 This video is called Area and Perimeter of Parallelograms 1. The instructions say to find the exact perimeter and area of the parallelogram. So if you look at this picture, you can see a parallelogram and to find the perimeter and area of the two things, well the perimeter, remember we just need to add up all four sides of my parallelogram and put a label on it and then area we learned a formula base times height and the key to this is that the base and the height are perpendicular to each other. So let's get to that. So now we kind of know what we have to do. We have to spend some time in our picture to figure everything out because in order to find the perimeter, I need to know what... I need to have more of my picture labeled. I remember back in chapter six at a parallelogram, both pairs of opposite sides are not only parallel but they're congruent. So since the diagonal on the left is... on the right is 12, the diagonal on the left also gets to be 12. Now I look at the top and the bottom and I don't have anything there so I'm going to have to do some work. Luckily, the person who made up this problem drew an imaginary line from the top and the bottom of my parallelogram where it meets the base at a 90 degree angle. So I think that imaginary line is going to be important because it's going to become the height of my whole shape. But before we even get to that, I think that imaginary line is important because it made a 45-45-90 triangle because this is marked as 45, here's 90, this is also 45. So I can use that to find the bottom side of this triangle. If I can find the bottom side, I'll also have the top. So I go back to chapter eight and I remember my 45-45-90. Opposite of my 45s are both n, opposite of my 90 is n root 2. So it looks like I have a little bit of work to do to solve for n because I have 12 equals n root 2. I'm going to start by getting that n alone. So I divide both sides by root 2. The root 2's cancel. I'm left with 12 over the square root of 2. Well, that is not a good answer because it's not proper form to have 2 in the denominator. So I'm going to rationalize the denominator. I'm going to make a new fraction that I'm going to multiply. I'm going to take root 2 and root 2. So now I can multiply across. 12 times root 2 is 12 root 2. And on the bottom, root 2 times root 2 is 4, which simplifies to 2. So now what's one more thing I can do to simplify this problem? The 2's and the 2's can't simplify because they're not like terms. One is underneath the square root, one is not. But the 12 and the 2 can to become 6 root 2. So I just found a really valuable piece of information and that is that n equals 6 root 2. So let's plug that into our picture. This bottom side became 6 root 2. And my imaginary line, the height of my shape, also became 6 root 2. Well, I think there's one other side that's 6 root 2. Remember how my left and the right side of the parallelogram were both 12? Well, the top and the bottom are going to be the same as well. Since the bottom is 6 root 2, the top is going to be 6 root 2. Now I have everything I need to find the perimeter and the area of this shape. So let's go to it. Remember, let's start with perimeter. Perimeter, you're just simply adding up all four of the outside, the distance around your shape. So it looks like we have a 12 and a 12. That's the left and the right diagonal sides and a 6 root 2 and a 6 root 2. So now I have to figure out how to combine like terms to simplify this up a little bit. Well, 12 and 12 are like terms. They're just constants. So that becomes 24. And the 6 root 2 and the 6 root 2 are like terms because both of the sixes have root 2s after them. This is where students have a little trouble is to remembering how to do 6 root 2 plus 6 root 2. I'm going to go over here. Let's practice that. If I asked you what 6x plus 6x is, most of you would easily be able to say, oh, it's 12x. So now just think of the x's as being root 2s. 6 root 2 plus 6 root 2, what does that become? Well, it becomes 12 root 2. So I've got 24 plus 12 root 2. And I would label this with units. Now be very careful here. A lot of students would go one step further and say your answer is 36 root 2. Sorry about that. A lot of students would go one step further and say their answer is 36 root 2. You cannot do that. 24 and 12 aren't considered like terms because the 24 is a constant. It doesn't have anything attached onto it or behind it. Or the 12 root 2, the 12 has a root 2 attached onto it or behind it. And remember, the rules of adding are strict. If they're not like terms, you can't combine them. Now keep in mind, if there's a multiplication sign right here, then you could. The rules of multiplication aren't as strict. So your answer for the perimeter is this 24 plus 12 root 2 units. All right, we're almost there. Now let's find the area. Remember, area formula is simply base times height where the base and the height have to be perpendicular. Well, if my base is along the bottom, it's 6 root 2. That would make my height be this imaginary vertical line because that's where it makes the 90 degrees. My height is also 6 root 2. So it looks like we have to do 6 root 2 times 6 root 2. You've gotten a lot of practice on solving things like this, but students still tend to have trouble. So let's be careful. The 6 and the 6 can multiply because they're both outside of the square root. So it's kind of like they're like terms. If you want to think about it that way, so it becomes 36. The 2 and the 2 are both underneath the square root. So they can multiply to become 4. Well, the square root of 4 is 2 and 36 times 2 is 72. So you get an answer of 72 and then we put a label on it, units squared. So this problem takes a little while. It took some thinking. We had to spend a little time using 45, 45, 90 to fill in everything for our shape. So we had to spend some time kind of finding our sidelines before we could actually find the perimeter and actually find the area. So this would be a problem on a test with quite a few points. You have to show a lot of work. Be very, very careful, but you are all very capable of doing a problem like this. You will get some practice in class tomorrow.