Question 586501
Inside a circle with a radius of 2mm, a 20-sided regular polygon is inscribed.
Each of the 20 vertices is on the circuference of the circle, at 2 mm from the center.
Since the polygon is a regular 20-gon, all the sides an angles have the same measure, and you could connect all the vertices to the center and split that 20-gon into 20 isosceles triangles with 2mm legs.
As we are splitting, let's draw a line from the middle of each side to the center, spiltting each isosceles triangle into 2 right triangles.
{{{drawing(300,100,-290,10,-50,50,
triangle(-279.1,0,0,22,0,-22),rectangle(0,0,-5,5),
red(line(-279.1,0,0,0)), locate(-140,-11,2mm)
)}}} This is one of those little isosceles triangles, magnified.
Each of those 40 right triangles has a {{{9^o}}} angle at the center, and a 2mm long hypotenuse.
We can calculate the length of the legs of those triangles with trigonometric functions.
The short leg, opposed to the {{{9^o}}} angle is half of the side of the 20-gon. We can consider it to be the base of the tiny right triangle, and its measure, in mm, is
{{{2*sin(9^o)}}}.
The length of the long leg, the height of the tiny right triangle, can be calculated as
{{{2*cos(9^o)}}}.
The area of each of those 40 right triangles, in square millimeters, would be
{{{(1/2)(2*sin(9^o))(2*cos(9^o))=2*sin(9^o)*cos(9^o)=sin(2*9^o)=sin(18^o)}}}
That's approximaterly 0.309 (((mm^2}}}
The area of the 20-gon, made up of 40 of those little right triangles, will be
{{{40*0.309mm^2=highlight(12.36mm^2)}}}