 Hi, this video is called Find the Area of the Shaded Region 1. You have already practiced some problems where you are trying to find the area of the shaded region. The approach will be the same. It's just going to get a little more complicated, because some of our shapes are a little more complicated. They have a greater number of sides, and we have to do things like finding the epithym and the perimeter. And that just will take a little while. It'll take some room on your sheet. So as you do this problem, be careful of your space. You're going to use every last inch of what we give you. In fact, I'm going to use two slides to get this one done. So looking at this picture, to find the area of the shaded region, it looks like if you took the area of an entire circle and then you subtract it out. So if you had a piece of paper that was a circle, and then someone came and cut out a pentagon, what you would have left is the shaded region. So we're going to do area of a circle minus the area of the pentagon. So there's really like two parts to this problem. We have to find the area of the circle, and we have to find the area of the pentagon, and then we will combine them. All right, so on the next slide, you see the same picture. So you just continue to take notes in that first slot that you have. We have two things we're going to try to figure out. The area of a pentagon, which we know has a formula of 1 half perimeter times apothem, and we are going to try to find the area of a circle. I'll write that up over here. We know that formula is pi r squared. In fact, I think I'll rewrite this up here. So we have two different problems going on. For the pentagon, I have to figure out the perimeter and the apothem. For the circle, I'm going to have to figure out what the radius is. OK, so now we're going to spend some time with our picture to get started. When I look at this picture, I see this marked as my apothem. I see a dotted line that looks like it's the radius of my pentagon, and also the radius of the circle. So this right here will become very important in finding, because I'm going to need that radius of the circle. This six is representing the edge of the pentagon. So I know all of my side lengths are going to be 6. And I actually think that helps me out a lot, because in the formula, for finding an area of the pentagon, I need the perimeter right here. So 6 times 5 will give me 30. It looks like I have a perimeter of 30. All right, the next thing I'm going to do in trying to find the area of the pentagon is find that apothem length. So I have it marked on my picture with an A. And in order to find that apothem, I've got to find my central angle measure. So I did 360 divided by 5. It got 72. So if I draw out all the radii of my pentagon, I know all these central angle measures are going to be 72 degrees. But before I do this bottom triangle of 72 degrees, I remember that my apothem cuts it in half. So I end up getting this to be 36, as well as this guy. So at this point, I'm going to take a second and redraw this triangle just a little bit bigger so it's easier for me to work with. So when I drew that right triangle, I remembered that if this angle's 36, this one would have to be 54. Because the 90, the 36, and the 54, I'll have to add up to 180. I put an A here, because that is my apothem. And then I remembered that when I dropped my apothem down, not only does the angle divide by 2, but so does the side length. So that became 3 and 3. So when I look at this triangle, I have to figure out how to solve for my apothem right here. Because once I have my apothem, I can put it in my formula. So I look at this. And I see it's not a special right triangle. It's not a 30, 60, 90. It's not a 45, 45, 90. So all I have left is Soka Toa. I'm going to decide I want to go from the perspective of the 54 degrees, which would make my apothem the opposite and the 3 my adjacent. Well, opposite and adjacent is definitely a tan. So I've got the tangent of 54 degrees equals the opposite, which is the apothem over the adjacent, which is 3. So now I have to solve this. I'm going to make the tangent of 54 into a fraction. So it's really easy to cross multiply. A times 1 is A. And 3 times the tangent of 54, I'll need to do into my calculator. Because the A is alone, I'm allowed to simply do 3 times the tangent of 54. Now when you plug that in, you're going to get a big long decimal. I got it to be, let me see what was it, 4.129145, is all I could see on my calculator. You're not allowed to round this yet. We're actually not allowed to round till the very, very end. So what are we going to do here? If the apothem is this big long number, just keep it in your calculator. You can write it down like I did, but technically you wouldn't have to. You just take this big number and put it up here for the apothem, 4.129145. So let that sit in your calculator for a minute, because you're ready to simplify. You've got the 1 half, you've got the 30, and you have the apothem. Well, you know that 1 half times 30 is the same thing as 30 divided by 2, which gives you 15. So now what you can do is since you have this big number, 4.12914 in your calculator, just hit times 15, press equal, and you should get, let me see, 61.937186. And at this point, because it's the area of my pentagon, at this point I will allow you to round. Let's go ahead and round to the nearest tenth. So it looks like it's going to stick at 61.9. So that 61.9, what did we find? It was the area of the pentagon. So in my overall problem, I'm going to replace area of the pentagon with 61.9. So it's good to remember what you started off doing. So now we're almost there. If we can just find the area of the circle, I can plug that in and we will be all set. So let's think about it. Back to our picture. Maybe I'll clean this up a little bit. Back into our picture, the radius of my circle was this dotted line. So I think I'm going to use that exact same triangle that I used before, this one right here. But now, which side do I care about finding? I care about finding that radius, which would be the hypotenuse of my circle. So I'm going to just redraw it where I have my 90 degrees. I have my apathome, my 3, the hypotenuse, which I want to find. This was 36, and this was 54. Well, now, let's, I don't know, go from the perspective of the 54 again. The 3 is your adjacent, and the radius is your hypotenuse. Well, in Sokotoa, adjacent and hypotenuse will be a cosine. So you'll do the cosine of 54 equals 3 over h. If I can find that hypotenuse, I've got my radius. All right, so let's make the cosine of 54 into a fraction by putting it over 1. So now it is really easy to cross multiply. 3 times 1 is 3. All right, so I jumped ahead a little bit. Let me back up. We said 3 times 1 is 3, and the cosine of 54 times h is the cosine of 54 times h. Since the h wasn't alone yet, I had to do a little bit of work. The cosine of 54 was multiplying the h, so I divided both sides by the cosine of 54. That makes the cosine of 54 on the right to cancel, and so it looks like we have a division problem this time. In your calculator, you do 3 divide by the cosine of 54, and I wrote 51.03. That decimal's in the wrong spot, I'm sorry. It's 5.10390485. That's as much as my calculator showed me, and at this step, it's too early to round. So remember, this big number right here, since it was the hypotenuse of my triangle, or I'm sorry, it was the hypotenuse of the triangle, it was also the radius of my circle. So that big number can go right up here. So you have pi, and then you've got 5.10390485 square. So you haven't rounded yet, so that big number should just be sitting in your calculator waiting for you to do something with it. Just hit the exponent of two button, press equal, and that is going to give you 26.0498 times pi. So again, it's still too early to round, so you've got this 26.0498 in your calculator. Keep it there. Now hit times, press the pi button, press equal, and you should get, let me see, 81.838. And that is where it would be appropriate to round. So this will become 81.8, that is the area of the circle. So I go back and remember that the area of the circle was 81.8, so that I'm ready to subtract, and I get 19.9 units squared. So the area of the shaded region is 19.9 units squared.