67 058
Assignments Done
99,3%
Successfully Done
In November 2018

Answer to Question #5329 in Calculus for Jessica

Question #5329
(a) Let A sub n be the area of a polygon with n equal sides inscribed in a circle with radius r. By dividing the polygon into n congruent triangles with central angle 2*(pi)/n, show that A sub n=1/2*(n/r^2)*sin*(2pi/n)
Expert's answer
Let S be the area of each of n triangles, then A_n = n*S.
So we have to find
the area of S.
Notice that S is an isosceles trianlge with two sides of
length r.
The angle betwenn these sides is 2*pi/n.
It is known that the
area of a triangle with sides a, b, and angle phi between them is
a*b*sin(phi)/2.
In our case
a=b=r
phi=2*pi/n.
Hence
S = 1/2
* r^2 * sin(2*pi/n)

Therefore
A_n = n*S = 1/2 * n * r^2 *
sin(2*pi/n)

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

Submit
Privacy policy Terms and Conditions