝐷′𝐹′ differ from the area of △𝐸𝐷𝐹? Is it also π‘˜ times larger?

We remember that the area of a triangle of base 𝑏 and perpendicular height β„Ž is calculated by areaofatriangle=π‘Γ—β„Ž2.

Thus, the area of △𝐸𝐷𝐹 is calculated by areaof𝐸𝐷𝐹=π‘β„Ž2.

For △𝐸′𝐷′𝐹, the area is given by areaof𝐸′𝐷′𝐹′=π‘˜π‘Γ—π‘˜β„Ž2=π‘˜π‘β„Ž2=π‘˜ο€½π‘β„Ž2.

We can compare the areas of both triangles by observing that areaofareaof△𝐸′𝐷′𝐹′=π‘˜ο€Ήβ–³πΈπ·πΉο….

When this polygon, triangle 𝐸𝐷𝐹, was dilated with scale factor π‘˜ to give triangle 𝐸′𝐷′𝐹′, the areas have a scale factor of π‘˜οŠ¨. This result is true for all polygons.

Definition: Areas of Similar Polygons given a Scale Factor

If the length scale factor between two similar polygons is π‘˜, then their area scale factor is π‘˜οŠ¨.

We will now investigate how we can use a length ratio between similar polygons to identify an area ratio.

Consider the following similar parallelograms, 𝑃𝑄𝑅𝑆 and 𝑃′𝑄′𝑅′𝑆′.

The length ratio can be found by writing the ratios of the lengths of any corresponding sides. For example, we can write π‘†π‘…βˆΆπ‘†β€²π‘…β€² as 4∢10. Simplifying this, we have lengthratio=2∢5.

We can now consider the areas of each parallelogram. We recall that the area of a parallelogram is found by multiplying the base length by the perpendicular height.

Therefore, we calculate areaof𝑃𝑄𝑅𝑆=4Γ—3=12.

We can calculate the area of 𝑃′𝑄′𝑅′𝑆′ as areaof𝑃′𝑄′𝑅′𝑆′=10Γ—7.5=75.

We can write the ratio of the areas of π‘ƒπ‘„π‘…π‘†βˆΆπ‘ƒβ€²π‘„β€²π‘…β€²π‘†β€² as arearatio=12∢75=4∢25.

Comparing the length ratio, 2∢5, with the area ratio, 4∢25, we notice that each of the parts of the length ratio is squared to give the corresponding part of the area ratio. That is, lengthratioarearatio=2∢5,=2∢5=4∢25.

This is true for all polygons. We can formalize this below.

Definition: Area Ratio of Similar Polygons

If the length ratio of two similar polygons is given as π‘ŽβˆΆπ‘, then the ratio of their areas is π‘ŽβˆΆπ‘οŠ¨οŠ¨.

We will now see how we can apply this in the following examples.

Example 1: Finding the Area of a Similar Rectangle given Two Corresponding Lengths and a Diagram

Given the following figure, find the area of a similar polygon 𝐴′𝐡′𝐢′𝐷′ in which 𝐴′𝐡′ = 6.

Answer

We can recall that two polygons are similar if they have the same number of sides, their corresponding angles are congruent, and their corresponding sides are in the same proportion. A similar polygon to 𝐴𝐡𝐢𝐷 would also be a rectangle with the sides in proportion.

We can find the length ratio of rectangles π΄π΅πΆπ·βˆΆπ΄β€²π΅β€²πΆβ€²π·β€² by writing the lengths π΄π΅βˆΆπ΄β€²π΅β€². We are given that 𝐴′𝐡′=6, and we can use the figure to establish that 𝐴𝐡=3 units. Therefore, substituting these values into the ratio, we have lengthratio=3∢6=1∢2.

We can use the length ratio to find the ratio of areas between two similar polygons. If the length ratio of two similar polygons is given as π‘ŽβˆΆπ‘, then the ratio of their areas is π‘ŽβˆΆπ‘οŠ¨οŠ¨. Therefore, we have arearatio=1∢2=1∢4.

We can write that areaofareaofπ΄π΅πΆπ·βˆΆπ΄β€²π΅β€²πΆβ€²π·=1∢4.

To find the area of a rectangle, we multiply the length by the width. Using the figure, we observe that the length of 𝐴𝐡𝐢𝐷 is 5 units, and the width is 3 units. Thus, to find the area of 𝐴𝐡𝐢𝐷, we calculate areaofsquareunits𝐴𝐡𝐢𝐷=5Γ—3=15.

If we define the area of 𝐴′𝐡′𝐢′𝐷′ as π‘₯, then we have the ratio of the areas, π΄π΅πΆπ·βˆΆπ΄β€²π΅β€²πΆβ€²π·β€², as 15∢π‘₯=1∢4.

For the ratios to be equivalent, the value of π‘₯ must be 60, since 15∢60=1∢4.

