Inradius, perimeter, & area

We're told the triangle ABC has perimeter P and inradius r and then they want us to find the area of ABC in terms of P and r. So we know that the perimeter is just the sum of the sides of the triangle, or how long a fence would have to be if you wanted to go around the triangle. And let's just remind ourselves what the inradius is. If we take the angle bisectors of each of these vertices-- each of these angles right over here. So bisect that right over there and then bisect that right over there. This angle is going to be equal to that angle. This angle is going to be equal to that angle and then this angle is going to be equal to that angle there. And the point where those angle bisectors intersect, that right over there, is our incenter and it is equidistant from all of the three sides. And the distance from those sides, that's the inradius. So let me draw the inradius. So when you find the distance between a point and a line, you want to drop a perpendicular. So this length right over here is the inradius. This length right over here is the inradius and this length right over here is the inradius. And if you want, you could draw an incircle here with the center at the incenter and with the radius r and that circle would look something like this. We don't have to necessarily draw it for this problem. So you could draw a circle that looks something like that. And then we'd call that the incircle. So let's think about how we can find the area here, especially in terms of this inradius. Well, the cool thing about the inradius is it looks like the altitude-- or this looks like the altitude for this triangle right over here, triangle A. Let's label the center. Let's call it I for incenter. This r right over here is the altitude of triangle AIC. This r is the altitude of triangle BIC. And this r, which we didn't label, that r right over there is the altitude of triangle AIB. And we know-- and so we could find the area of each of those triangles in terms of both r and their bases and maybe if we sum up the area of all the triangles, we can get something in terms of our perimeter and our inradius. So let's just try to do this. So the area of the entire triangle, the area of ABC, is going to be equal to-- and I'll color code this-- is going to be equal to the area of AIC. So that's what I'm shading here in magenta. It's going to be equal to the area of AIC plus the area of BIC, which is this triangle right over here. Actually let me do that in a different color. I've already used the blue. So let me do that in orange. Plus the area of BIC. So that's this area right over here. And then finally plus the area-- I'll do this in a, let's see, I'll use this pink color-- plus the area of AIB. That is the area AIB. Take the sum of the areas of these two triangles, you got the area of the larger triangle. Now AIC, the area of AIC, is going to be equal to 1/2 base times height. So this is going to be 1/2. The base is the length of AC, 1/2 AC times the height-- times this altitude right over here, which is just going to be r-- times r. That's the area of AIC. And then the area of BIC is going to be 1/2 the base, which is BC, times the height, which is r. And then plus the area of AIB, this one over here, is going to be 1/2 the base, which is the length of this side AB, times the height, which is once again r. And over here, we can factor out a 1/2 r from all of these terms and you get 1/2 r times AC plus BC plus AB. And I think you see where this is going. Plus-- now that's a different shade of pink-- plus AB. Now what is AC plus BC plus AB? Well that's going to be the perimeter, P, if you just take the sum of the sides. So that is the perimeter of P and it looks like we're done. The area of our triangle ABC is equal to 1/2 times r times the perimeter, which is kind of a neat result. 1/2 times the inradius times the perimeter of the triangle. Or sometimes you'll see it written like this. It's equal to r times P over s-- sorry, P over 2. And this term right over here, the perimeter divided by 2, is sometimes called the semiperimeter. And sometimes it's denoted by s so sometimes you'll see the area is equal to r times s, where s is the semiperimeter. It's the perimeter divided by 2. I personally like it this way a little bit more because I remember that P is perimeter. This is useful because obviously now if someone gives you an inradius and a perimeter, you can figure out the area of a triangle. Or if someone gives you the area of the triangle and the perimeter, you can get the inradius. So if they give you two of these variables, you can always get the third. So for example, if this was a triangle right over here, this is maybe the most famous of the right triangles. If I have a triangle that has lengths 3, 4, and 5, we know this is a right triangle. You can verify this from the Pythagorean theorem. And if someone were to say what is the inradius of this triangle right over here? Well we can figure out the area pretty easily. We know this is a right triangle. 3 squared plus 4 squared is equal to 5 squared. So the area is going to be equal to 3 times 4 times 1/2. So 3 times 4 times 1/2 is 6 and then the perimeter here is going to be equal to 3 plus 4, which is 7, plus 5 is 12. And so we have the area. So let's write this area is equal to 1/2 times the inradius times the perimeter. So here we have 12 is equal to 1/2 times the inradius times the perimeter. So we have-- oh sorry, we have 6. Let me write this in. The area is 6. We have 6 is equal to 1/2 times the inradius times 12. And so in this situation, 1/2 times 12 is just 6. We have 6 is equal to 6r. Divide both sides by 6, you get r is equal to 1. So if you were to draw the inradius for this one, which is kind of a neat result. So let me draw some angle bisectors here. This 3, 4, 5 right triangle has an inradius of 1. So this distance equals this distance, which is equal to this distance, which is equal to 1, which is kind of a neat result.