document.write( "Question 586501: What is the area of a 20-gon with a radius of 2 mm? Please explain. \n" ); document.write( "
Algebra.Com's Answer #374097 by KMST(5328)\"\" \"About 
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Inside a circle with a radius of 2mm, a 20-sided regular polygon is inscribed.
\n" ); document.write( "Each of the 20 vertices is on the circuference of the circle, at 2 mm from the center.
\n" ); document.write( "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.
\n" ); document.write( "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.
\n" ); document.write( " This is one of those little isosceles triangles, magnified.
\n" ); document.write( "Each of those 40 right triangles has a \"9%5Eo\" angle at the center, and a 2mm long hypotenuse.
\n" ); document.write( "We can calculate the length of the legs of those triangles with trigonometric functions.
\n" ); document.write( "The short leg, opposed to the \"9%5Eo\" 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
\n" ); document.write( "\"2%2Asin%289%5Eo%29\".
\n" ); document.write( "The length of the long leg, the height of the tiny right triangle, can be calculated as
\n" ); document.write( "\"2%2Acos%289%5Eo%29\".
\n" ); document.write( "The area of each of those 40 right triangles, in square millimeters, would be
\n" ); document.write( "
\n" ); document.write( "That's approximaterly 0.309 (((mm^2}}}
\n" ); document.write( "The area of the 20-gon, made up of 40 of those little right triangles, will be
\n" ); document.write( "\"40%2A0.309mm%5E2=highlight%2812.36mm%5E2%29\" \n" ); document.write( "
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