Let an equilateral triangle have sides of length 1.
Drop a perpendicular from the apex to the base, cutting the base in half.
Use the pythagorean theorem to compute its length.
Thus 12 - ½2 = x2, and x = sqrt(3)/2, or 0.866.
The area, bh/2, is sqrt(3)/4.
If each side has length s, the base and height are multiplied by s, and the area is multiplied by s2.
Connect the six vertices of a regular hexagon to its center, creating 6 equilateral triangles. Thus the area of a regular hexagon, whose sides have length 1, is 3×sqrt(3)/2. If each side has length s, the area is multiplied by s2. |

Let the sides of a regular octagon have length 1 and draw two vertical segments connecting the bottom and top edges. This creates a rectangle of width 1, and height still to be determined. Draw horizontal segments connecting the left and right edges, creating another rectangle that overlaps the first in a square. Now there are four triangles at the lower left, lower right, upper left, and upper right. These are right isosceles triangles with hypotenuse 1. If the leg has length x, solve x2 + x2 = 1, hence x = sqrt(2)/2. The four triangles have a combined area of 1, the north south east and west rectangles have a combined area of 2sqrt(2), and the central square has area 1. Thus the octagon has area 2+2sqrt(2). If each side has length s, the area is multiplied by s2. |