formula for working out the area of any regular shape with 1000m perimeter?.?

Question

now is this actually possibe?
is it soemthing like -
1000/ no. sides = side lengths
360 / no.of sides equals exterior angles.

and can you work out the area from this?
is there a formula i can apply to get the area of regular 4, 5, 6, 7, 8, 9, 10 (etc) sided shapes?

please explain how the formula works!

Dr Ivan your formula is pilliocks.

Answer 1

Dear sweetmango! Correct me if I'm wrong. Why insulting each other?

A(n)=250000cot(π/n)/n

Here n = no. of sides

An n-sided regular polygon consists of n isosceles triangles. The base length of each triangle is 1000/n, and its height is
[1000/2n]cot(2π/2n)
=[500/n]cot(π/n)
(I need to draw a picture to explain).

Hence the area of each triangle is 1/2 times base times height
= 1/2[1000/n][500/n]cot(π/n)
=250000cot(π/n)/n^2

We have n such triangles, so multiply the above number by n. Thus:
A(n)=250000cot(π/n)/n

Answer 2

Create a set of triangles by drawing lines from each vertex to the center of the polygon. If there are n sides then there will be n such triangles. If T is the area of each triangle then the area of the polygon A will be nT

A = nT

If S is the side of the polygon then S = 1000/n
If H is the height of the triangle (the length of the line connecting a vertex to the center) then:

tan (Angle/2) = H/(S/2) = 2H/S

Where Angle is the interior angle at each vertex.
The angles at the center for each triangle are 2pi/n. The sum of the other two angles in any of the triangles is Angle since each is one half the interior angle. Sonce the angles of a triangle must sum to pi radians, this gives:

pi = Angle + 2pi/n
Angle = (n - 2)pi/n

So:
tan ((n-2)pi/2n) = 2H/S .... from above equation
H = S* tan ((n-2)pi/2n) / 2

T = (1/2)base * height = (1/2)S * H
T = (1/2)(1000/n)^2(S* tan ((n-2)pi/2n) / 2)
T = (1000/n)^2 tan ((n-2)pi/2n))/4
T = (250000/n^2)tan((n-2)pi/2n)

A = nT = (250000/n)tan((n-2)pi/2n)

tan((n-2)pi/2n) = tan(pi/2 - pi/n) = cot(pi/n)

A = (250000/n)cot(pi/n)

Check:
square: n = 4, S = 250 and A = 250^2 = 62500
from formula: 62500cot(pi/4) = 62500

Answer 3

The regular plygon must have all the sides equal and its points on a circle with the center in the intersection of all angles' bisectors. Theese bisectors also form the sides of N 2 equal sides triangles formed with the sides. The 2 equal sides are equal to the radius of the circle and the total aria of the polygon is the sum of the arias of the N triangles.
Perimeter = P =1000
Ap = N*At
At = sqrt(R*R-(P/N)*(P/N)) * P/(*N)
now u have to work out the relation between the radius and perimeter
u do this by writing the radius in relation to the angle between the 2 equal sides and base
sin(a/2)=((P/N)/2)/R
a = 360/N
thus R = ((P/N)/2)/sin(a/2)
now u have a complete formula based on the perimeter and number of sides

a = 360/N
R = ((P/N)/2)/sin(a/2)
At = sqrt(R*R-(P/N)*(P/N)) * P/(*N)
Ap = N*At

Ap = N*sqrt((((P/N)/2)/sin(360/(N*2)))*(((P/N)/2)/sin(360/(N*2)))-(P/N)*(P/N)) * P/(*N)

for further questions please contact me on yahoo messenger [e-mail]

Answer 4

If you are talking about equilateral polygons, then yes, you can develop a fomula.

It is trivial to show the sum of the interior angles of a n-sided, equilateral polygon is (n-2)pi (or (n-2)180 for degrees). Each angle is then [(n-2)/n]pi.

The area of such an n-sided, equilateral polygon of side length s is [(n/4)tan((n-2)pi/2n)]s^2 = [(n/4)cot(pi/n)]s^2.

For your question, s = 1000/n (meters) so:

A(n) = 1,000,000 * tan((n-2)pi/2n) / 4n square meters.

Using difference formulas for the tangent:

A(n) = 1,000,000 * cot(pi/n) / 4n square meters.

If you want to work in degrees, instead of radians, use 180 for pi. However, the level of your question suggests you are likely working in radians.

Answer 5

lol no. there is no such formula.