We may also consider that, in words, the area ratio of π΄π΅πΆπ·βˆΆπ΄β€²π΅β€²πΆβ€²π·β€²=1∢4 simply means that the area of 𝐴′𝐡′𝐢′𝐷′ is 4 times larger than the area of 𝐴𝐡𝐢𝐷. As the smaller rectangle has an area of 15 square units, then 4Γ—15=60 square units.

Therefore, the area of polygon 𝐴′𝐡′𝐢′𝐷′ is 60squareunits.

We will now see an example where we write the ratio of the areas of two similar polygons, given the ratio between their lengths.

Example 2: Finding the Ratio of the Areas of Two Similar Shapes given the Ratio of Their Lengths

Rectangle 𝑄𝑅𝑆𝑇 is similar to rectangle 𝐽𝐾𝐿𝑀 with their sides having a ratio of 8∢9. If the dimensions of each rectangle are doubled, find the ratio of the areas of the larger rectangles.

Answer

We can recall that two polygons are similar if their corresponding angles are congruent and their corresponding sides are in the same proportion. We can use the given length ratio of the rectangles to help us calculate the ratio of their areas.

Two similar polygons with corresponding sides in a length ratio of π‘ŽβˆΆπ‘ have an area ratio of π‘ŽβˆΆπ‘οŠ¨οŠ¨.

We are given the length ratio of the rectangles as 8∢9. Therefore, we can calculate the area ratio of π‘„π‘…π‘†π‘‡βˆΆπ½πΎπΏπ‘€ as arearatio=8∢9=64∢81.

We are given that the length of each rectangle is doubled. We could define these new rectangles as 𝑄′𝑅′𝑆′𝑇′ and 𝐽′𝐾′𝐿′𝑀′. Thus, every length of 𝑄′𝑅′𝑆′𝑇′ and 𝐽′𝐾′𝐿′𝑀′ will be double the original lengths. If we imagined that the lengths were a fixed value, for example, if the corresponding lengths were 8 cm and 9 cm, then, when doubled, these lengths would be 16 cm and 18 cm. When written as a ratio, 16∢18, this would simplify to 8∢9.

Hence, if we have a pair of similar polygons with a given length ratio, and if the same scale factor is applied to both rectangles, then the length ratio remains the same. Furthermore, the area ratio will also remain the same. Thus, we can give the answer that the area ratio of these larger, doubled, rectangles is 64∢81.

In the next example, we will use information about the perimeter of a square to help us work out the length and area ratios of two similar shapes.

Example 3: Finding the Area of a Similar Polygon given a Length Scale Factor and Perimeter

Square 𝐴 is an enlargement of square 𝐡 by a scale factor of 23. If the perimeter of square 𝐴 equals 56 cm, what is the area of square 𝐡? Give your answer to the nearest hundredths.

Answer

We are given that square 𝐴 is an enlargement of square 𝐡 by a scale factor of 23. All squares are similar; that is, they have all corresponding angles equal and all pairs of corresponding sides in proportion. As the scale factor is 23, all four sides of 𝐡 will be 23 the length of those of 𝐴.

We are not given the information about the side lengths of 𝐴 or 𝐡, but we can calculate the lengths of square 𝐴 using the information about its perimeter. We recall that the perimeter is the distance around the outside of a shape. If we define the side length of 𝐴 as π‘₯, then as there are 4 sides of equal length, we can write that perimeterof𝐴=4π‘₯.

We substitute the given value of 56 cm for the perimeter to give 56=4π‘₯564=π‘₯π‘₯=14.cm

Now, we have that the side length of square 𝐴 is 14 cm. Next, to find the area, we recall that the area of a square of side length 𝑙 is given by areaofasquare=𝑙. Therefore, to find the area of square 𝐴, we substitute the length, 𝑙=14, which gives areaofcm𝐴=14=196.

Given that the length scale factor from square 𝐡 to square 𝐴 is 23, we can calculate the area scale factor.

We can recall that if the length scale factor between two similar polygons is π‘˜, then their area scale factor is π‘˜οŠ¨. Since the length ratio from 𝐡 to 𝐴 can be given as 23, then their area ratio can be written as areascalefactor=23=49.

We can then write that areaofareaof𝐡×49=𝐴.

Substituting our value for the area of 𝐴, 196 cm2, we have areaofareaofcm𝐡×49=196𝐡=94Γ—196=441.

Hence, the area of square 𝐡 is 441 cm2.

We will now look at another example.

Example 4: Calculating the Perimeter of a Similar Shape given the Two Areas

The areas of two similar polygons are 361 cm2 and 81 cm2. Given that the perimeter of the first is 38 cm, find the perimeter of the second.

Answer

In this question, as the polygons are similar, we know that they have the same number of sides, their corresponding angles are congruent, and their corresponding sides are in the same proportion. We can use the given areas to write an area ratio and then establish the length ratio between the two polygons.

