Lesson Explainer: Areas of Regular Polygons Mathematics

In this explainer, we will learn how to find areas of regular polygons given their side lengths using a formula.

We recall that a regular polygon is a shape made of straight edges where all of the side lengths are equal and the internal angles are also equal. To find a formula to calculate the area of any regular polygon, we first note that we can split any regular 𝑛-sided polygon into 𝑛 congruent triangles by connecting the vertices to the center. For example, in the following diagram, we connect the center of a regular pentagon of side length 𝑥 to each of its vertices.

To show that each of these triangles is congruent, we note that the line from the center to each vertex bisects the internal angle of the pentagon, and the center is equidistant from all of the vertices of the regular polygon. Hence, by the SSS criterion, the triangles are all congruent.

Next, since each triangle has two equal angles, they are isosceles triangles, and given that we have congruent triangles, the final angles in each of the five shapes (at the center of the pentagon) must all be equal to each other. Finally, we know that each triangle has a corresponding side of length 𝑥 as all of the triangles are congruent.

The area of this regular pentagon is 5 times the area of one of the triangles. To find the area of one of the triangles, we recall that the area of a triangle with a base of length 𝑏 and perpedincular height ℎ is given by area=12𝑏ℎ.

We can use the side with length 𝑥 as the base, which means we then need to find the height of the triangle to determine its area.

To find the value of ℎ, we will find the angle at the center. The angles in the center are all corresponding angles in congruent triangles, and so they are equal. They also sum to give 360∘. Hence, in the example of the regular pentagon, the central angle is 3605=72∘∘.

Since this is an isosceles triangle, the angle bisector at the central angle will bisect the base, giving us an angle of 722=36∘∘ and the length 𝑥2, as shown in the diagram.

We can determine the value of ℎ by applying trigonometry to the following right triangle.

We know tan(36)∘ will be equal to the ratio of the opposite side’s length divided by the adjacent side’s length, giving tantan(36)=ℎℎ=𝑥2(36).∘∘

This means our 5 isosceles triangles are as shown.

The area of this triangle is given by areaoftriangletantan=12𝑥𝑥2(36)=𝑥4(36).∘∘

Multiplying this by 5 and simplifying by using the reciprocal trigonometric identity cottan𝜃=1𝜃, we can find the area of the regular pentagon as follows: areaofpentagontancot=5𝑥4(36)=5𝑥(36)4.∘∘

This method can be generalized to any regular 𝑛-sided polygon of side length 𝑥. We would still be able to split the regular polygon into right triangles in this form. There would be 𝑛 isosceles triangles, and the central angle would be 3602𝑛=180𝑛∘∘. We can use this to find the height of the isosceles triangles, ℎ, from the following right triangle.

By applying trigonometry, we have ℎ=𝑥2,tan∘ and then the area of the isosceles triangles is then given by areaoftriangletancot=12𝑥𝑥2=𝑥4.∘∘

Finally, the regular polygon is constructed of 𝑛 of these congruent isosceles triangles, so its area is given by areaofregularpolygoncotcot=𝑛×𝑥4=𝑛𝑥4180𝑛.∘∘

We can summarize this result as follows.

Theorem: Area of a Regular 𝑛-Sided Polygon

The area of a regular 𝑛-sided polygon of side length 𝑥 is given by 𝑛𝑥4180𝑛.∘cot

For example, if 𝑛=3, we have an equilateral triangle, and its area would be given by areaofequilateraltrianglecotcot=(3)𝑥41803=𝑥(60)=3𝑥41√3=√34𝑥.∘∘

We can also find the area of a regular polygon when the angles are measured in radians; in this case, we have that 180∘ is 𝜋 radians, giving us the following.

Theorem: Area of a Regular 𝑛-Sided Polygon

The area of a regular 𝑛-sided polygon of side length 𝑥 is given by 𝑛𝑥4𝜋𝑛.cot

Let’s see an example of applying this formula to find the area of a regular hexagon.

Example 1: Finding the Area of a Regular Hexagon

Find the area of a regular hexagon with a side length of 35 cm giving the answer to two decimal places.

