Question 409846
The indirect way would be to assume that the area is not equal to {{{pi*r^2}}} and find a contradiction. However this would involve finding the actual area of a circle, so an indirect proof is unnecessary.


Instead, construct a regular n-gon with n isosceles triangles (similar to "Theorem 11.8 which I recently solved, you can find it under the solutions I posted). Instead of finding the length of the base, we find the area, and multiply by n.

{{{drawing(200, 200, 0, 6, 0, 10,
triangle(2, 2, 4, 2, 3, 8),
locate(2, 5, r),
locate(2.9, 7, theta)

)
}}}


Recall that the area of the triangle is {{{r^2*sin(theta)/2}}} where {{{theta = 2*pi/n}}}. We have n of these triangles, so the area of the n-gon is


{{{A = nr^2*sin(2*pi/n)/2}}} (this may or may not be a good formula to memorize, but it's more important to know where it came from). Take the limit as n goes to infinity:


{{{A(circle) = lim(n->infinity, nr^2*sin(2*pi/n)/2)}}}. Just like the last problem, if we let {{{u = 1/n}}},


{{{A(circle) = lim(u->0, r^2*sin(u*2pi)/2u) = r^2*lim(u->0, sin(u*2pi)/2u)}}}. Applying L'Hopital's rule,


{{{A(circle) = r^2*lim(u->0, sin(u*2pi)/2u) = r^2*lim(u->0, (2pi*cos(u*2pi))/2)}}}. Since {{{lim(u->0, (2pi*cos(u*2pi))/2) = pi)}}} (by direct substitution and noting we obtain a finite and determinate number),


{{{A(circle) = r^2*pi}}}, hence the area of a circle.