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Area and Perimeter

Up to this point, we've learned many ways to solve quadratic equations. In this lesson, we'll use those same techniques to solve problems involving area and perimeter. Now, remember area is the amount of space in square units that a closed plane figure takes up, whereas the perimeter is the distance measured around the object in units. We could be measuring in feet, in miles or even in inches. As a quick review, what are the equations for the area and the perimeter for these two rectangles? Write the expression for the area here, and write the expression for the perimeter here. Also, be sure you use the variables w and l in each of the expressions. Good luck.

Area and Perimeter

The area of this rectangle is the number of square units that occupies this space. So we could find that by multiplying the length times the width, l times w. For the perimeter, we're finding the distance around this rectangle, so we'll want to have 2w plus 2l, or I could write that in reverse order as 2l plus 2w. Either one would be correct. Nice work if you found these two expressions.

System of Equations using Area and Perimeter

These two equations can be solved as a system of equations if we know the values of the area and the perimeter. Throughout this lesson, we'll usually see the equations written like this. Here A just stands for area, which is usually a number. And P stands for perimeter, which is also a number. But this equation is pretty important, it turns out P will usually be an even number. And that usually happens, since we have two ws and two ls. So if w is odd, or if l is odd, this number will still wind up being even. So, let's take every number and divide it by 2, since the perimeter will usually be even. So if we divide p by have to work with fractions here. But keep in mind that P will usually be an even number, so P divided by two will just be an integer.

Fencing for a Patio

Let's see if we can use our new system of equations to problem solve. Here's our first problem. It would take 42 feet of fencing to enclose the rectangular patio. If the area of the patio is 108 square feet, what are the dimensions of the patio? So, here, this is my patio in black and then, this orange part is my fencing. Now, this view isn't so helpful. We're going to take a bird's eye view, and look down into our patio to figure this problem out. Essentially, our pattern looks like this. It's a rectangle in shape, and it has a length and it has a width. What we want to do is take our equations for perimeter and area. And make them into two equations with a system of two variables. In other words, we just want to have two equations with only l and w in them. Since we know it would take 42 feet of fencing to enclose our rectangular patio We know that the perimeter equals 42 feet. Notice that eh perimeter is even, so we can divide each of these by two to get a new equation that is 21 equals l plus w. We know the area of the patio is 108 square feet, so we can replace this area with the value of 108. Now, notice that we have two equations with two variables, l and w. This is a system of equations, and we've solved something like this before. We're going to go ahead and solve this system using substitution. First I'm going to subtract w from both sides of this equation to get 21 minus w equals L. So, now I want you to use substitution to solve this system of equations. We know the length equals 21w, so we can replace this length with 21 minus w. I want you to make this substitution and then solve for the width and the length. After you solve, the length should be the longer side between the two. As a hint, when you substitute 21 minus w in for l, you should wind up with something that's quadratic. Use what you know about factoring, and try and find the length and the width. Good luck here.

Fencing for a Patio

For our patio, the dimensions are 12 by 9. The length is 12 feet long, and the width is 9 feet long. Great thinking if you found these two numbers. Now, this required us to solve a quadratic equation. So don't worry if they give you trouble. We haven't even done a problem like this yet. We're going to start by taking our area equation, and we're going to replace this l with the expression replacing this l with this expression. We'll need to multiply w to each of these terms inside the parentheses. So, 21 times w equals 21w, and negative w times w is negative w squared. We need to set this equation equal to 0. So, we arrange these two terms and move them to the other side. So, we'll subtract 21w from both sides, and add w squared to both sides. This leaves us with w squared, minus 21w, plus 108 equals 0. Now we're ready to try to factor our quadratic. We want to find factors of 1 times 108. Or just 108 that sum to negative 21. Those two factors are negative 9 and negative 12. If we set each factor equal to 0, then we know that w could equal positive 9, or w can equal positive 12. And maybe this is where you might have gotten confused. We have the width can equal 9, or the width can equal 12. If the width does equal 9, we know that the entire perimeter should be 42. So doing some mental math, we know that the length would have to be 12 in order for the entire perimeter to be 42 feet. We could get that same result by using our area formula. We know this area is 108. And now we know the width. The width is 9, so we divide both sides by 9 to get the length, is equal to 12 or 12 feet. Now if we would have let the width equal 12 feet, we would have gotten this rectangle. It's really the same exact rectangle, it's just been rotated 90 degrees. The longest dimension of a closed figure or any three dimensional shape is considered to be the length. So instead of making the width equal 12, we'll make the length equal 12, and the width equal 9. Finally it's always best to check your result, to make sure that the perimeter and the area are what we expect it to be. If we add up all the sides together, we get a perimeter of 42, just like what was stated in our problem. And if we multiply the length and the width, we get an area of 108 square feet. This is why we know these two answers are correct.