Answer

There are a number of different methods of solving this problem. For example, a regular hexagon can be constructed from 6 equilateral triangles as shown.

We could then find the area of any one these equilateral triangles as 12(35)(3560)=35√34sin∘. Since there are 6 of these triangles, the area of the hexagon is areacm=6×1225√34=3182.643…, which to two decimal places is 3‎ ‎182.64 cm2.

We can also find this area by using the formula for the area of a regular polygon. We recall that the area of a regular 𝑛-sided polygon of side length 𝑥 is given by 𝑛𝑥4180𝑛.∘cot

A hexagon has 6 sides, so we set 𝑛=6 and 𝑥=35cm. This gives us areacotcottan=6(35)41806=36752(30)=36752(30).∘∘∘

We know that tan(30)=√33∘, which gives us 36752(30)=36752=3675√32=3182.643…,tancm∘√ which to two decimal places is 3‎ ‎182.64 cm2.

Hence, the area of a regular hexagon of side length 35 cm is 3‎ ‎182.64 cm2 to two decimal places of accuracy.

In our next example, we are tasked with finding the area of a regular 14-sided polygon. We could solve this with triangles; however, this would be much more cumbersome. Instead, we will just apply the formula to find the area of a regular 𝑛-sided polygon.

Example 2: Finding the Area of a Regular 14-Sided Polygon

Find the area of a regular 14-sided polygon given the side length is 21 cm. Give the answer to two decimal places.

Answer

We recall that the area of a regular 𝑛-sided polygon of side length 𝑥 is given by 𝑛𝑥4180𝑛.∘cot

In our case, since this is a regular 14-sided polygon of side length 21 cm, the value of 𝑛 is 14 and that of 𝑥 is 21 cm. We substitute these values into the formula to get areacotcot=14(21)418014=30872907.∘∘

Recall that cottan907=1∘∘, which gives us 30872907=30872=6762.515…,cottancm∘∘ which to two decimal places is 6‎ ‎762.52 cm2.

Hence, the area of a regular 14-sided polygon with a side length of 21 cm is 6‎ ‎762.52 cm2 to two decimal places of accuracy.

In our next example, we will use the perimeter of a regular polygon to determine its area.

Example 3: Finding the Area of a Regular Pentagon given Its Perimeter

The perimeter of a regular pentagon is 85 cm. Find the area giving the answer to the nearest square centimetre.

Answer

We recall that the area of a regular 𝑛-sided polygon of side length 𝑥 is given by 𝑛𝑥4180𝑛.∘cot

We can find the value of 𝑥 by recalling that the perimeter of a polygon is the sum of its side lengths. Since this is a regular pentagon, there are five sides all of the same length, 𝑥; hence, the perimeter is 5𝑥, giving us the equation 85=5𝑥855=𝑥𝑥=17.cm

We substitute 𝑥=17cm and 𝑛=5 into the formula for the area of a regular polygon to get areacottancm=5(17)41805=14454(36)=497.217…,∘∘ which to the nearest square centimetre is 497 cm2.

Hence, the area of a regular pentagon whose perimeter is 85 cm, to the nearest square centimetre, is 497 cm2.

In our next example, we will use the area of a regular hexagon to find the length of its sides.

Example 4: Finding the Side Length of a Regular Hexagon given Its Area

A flower bed is designed as a regular hexagon with an area of 54√3 m2. Find the side length of the hexagon giving the answer to the nearest metre.

Answer

We recall that the area of a regular 𝑛-sided polygon of side length 𝑥 is given by 𝑛𝑥4180𝑛.∘cot

A hexagon has 6 sides, so we set 𝑛=6 and then the area must equal 54√3, giving us 54√3=6𝑥41806.∘cot

We can then solve for 𝑥: 54√3=3𝑥2(30)2×54√33(30)=𝑥2×54√3×(30)3=𝑥.mcotcottan∘∘∘

Then, we cancel the shared factor of 3 and use the fact that tan(30)=√33∘ to get 36√3×√33=𝑥, which simplifies to 36=𝑥.

