# Area of triangle/polygon

I am trying to find total area, so far- I was able to determine area of the semi-circles but I cannot figure out the area of the triangle in the middle.

The variables for the circle are L (arc length) and R (radius).

I found Area of each semi-circle is A= L*R/2

On a similar vein, this could happen, and so on.

I found that all I need to figure out is the length of one side as a function of R and L of the circle. All sides are equal since all circles are similarly shaped and mirror symmetrical (drawings may not show that exactly).

Any help?

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When you say "all circles are similarly shaped and mirror symmetrical" do you mean that the circles are congruent (i.e. exactly the same size)? –  Duncan Jul 24 '14 at 19:41
So you're looking to find the total area of all of the parts covered by at least one circle? Also, if you have $n$ circles, the diagram is meant to be $n$-fold symmetric about the center? –  John Jul 24 '14 at 19:52
@Duncan 1) Yes, they are congruent. 2) I am looking to find the area of the n-gon made from the intersection of the circles. 3) Not sure what you mean, but yes, it is symmetrical about the center. Thanks –  Koopa Jul 24 '14 at 20:39
Intersection of the circles?? Really? In the square case the circles don't intersect at all, at least not all together. I guess you might have meant union. In that case, do you want to also include the central area which is not covered by any circle, or just the circle areas? –  MvG Jul 24 '14 at 23:19
I presume that the "variable" L and R are the givens. There are 2 types of arcs in your picture, which one is L? The major or the minor? Also, in your picture, I can see major and minor segments but I don't see any semi-circle. –  Mick Jul 25 '14 at 5:01

Let $d$ be the length of one side of the polygon, which is what we want to find (e.g. $EH$ in the second picture). Also let $\theta = \angle{EAH}$.
Then, with $R$ the radius and $L$ the inner arc length ($EH$ for example), we find $\theta$ by dividing $L$ by the circumference of the circle:
$\theta = \dfrac{L}{2\pi R} 2\pi = \dfrac{L}{R}.$
$d = 2R \sin{\left(\dfrac{\theta}{2}\right)} = 2R \sin{\left(\dfrac{L}{2R}\right)}.$