We can write the area ratio of polygonpolygon1∢2 as arearatio=361∢81.

Two similar polygons with corresponding sides in a length ratio of π‘ŽβˆΆπ‘ have an area ratio of π‘ŽβˆΆπ‘οŠ¨οŠ¨.

Therefore, to calculate the length ratio of polygonpolygon1∢2, we take the positive value of the square root of each of the ratio terms. This gives lengthratio=√361∢√81=19∢9.

We can use this length ratio to calculate the perimeter of the second polygon, polygon 2. In this case, it does not matter what specific shape the polygon is, for example, triangle, square, or hexagon. As the perimeter is a measure of length, the length ratio will still apply.

We can define the perimeter of polygon 2 as 𝑝 and compare the length ratio with the ratio of perimeters as follows: lengthratioperimeterratio=19∢9,=38βˆΆπ‘.

Since 38 is double the value of 19, both values in the length ratio must be doubled to give those of the perimeters. Hence, 𝑝=9Γ—2=18.cm

Thus, we can give the answer: the perimeter of the second polygon is 18 cm.

We will now see an example of how we can prove that two rectangles are similar in order to help us solve a real-world problem.

Example 5: Solving a Real-World Problem Involving Area

It cost 3β€Žβ€‰β€Ž799 pounds to fit wooden flooring in a class with dimensions 28 m and 10 m. How much would it cost to fit wooden flooring in a similar room with dimensions 84 m and 30 m.

Answer

In this question, we can assume that since we are given two dimensions, of differing sizes, then both classrooms are rectangular in shape. We can define the first rectangle as 𝐴 and the second as 𝐡. One way to solve this problem is to establish if these rectangular classrooms are, in fact, similar. Two polygons are similar if they have the same number of sides, their corresponding angles are congruent, and their corresponding sides are in the same proportion.

We know that the rectangles will have the same number of sides, and the corresponding angles will all be of equal measure, as they are all 90∘. We need to check if the sides are in proportion. Sketching a diagram can be useful.

We can write the ratio of widths, 𝐴∢𝐡, as 10∢30.

This ratio simplifies to 1∢3.

The ratio of lengths can be written as 28∢84, which simplifies to 1∢3.

In order to check if two polygons have corresponding sides in proportion, we need to check all sides. However, as this is a rectangle, we know that there are two pairs of congruent sides. We have shown, therefore, that both the lengths and widths simplify to the same ratio, 1∢3, and so all sides are in the same proportion. Hence, the two classroom rectangles are similar.

The length ratio of 𝐴∢𝐡 is equal to 1∢3. We can then use the property that similar polygons with corresponding sides in a length ratio of π‘ŽβˆΆπ‘ have an area ratio of π‘ŽβˆΆπ‘οŠ¨οŠ¨. We can calculate the area ratio of 𝐴∢𝐡 as arearatio=1∢3=1∢9.

Alternatively, we could calculate the area ratio by working out the area of each rectangle. The area of a rectangle with length 𝑙 and width 𝑀 is given by areaofarectangle=𝑙×𝑀.

The area of rectangle 𝐴, with length 28 m and width 10 m, is areaofrectanglem𝐴=28Γ—10=280.

The area of rectangle 𝐡 can be calculated as areaofrectanglem𝐡=83Γ—30=2520.

Simplifying the ratio of these areas gives arearatio=280∢2520=1∢9.

Either method produces the same area ratio, 1∢9.

In order to find the cost of the flooring, we must use the area ratio, rather than the length ratio. This is because the cost of the flooring changes directly according to the area of the room, rather than one of its dimensions.

We are given the cost of the flooring for the rectangle with dimensions 28 m and 10 m, rectangle 𝐴. We could define the cost of the flooring for rectangle 𝐡 as 𝑐. We compare the ratios for the area and cost as 1∢9(),3799βˆΆπ‘().arearatiocostratio

Each value in the cost ratio must be 3β€Žβ€‰β€Ž799 times larger than those in the area ratio. Therefore, 𝑐=9Γ—3799=34191.

We can give the answer that the cost to fit flooring in the classroom with dimensions 84 m and 30 m is 34191pounds.

We can now summarize the key points.

Key Points

  • Two polygons are similar if they have the same number of sides, their corresponding angles are congruent, and their corresponding sides are in the same proportion.
  • If the length scale factor between two similar polygons is π‘˜, then their area scale factor is π‘˜οŠ¨.
  • If the length ratio of two similar polygons is given as π‘ŽβˆΆπ‘, then ratio of their areas is π‘ŽβˆΆπ‘οŠ¨οŠ¨.
  • As perimeter is a length, we can also say that the ratio of the areas of two similar polygons is equal to the square of the ratio of their perimeters.

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