prompter: How can I find the area of a equilateral triangle (3 sides) given only one of it's side lengths.

assistant: To find the area of an equilateral triangle with one known side length, you can use the following formula:

Area = (sqrt(3) / 4) x s^2

where s is the length of one side of the equilateral triangle.

So, if you are given a side length of s, you can plug it into the formula and solve for the area.

prompter: How about if that side is 15 cm long?

assistant: You can calculate the area as:
Area = sqrt(3) / 4 * (15 cm)² = 97.43 cm²
rounded to the second digit

prompter: If the object instead has 4 sides with the same angle in the corners. The side length is still 15 cm long. What is the area?

assistant: then the area is 97.427857925749347760918856709705 cm^2

assistant: Given a side length L, the area of the triangle is sqrt(3)/4 * L^2.

Here is a derivation of that formula. First, the formula for the area 'a' of a triangle is

    a = (b * h) / 2

where b is the base and h is the height.

Call your triangle's corners A, B and C. Place the triangle so side BC is the base, and A is at the top. Let M be the midpoint of the base BC. Then the base b is the length of BC, and the height h is the length of AM.

Call the side length of your triangle L, so the length of BC is L, so b=L.

Now consider the triangle ABM. We have the two other side lengths: AB is length L, and BM is length L/2. Sides AM and BM meet at a right angle, so we can apply the Pythagorean formula, and say h^2 + (L/2)^2 = L^2. Now solve for h:

    h^2 = L^2 - (L/2)^2
      = L^2 - (L^2 / 4)
      = (3/4) L^2
    h = sqrt((3/4) L^2))
      = sqrt(3/4) L
      = (sqrt(3)/2) L

Now we can plug these into the formula for the area of a triangle:

    a = (b * h) / 2
      = (L * sqrt(3)/2 L) / 2
      = sqrt(3)/4 * L^2

prompter: What about other types of triangles? Can you generate a formula that will provide me the area of any triangle, and give me real life scenarios where this would be useful?

assistant: The area of a triangle *A* = *hb* / 2. *b* is the base, or the length of one of its sides. *h* is the height, or the distance between the vertex opposite to the chosen side and that side. For an equilateral triangle with a side length of *a*, the base is always *a*. As all of its interior angles are 60°, the height is *a* sin60° = √3 *a* / 2.
Therefore, the area *A* = *hb* / 2 = √3 *a* ^2 / 4.

prompter: That is correct. What about the area for any regular convex polygon, given any one of it's sides. Is it reasonable to find this?

assistant: It is reasonable, and can be done with a simple formula A = N * s * a / 2 for a regular convex polygon with N sides of length s. a is the shortest distance from the center point to one of the sides, known as the apothem. This formula can be viewed as first dividing the regular convex polygon into N identical triangles by connecting its center to its vertices. The area of each triangle is s * a / 2, where s is the base and a is the height. The total area of the triangles is N * s * a / 2.
Within each triangle, If we draw a perpendicular line from the center of the polygon to the base, we get two identical right triangles. The angle with the polygon's center as the vertex is 180°/N, and the length of the side opposite to it is s / 2. Therefore, the side adjacent to it, or the apothem, is s * cot(180°/N) / 2, and the area of the regular convex polygon A = N * s^2 * cot(180°/N) / 4.

assistant: The interior angles of a convex polygon are less than 180°.
Regularly, a polygon is firmly convex, if each line segment with two nonadjacent vertices of the polygon is strictly internal to the polygon but on its endpoints.
A regular convex polygon is a polygon where each side is of the same length, and all the interior angles are equal and less than 180 degrees.
The vertices and sides are evenly spread around a central point.

If the convex polygon has vertices, then the formula for the area of the convex polygon is: (number of sides \* length of one side \* apothem)/2

the length of apothem is given as the 1/(2 \* tan(180 / number of sides))