Pg 24, q51. How do I figure this out? thanks

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2 Answers
Jan 24, 2018

#A_n = piR^2 sin((2pi)/n)/((2pi)/n)#

Explanation:

Consider for the #n#-sided regular polygon the triangle limited by one side and the radii joining the center with the two consecutive vertices limiting the side.

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For symmetry reasons, the area of the polygon is #n# times the area of the triangle.

Moreover the triangle is isosceles and the angle at the top is:

#alpha = (2pi)/n#

so that we can see that the height of the triangle is:

#h= R cos(alpha/2) = R*cos(pi/n) #

while the length of the side is:

#l= 2Rsin(alpha/2) = 2Rsin(pi/n) #

The area of the triangle is then:

#a_n = (l*h)/2 = (2R^2sin(pi/n)cos(pi/n))/2 = R^2/2 sin((2pi)/n)#

and the area of the polygon is:

#A_n= (nR^2)/2 sin((2pi)/n)#

If we write this as:

#A_n = piR^2 sin((2pi)/n)/((2pi)/n)#

We can see that as #n# increases, #A_n# tends to the area of the circumscribed circle:

#lim_(n->oo) A_n = piR^2 lim_(n->oo)sin((2pi)/n)/((2pi)/n) = piR^2#

Jan 24, 2018

See below

Explanation:

The regular polygons are able to be split into equal isosceles triangles.

The number of these triangles is equal to the number of sides in the polygon, so for #n=5# there are 5 equal isosceles triangles.

The area for one of these triangles is equal to #1/2ab sinC#, however, since #a=b=1#, the area is #sinC/2#

#C# is easy to find, by doing #360^circ/n# or #(2pi)/n#. This is because for 1 full circle, it is #360^circ = 2pi#. And in terms of full triangles, these triangles fit into the circle #n# times.

#360^circ = 2pi:"full circle"#

#"full circle"/n:360^circ/n = (2pi)/n#

N.B. If an equation uses #pi# then it is in radians, if it uses #""^circ# then it is in degrees. The equations work either way

So, we know have an area for one triangle, #sin((2pi)/n)/2=sin(360^circ/n)/2#

However, we need to total area.

#"Total area"=Sigmasin(360^circ/n)/2=nsin(360^circ/n)/2#
#"Total area"=Sigmasin((2pi)/n)/2=nsin((2pi)/n)/2#

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n=5:
#"Total area"=5sin(360^circ/5)/2~~2.38#
#"Total area"=5sin((2pi)/5)/2~~2.38#

n=6:
#"Total area"=6sin(360^circ/6)/2=(3sqrt(3))/2#
#"Total area"=6sin((2pi)/6)/2=(3sqrt(3))/2#

n=7:
#"Total area"=7sin(360^circ/7)/2~~2.74#
#"Total area"=7sin((2pi)/7)/2~~2.74#

n=8:
#"Total area"=8sin(360^circ/8)/2=2sqrt(2)#
#"Total area"=8sin((2pi)/8)/2=2sqrt(2)#

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In general, as the value of #n# increases to #oo#, the size of the triangles will become much smaller, and will be more likely to take on the shape of a circle. Therefore, the combined are will slowly reach #pi#.

With radians: #lim_(n->+oo)(nsin((2pi)/n))/2=pi#
With degrees: #lim_(n->+oo)(nsin(360^circ/n))/2=pi#

This effect can be explored further, using just a value for the angle, here. Keep in mind that the notes sections don't allow for formatting, i.e. fractions.