Area of a Segment - Concept
The area of a segment in a circle is found by first calculating the area of the sector formed by the two radii and then subtracting the area of the triangle formed by the two radii and chord (or secant). In segment problems, the most challenging aspect is often calculating the area of the triangle. Related topics include area of a sector, area of a circle and area of an annulus.
If you want to find the area of a segment you're going to have to do a little bit of problem solving first. The segment is the shaded region right up here, so let's start off by what do we know, well I know how to calculate the area of a sector so I'm going to write area of a sector down below and I'm going to sketch the area of a sector here. So I'm going to draw my circle and I'm going to draw in that sector, so the sector will calculate this whole region right here and then I can take out the part of this triangle. So if I can calculate the area of a triangle which I know how to do, so I'm going to say subtract in the same circle the area of that triangle. So I'll say area of the triangle, so if I start with this whole sector and I take out the part that's the triangle what's left is the segment. So we can say that the area of the segment is equal to the area of the sector minus the area of the triangle. So now let's write this in terms of what we know, well r is our radius, x is our central angle or the intercepted arch since those are always congruent. So the area of the segment is equal to I'll write it down here the area of the sector which we said was the area of the whole triangle pi r squared times x out of 360 and then we're going to subtract our area of our triangle which is going to be the base times the height divide by 2. So in this triangle if I drew in my height which is h and I said that we had some base of that triangle then we'll have the formula for the area of our segment. So when you think about the segment calculating its area you're going to calculate the sector and then subtract the area of that triangle.