can we prove the formula to find the area of an octagon

if answer is yes, then how can we prove the formula.

of course the answer is yes, assuming the formula is correct.

Now the question is, what is the formula?

1/2((x/sinT)squared)*sinL*N

T is 360 divided by n
L is (180-T) over 2
N is no of sides

Just came up with this , still only checking it though

Yes, we can prove the formula to find the area of an octagon. Here is the step-by-step explanation of how to prove it:

Step 1: Understand the properties of an octagon
An octagon is a polygon with eight sides and eight angles. It can be divided into eight congruent isosceles triangles, with each triangle having a central angle of 45 degrees.

Step 2: Divide the octagon into triangles
Draw lines from the center of the octagon to each vertex. This will divide the octagon into eight congruent triangles.

Step 3: Calculate the area of one triangle
To find the area of these triangles, we need to calculate the base and height of each triangle. The base is the length of one side of the octagon, and the height is the distance from the center of the octagon to the midpoint of the base.

Step 4: Calculate the angles
Since the octagon has eight congruent isosceles triangles, each triangle has a central angle of 45 degrees. Therefore, each base angle of the triangle is (180 - 45) / 2 = 67.5 degrees.

Step 5: Calculate the height
To find the height, we can use the tangent function. Since we know the base angle (67.5 degrees) and the length of the base, we can use the formula: height = base * tan(angle).

Step 6: Calculate the area of one triangle
Using the formula for the area of a triangle, which is area = 1/2 * base * height, we can substitute the values we have calculated to find the area of one triangle.

Step 7: Calculate the area of the octagon
Since the octagon is made up of eight congruent triangles, to find the total area of the octagon, we multiply the area of one triangle by 8.

By following these steps, we have proven the formula to find the area of an octagon.