+129845 how can I prove that pi times radius squared is the area of a circle
Mmmmmm....this is an interesting one.
Let's start by inscribing a polygon of some "n" sides in a circle. It can be shown that, as the number of sides increase (to some "large" value), the perimeter of this polygon is almost the same as the circle's circumference ≈ 2*pi*r. And since we have n sides, the length of each side is just ≈ (2*pi*r) / (n)
Now, if from the center of the circle, we draw a radial line (r) to each vertex of the polygon, we will form "n" triangles. The base of each triangle is just the side length of the polygon (given in the previous paragraph). And we can also show that, as the number of sides of the polygon increase to some "large number," the height of each triangle aproaches "r" (the radius of the circle).
Therefore, the area of each triangle is (1/2)*(b)*(h) = (1/2)*[(2*pi*r) / (n)]*(r) =
pi*r^2 / (n).
And since we have n triangles, then the total area of all the triangles = (the number we have) * (the area of each one) =
(n)*(pi*r^2) / (n) =
(pi)*r^2........
+129845 how can I prove that pi times radius squared is the area of a circle
Mmmmmm....this is an interesting one.
Let's start by inscribing a polygon of some "n" sides in a circle. It can be shown that, as the number of sides increase (to some "large" value), the perimeter of this polygon is almost the same as the circle's circumference ≈ 2*pi*r. And since we have n sides, the length of each side is just ≈ (2*pi*r) / (n)
Now, if from the center of the circle, we draw a radial line (r) to each vertex of the polygon, we will form "n" triangles. The base of each triangle is just the side length of the polygon (given in the previous paragraph). And we can also show that, as the number of sides of the polygon increase to some "large number," the height of each triangle aproaches "r" (the radius of the circle).
Therefore, the area of each triangle is (1/2)*(b)*(h) = (1/2)*[(2*pi*r) / (n)]*(r) =
pi*r^2 / (n).
And since we have n triangles, then the total area of all the triangles = (the number we have) * (the area of each one) =
(n)*(pi*r^2) / (n) =
(pi)*r^2........
+129845 Not to beat this question to death....(maybe Rom and I already have!!!)...but here's something you might want to look at......
Here's a "formula" to calculate the side length (s) of an "2^n-sided" regular polygon inscribed in a circle with a radius of "1."
(Really, the radius length doesn't matter....all circles are "similar." I've just chosen "1" because it's convenient!! By using a "scaling factor," we could show the same proof for a circle of any radius!!!)
s = (sin(360/2^n)/sin((180-(360/2^n))/2),
where the (2^n) represents the number of sides of the polygon. Note something interesting.....as "n" increases, s "decreases." (Try it for yourself using the on-site calculator!!)
Notice something else.....If you take the output for any particular "s" and multiply it back by (2^n), you will get the perimeter of the polygon. You will also find that, as n increases, this perimeter approaches 2pi (r), when r = 1.
For example, let n be "8." Then the number of sides = 2^8 = 256. And calculating the length of one side of this 256 sided polygon gives us......0.0245430765715311. Now, multiplying this by 256, we get the polygon's perimeter.......6.2830276023119616.......which is > 99.99% of 2*(pi)* (r) , where r = 1.
So, it appears that, as the sides increase to some "large number," the perimeter of the 2^n-sided polygon approaches the perimeter of the circle !!!
If, from the center of the circle, we draw a radial line (r) to each vertex of the polygon, we will form "2^n" triangles............ We can also show that, as the number of sides of the polygon increases to some "large number," the height of each triangle aproaches "r" (the radius of the circle ).
And there's a "formula" to calculate the height of each such triangle....To see how this is dervived, note, that in each triangle, we can draw a line from the"apex" of the triangle such that it interesects the side of the polygon at right angles.... (this is the height, (h), of the triangle). And by the Law of Sines, we have
r /sin (90) = h /sin((180-(360/2^n))/2).......the angle represented by (180-(360/2^n))/2) is just the angle formed by a radial line and the side of the polygon. Note that r =1 and sin(90) = 1. Therefore, the height of the triangle, (h), is just = sin((180-(360/2^n))/2)
As before, let n be = 8. (This, again, represents a polygon with 256 sides.)
Putting this into the on-site calculator, we get.... (.999924701839). Note how close this is to the radius (1)!!! Letting n = 10, (a polygon with 1024 sides), we have, h = .99999529381......even closer to the radius of the circle !!
Putting this all together......by inscribing some "2^n-sided" polygon in a circle (where 2^n is "large"), and drawing radial lines to each vertex of the polygon, we form "2^n" triangles. Each triangle will have a side length of "s" and a height of "h." Then, the area of each such triangle will be (1/2) * (s) * (h). And multiplying the number of triangles we have, (2^n), by the area of each, we get the area of the polygon .... (2^n)* (1/2) * (s) * (h) = (1/2) * (2^n) * (s) * (h) . But note, (2^n) * s equals the perimeter of the polygon which ≈ 2(pi)(r). And h ≈ r..... (as we've shown).
So, the area of the polygon = (1/2) * 2(pi)(r) * (r) = pi * r^2. (For all practical purposes.)
