Regular Polygons with Equal Areas

This lesson has two versions. One is for higher grades and uses the TI-34 II calculator. The other one is for early grades and uses a four-operation calculator, the TI-108. The part that is common to both lessons is that students design, cut out from construction paper, and display a collection of regular polygons.

Task for higher grades.

Design, cut out from colored construction paper, and display in the classroom a collection of regular polygons.All the polygons must have the same area.The suggested area is A = 20 square inches, but other values are possible, and their choice may depend on the supplies that are available.

1. Analysis of the problem.

It is not difficult to construct a regular n-gon inscribed in a circle of radius r.We only need to compute the "central" angle c = 360°/n, and use a protractor to divide the circle into n equal angles of size c°. In the picture below, c = 360°/5 = 72°.

The inscribed polygon is formed by n triangles; each one has area T = r²*cos(c/2)*sin(c/2). (See the picture below, for n = 5.)

This is so because half of the base of each triangle, which is half of the side of the polygon, is equal to r*sin(c/2), and the height of the triangle is equal to r*cos(c/2).Therefore, A = n*T = n*r²*cos(c/2)*sin(c/2) = n*r²*cos(180/n)*sin(180/n).

We need to solve this equation for r. So we have

r = √(A/(n*cos(180/n)*sin(180/n)).

2. Computation of r.

Variables:

>A : keeps the area (here 20 square inches)

B : keeps the radius r

Ans : stores the current display, which is the number of sides n.

Define:

OP1 = √(A/(Ans*cos(180/Ans)*sin(180/Ans))→B

This gives us r.

Since our ruler measures in 32^nds of an inch, we want to change the measurement to some whole number of inches plus some number of 32^nds of an inch.So we also define

OP2= round(fPart(Ans)*32,1)

(This will tell how many 32^nds of an inch we need.)

Set [2^nd][DR] to DEG (angles are measured in degrees).

Enter 20→A

Now, in order to compute r in inches, execute, for n = 3, 4, 5, ...

n [=]

OP1 computes r, displays it, and also stores it in B. Write down the whole part.

OP2 computes the number of32nds of an inch.

Here is a table of values:

n=no. of sides in regular polygon	radius r of circle whose inscribed regular polygon with n sides has area 20 square inches	radius r in inches and 32^nds of an inch
3	3.923774609 inches	3 ^29.6/32 inches
4	3.16227766	3 ^3.2/32
5	2.900292708	3 ^28.8/32
6	2.774527634	2 ^24.8/32
7	2.703487703	2 ^22.5/32
8	2.659147948	2 ^21.1/32
9	2.629511004	2 ^20.1/32

3.Draw and cut out the required n-gons.(See pictures, for n = 3, 4, 5, 6, 7, 8, and 9.)

Click here if the animation above is not showing

A task for lower grades.

Students compute the "central" angle mentally or with a TI-108 calculator, draw a circle of any size, and then craw and cut out the inscribed polygon. Dividing a circle into equal sectors using a protractor is a challenging task, which is the center of this activity.

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Last Modified: March 12, 2007

	Regular Polygons with Equal Areas

This lesson has two versions. One is for higher grades and uses the TI-34 II calculator. The other one is for early grades and uses a four-operation calculator, the TI-108. The part that is common to both lessons is that students design, cut out from construction paper, and display a collection of regular polygons.

Task for higher grades.

Design, cut out from colored construction paper, and display in the classroom a collection of regular polygons.All the polygons must have the same area.The suggested area is A = 20 square inches, but other values are possible, and their choice may depend on the supplies that are available.

1. Analysis of the problem.

It is not difficult to construct a regular n-gon inscribed in a circle of radius r.We only need to compute the "central" angle c = 360°/n, and use a protractor to divide the circle into n equal angles of size c°. In the picture below, c = 360°/5 = 72°.

The inscribed polygon is formed by n triangles; each one has area T = r²*cos(c/2)*sin(c/2). (See the picture below, for n = 5.)

This is so because half of the base of each triangle, which is half of the side of the polygon, is equal to r*sin(c/2), and the height of the triangle is equal to r*cos(c/2).Therefore, A = n*T = n*r²*cos(c/2)*sin(c/2) = n*r²*cos(180/n)*sin(180/n).

We need to solve this equation for r. So we have

r = √(A/(n*cos(180/n)*sin(180/n)).

2. Computation of r.

Variables:

>A : keeps the area (here 20 square inches)

B : keeps the radius r

Ans : stores the current display, which is the number of sides n.

Define:

OP1 = √(A/(Ans*cos(180/Ans)*sin(180/Ans))→B

This gives us r.

Since our ruler measures in 32^nds of an inch, we want to change the measurement to some whole number of inches plus some number of 32^nds of an inch.So we also define

OP2= round(fPart(Ans)*32,1)

(This will tell how many 32^nds of an inch we need.)

Set [2^nd][DR] to DEG (angles are measured in degrees).

Enter 20→A

Now, in order to compute r in inches, execute, for n = 3, 4, 5, ...

n [=]

OP1 computes r, displays it, and also stores it in B. Write down the whole part.

OP2 computes the number of32nds of an inch.

Here is a table of values:

n=no. of sides in regular polygon	radius r of circle whose inscribed regular polygon with n sides has area 20 square inches	radius r in inches and 32^nds of an inch
3	3.923774609 inches	3 ^29.6/32 inches
4	3.16227766	3 ^3.2/32
5	2.900292708	3 ^28.8/32
6	2.774527634	2 ^24.8/32
7	2.703487703	2 ^22.5/32
8	2.659147948	2 ^21.1/32
9	2.629511004	2 ^20.1/32

3.Draw and cut out the required n-gons.(See pictures, for n = 3, 4, 5, 6, 7, 8, and 9.)

A task for lower grades.

Students compute the "central" angle mentally or with a TI-108 calculator, draw a circle of any size, and then craw and cut out the inscribed polygon. Dividing a circle into equal sectors using a protractor is a challenging task, which is the center of this activity.

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