# Area of a circle, analyzing with trigonometry and geometry

• December 25th 2009, 02:11 PM
Boyrog
Area of a circle, analyzing with trigonometry and geometry
I was trying to find the area of a circle decomposing it into polygons. So looking at the parametric equations of a circumference and the formula to define the area of a polygon having its coordinates i did the following calculus:

(I) Parametric equations of circle:
x= r * cos(t) I assume r=1 so I must get: x= cos(t) and area=pi
y= r * sin(t) y= sin(t)

(II) Area of a polygon having its coordinates:
A(poly)=1/2 * | Xa Xb Xc Xd ... Xn Xa|
| Ya Yb Yc Yd ... Yn Ya|

Giving any value of 't' i will get a point of the circle, so i imagined a polygon of 360 sides (I used t as degrees) that would have an area very near the area of the circle.
A(circle)=1/2*| cos(1) cos(2) cos(3) ... cos(360) cos(1) |
| sin(1) sin(2) sin(3) ... sin(360) sin(1) |
Developing the determinant I get:
A(circle)= 1/2* | [cos(1)*sin(2)+cos(2)*sin(3)+...+cos(360)*sin(1)]-
[sin(1)*cos(2)+sin(2)cos(3)+...+sin(360)*cos(1)] |
And finally syntesis of the formula to get circle's area:
1/2*
{summation of t=1 to t=359} cos(t)* sin(t+1)- sin(t)* cos (t+ 1)
+cos(360)*sin(1)-sin(360)*cos(1)

Using the calculator On-Line Calculator to find the result of the sum a got a very big number and after some adjusts i got still a number very far away from pi. Could someone tell me what did i make wrong?
Sorry by mistakes in grammar, i am not native.
• December 25th 2009, 04:24 PM
Bacterius
You could have done it a lot simpler.
A circle can be roughly represented as lots of triangles (with the same area) put together. Like a pizza, yes.
Let's look at one of these triangles (they are all equal). The angle that touches the center of the circle can be obtained easily : for example, if you use $90$ triangles, then each triangle will have that angle equal to $4$, because $90 \times 4 = 360°$.
Now, you know two sides of the triangle easily, too, because they are equal to the radius of the circle (let's take $1$).

So, if we take a triangle ABC, with A pointing towards the center of the circle, we have : angle $BAC = \frac{360}{t}$, where $t$ is the number of triangles wanted, and $AB = AC = 1$ (radius of the circle).

Now, what was the goal of this, already ? Ah, yes, approximating the area of the circle. Thus, we need to know the area of the triangles. Note that since triangle ABC is isosceles, we can divide it into two equal rectangle triangles. Let $H$ be the middle of $BC$.

Therefore, angle $BAH = \frac{180}{t}$, and using trigonometry, $\cos{\frac{180}{t}} = \frac{AH}{AB}$, thus $AH = \cos{\frac{180}{t}}$.

You are still missing something ! You need the side $BH$ to calculate the area. Since you know two sides in the rectangle triangle, you can use trig :

$\sin{\frac{180}{t}} = \frac{BH}{AB}$, thus $BH = \sin{\frac{180}{t}}$.

From there you can calculate the area of the rectangle triangle. Let $A$ be the area of the rectangle triangle.

$A = \frac{AH \times BH}{2}$.

Now remember we divided our isosceles triangle into two equal rectangle triangles, therefore the area of the isosceles triangle must be twice the area of the rectangle triangle. Let $A'$ be the area of the isosceles triangle.

$A' = AH \times BH$.

Now that you know that, all you have to do is multiply this area by the quantity $t$ of triangles used. Let $Q$ be the area of the circle.

$Q = AH \times BH \times t$.

That is :

$Q = \cos{\frac{180}{t}} \times \sin{\frac{180}{t}} \times t$.

Note that this only works if $t > 2$ (can you put two triangles next to each other inside a circle ? :) and also, $t = 1$ and $t = 2$ fails).

(Remember that the area of a circle of radius $1$ is equal to $\pi$). Let's see how our formula goes :

$t = 8$, $Q = \cos{\frac{180}{8}} \times \sin{\frac{180}{8}} \times 8 = \cos{22.5} \times \sin{22.5} \times 8 \approx 2,83$. That is not awesome yet ...

$t = 10000$, $Q = \cos{\frac{180}{10000}} \times \sin{\frac{180}{10000}} \times 10000 = \cos{0.018} \times \sin{0.018} \times 10000 \approx 3,1415924$. Now this is great !

Provided you input enough triangles (a greater $t$), and that your computing device has good floating point precision, you can approximate your areas easily !

:)

PS : this formula only works with a circle of radius $1$. I think you can change it easily enough yourself to make it fit to any circle, by introducing a new parameter $r$ (radius). Okay, I'll give you a hint : it only involves multiplying the area of a $1$-radius circle by $r^2$ :)