Question

Have a look please. I made a Matlab code for giving solution to the following problem:

A regular polygon of n sides is inscribed in a circle of radius r=1.33. Considering this polygon to consist of identical triangles, calculate the area and perimeter of the polygon for n=6, n=60 and n=600. Also Calculate the perimeter and area of the circle for comparison.

Answer 1

Code: function [area] = ntri(n)

r=1.33;

circleP=2*pi*r
% area= (cos(pi-(pi)/n-degtorad(90))*(sin(pi-(pi)/n-degtorad(90)))/2
%
% perimeter=cos(pi-(pi)/n-degtorad(90))*r

%determination of hip adjacent angle
Had=pi-(pi)/n-degtorad(90);

%Determination of the base
base=cos(Had)*r;

%Hight
H=sin(Had)*r;

%area of one triangle

area1= base*H;

%area of all triangles

triangles_area=area1*n

Answer 2

It outputes this:

>> ntri(6)

circleP =

8.3566


triangles_area =

4.5957

>> ntri(60)

circleP =

8.3566


triangles_area =

5.5470

>> ntri(600)

circleP =

8.3566


triangles_area =

5.5571

Answer 3

you think the calculations are correct?

Answer 4

the area of the triangles and their perimeter should be much closer to the circles for n=600

Answer 5

to debug this, figure out by hand what the area of one triangle is for n=600
and compare that to what your program finds

Answer 6

I did by hand too. I'm just not sure if i got the calculation right...