Taking the square root of both sides, and noting that 𝑥 must be positive and measured in metres, we have 𝑥=√36𝑥=6.m

Hence, a regular hexagon with an area of 54√3 m2 has sides of length 6 m.

In our next example, we will use the area of a regular polygon to determine its perimeter.

Example 5: Finding the Perimeter of a Regular Decagon given Its Area

The area of a regular decagon is 155.8 cm2. What is the perimeter of the decagon rounded to one decimal place?

Answer

We recall that the area of a regular 𝑛-sided polygon of side length 𝑥 is given by 𝑛𝑥4180𝑛.∘cot

A decagon has 10 sides, so we can set 𝑛=10, and then this expression must be equal to the area, 155.8 cm2, giving us 155.8=10𝑥418010155.8=5𝑥2(18).∘∘cotcot

We can then solve this equation for 𝑥, the side length of the regular decagon. We start be rewriting cot(18)∘ as 1(18)tan∘ to give us 155.8=5𝑥2(18).∘tan

Then, we solve for 𝑥: 155.8(18)25=𝑥𝑥=1558(18)25.tantan∘∘

Finally, the perimeter is the sum of the side lengths of the polygon. Since this is a regular decagon, there are 10 sides all having the same length. We leave the side length as a surd to get an exact answer for the perimeter before rounding. So, the perimeter is 10𝑥, which we can calculate as 10𝑥=101558(18)25=44.998…,tancm∘ which to one decimal place is 45.0 cm.

Hence, the perimeter of a regular decagon of area 155.8 cm2, to one decimal place, is 45.0 cm.

In our final example, we will use the area formula for regular polygons to determine the area of a region represented in a diagram.

Example 6: Finding the Area of a Shaded Region given Multiple Regular Polygons

Find the total area of the shaded regions in the regular polygons below, giving your answer to the nearest tenth.

Answer

To determine the area of the shaded regions in the diagram, we first note that all of the polygons are regular polygons of side length 39. We can find the area of each region separately. Let’s start with the outer region.

The outer region is the area between the regular hexagon and the regular pentagon, so we can calculate this area by finding the difference in the areas of the two shapes. To do this, we recall that the area of a regular 𝑛-sided polygon of side length 𝑥 is given by 𝑛𝑥4180𝑛.∘cot

We can find the area of the hexagon as follows: areaofhexagoncottan=6(39)41806=3(39)2(30).∘∘

Then, we use the fact that tan(30)=1√3∘ to write this as areaofhexagon=3(39)2=3(39)√32.√

Similarly, we can find the area of the pentagon as follows: areaofpentagoncottan=5(39)41805=5(39)4(36).∘∘

Then, the area of the outer region is the difference between these areas: outerareatan=3(39)√32−5(39)4(36)=1334.827….∘

We can find the area of the inner shaded region in a similar manner.

It is the area between a square and an equilateral triangle both with side lengths 39, so the inner region is the difference between their areas.

We find the area of the square as areaofsquare=39.

We then find the area of the equilateral triangle as follows: areaoftrianglecottan=3(39)41803=3(39)4(60)=3(39)4√3=3(39)4√3×√3√3=39√34.∘∘

The difference between these values gives us the inner shaded area: innerarea=39−39√34=391−√34=862.387….

Finally, we need to find the sum of the inner and outer shaded regions. We will use the exact values and then round at the end: totalshadedareaouterareainnerareatanareaunits=+=3(39)√32−5(39)4(36)+391−√34=2197.21…,∘ which to one decimal place is 2‎ ‎197.2 area units.

Let’s finish by recapping some of the important points of this explainer.

Key Points

  • The area of a regular 𝑛-sided polygon of side length 𝑥 when working in degrees is given by 𝑛𝑥4180𝑛.∘cot
  • The area of a regular 𝑛-sided polygon of side length 𝑥 when working in radians is given by 𝑛𝑥4𝜋𝑛.cot
  • The area formula can be used to find the area of compound shapes constructed of one or more regular polygons.

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