# Area of a Heptagon Formula

TR Smith is a product designer and former teacher who uses math in her work every day.

A regular heptagon is a polygon with seven sides of equal length whose edges meet at equal angles. The interior angles at the vertices of a regular heptagon are approximately 128.5714 degrees. This is derived from the fact that the interior angles of a regular n-gon are 180 - 360/n degrees. Plugging n = 7 into this equation gives 128.5714.

The area of a regular heptagon with a side length of x is approximately 3.633912444*x^2. This can be derived from trigonometry by slicing the heptagon into seven equal triangles and finding their areas. Below is a proof of the formula and some example geometry problems involving heptagons.

## Deriving the Formula for the Area of a Heptagon

Take a regular heptagon with a side length of x. If you draw line segments from the center of the heptagon to each of the seven vertices, you will split the heptagon into seven equal isosceles triangles. The short side of each triangle has a length of x opposite an angle of 360/7 degrees. The area of each triangle is one seventh the area of the heptagon, so finding the area of one triangle and multiplying it by seven will give you the total area.

What is the area of one of these triangles? If you cut it in half along the longest altitude, you get a right triangle with a width of x/2 and a height of (x/2)/tan(360/14). The area of the half-triangle is (1/2)(x/2)(x/2)/tan(360/14) = (x^2)/[8*tan(360/14)]. The area of the full triangle is thus (x^2)/[4*tan(360/14)].

Multiplying this by 7 gives us the full area of the heptagon with a side length of x:

7(x^2)/[4*tan(360/14)]

Simplifying the constant 7/[4*tan(360/14)] to 3.633912444 gives us the alternative formula

3.63391244x^2

as the area of a regular heptagon.

## Example

A heptagonal coin is 30% gold and 70% copper by weight, not by volume. If the thickness of the coin is 3.5 mm and the side length is 11.7 mm, what is the weight of the coin?

The first things to know are the densities of gold and copper at room temperature. Gold weighs 19.3 grams per cubic centimeter, and copper weighs 8.96 grams per cubic centimeter.

Next, we need to find the volume of the coin in cubic centimeters. The volume is the area of the heptagonal face times the thickness. In cubic millimeters this is

3.63391244*(11.7^2)*3.5 = 1741.0619586906 cubic mm

Converted to cubic centimeters this is approximately 1.74106 cm^3. Of this volume, let's say K cubic cm are gold and 1.74106 - K cubic cm are copper. The ratio of gold to copper by weight is 3/7, therefore we need to find the value of K such that the ratio

[19.3K] / [8.96(1.74106 - K)]

equals 3/7. This is because weight equals density times volume. Solving the equation

[19.3K] / [8.96(1.74106 - K)] = 3/7

for K gives us the solution K = 0.288923 cubic cm. and 1.74106 - K = 1.452137. The total weight of coin is therefore

0.288923*19.3 + 1.452137*8.96 = 18.58736142 grams.

## Another Example

If the area of a regular heptagon is 78.105 square cm, what is the length of a side? For this problem, we need to solve backward using the area formula to find side length. The equation to solve is

78.105 = 3.63391244x^2

where x is the side length. Doing the algebra gives us

78.105/3.63391244 = x^2
sqrt(78.105/3.63391244) = x
x ≈ 4.636 cm

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• Judy Specht 2 years ago from California

You are my go to person when someone needs help with calculus. Excellent and easy to understand explanations.

• Author

TR Smith 2 years ago from Eastern Europe

Hey thanks. I never know if these tutorials are ever going to be useful to anyone, but occasionally someone comments that it helped them solve their homework.