Fencing for a Garden

Let's try our second problem using area and perimeter. This time we're going to fence in a garden. You want to enclose a rectangular garden with 36 feet of fencing. So, this would be the fencing around your garden. If you need 72 square feet of garden to grow all the vegetables you want to plant, what should be the dimensions of the garden? Now, I don't want to find the dimensions of the garden just yet. Let's do this problem in a couple steps. Start solving this problem by writing two equations for the perimeter and the area. Since you want to write two equations, these boxes need to have an equals sign in them. Some expression on the left, and then some expression on the right.

Fencing for a Garden

For the perimeter, we want to use the fact that we can enclose the garden with feet. So that's the perimeter. We know that 36 would equal 2l plus 2w. We have two lengths and then two widths. The other piece of information we know is that the garden is 72 square feet. That's the space we need to plant all of our vegetables. So, this area inside in the green. So we know 72 would equal the length times the width. This would be how we find the area, the product of these two dimensions. Excellent work if you found these two equations.

Fencing for a Garden Substitution

Now that we have an equation for the perimeter and an equation for the area, let's try and rearrange this equation to solve it for l. We want to solve this equation for the length, so that way, we can perform the substitution for the length in this equation. We'll be able to solve this system of equation by doing so. So take this equation and solve it for l. What do you think we would get here?

Fencing for a Garden Substitution

We should get the expression 18 minus w. So for our equation, we know 18 minus w equals l. We get this equation by taking our perimeter equation, and first, dividing each term by 2. Each of these is even, so we know we can divide by 2 to get 18 equals l plus w. And finally, we just subtract w from both sides to get 18 minus w equals l.

Fencing for a Garden Solved

So, we know this perimeter equation is really the same as 18 minus w equals l. And here, our area equation stays the same. Now we're ready to perform substitution. We know the length equals 18 minus w, so we can replace this length with this expression. Be sure you use parentheses when you replace the length with 18 minus w. And then, try and solve for the width and the length. Keep in mind that the length should be the longer side for whichever value you find. Good luck.

Fencing for a Garden Solved

The dimensions of our garden are 12 feet in length, and 6 feet in width. Nice solving if you found these two answers. Now, we're just getting warmed up, so if you didn't get it, that's okay. Let's see how we solve. First we replace this variable l with the equivalent expression 18 minus w. That goes inside parenthesis here. Next we multiply each term in the parenthesis by w. So, we'll have 18w minus w squared. Again, we have a quadratic equation here, since we have w squared. So we want to move all these terms to the left-hand side so our w squared will be positive. So we'll add w squared to both sides and we'll subtract 18w from both sides. This gives us w squared minus 18w plus 72 on the left. And zero on the right. Next, we want to factor this quadratic. So, we find factors of one times 72, equals 72, that sum to negative 18. These two factors are negative six and negative 12. We'll set each factor, equal to zero, so the width can equal six or the width can equal 12. We know the length should be 12 feet and the width should be 6 feet, since the length is the longer side. You'll notice that it doesn't matter which value that we use for w. The final result will be the same rectangular garden. It will be 12 feet by 6 feet. And as always, we can quickly check to make sure these dimensions are correct. If we go all the way around our garden, we'll have the perimeter, which is 12 plus what we expected the perimeter to be in our original problem. For the area, we multiple 12 by 6 to get 72 feet squared. This also checks, which is great, since 72 was the area of our garden originally.

Geometry Applications with Quadratics 1

Now, let's see if you have the hang of this. What do you think would be the dimensions of a rectangle that had a perimeter of 25 inches and an area of 32 inches squared? Write your answer for the length here and you answer for the width here. Notice also, that I've already included the units of inches for both dimensions. You won't need to type that in. And, keep in mind, the length is always the longer side. Good luck.

Geometry Applications with Quadratics 1

Here, the dimensions of the rectangle are 8 inches by 4 inches. Great thinking if you found these numbers. We begin solving by starting with an equation for the perimeter and the equation for the area. We know the value of the perimeter here. The perimeter is 24 inches, so we can replace P with 24. For the area equation, we know A equals l times w. We can replace it this value of A with its numerical value, 32. Now that we have these two equations, we want to solve this one for the value of l. So we can use substitution to solve. We'll divide each term by 2 to get 12 equals l plus w, and then we'll subtract w from both sides to get 12 minus w equals l. And now, here comes the substitution. We replace l with this expression, 12 minus w here. We multiply each term in the parentheses by w to get 32 equals 12w minus w squared. Then, to solve this quadratic, we move these terms to the opposite side of the equation to set it equal to 0. We factor this quadratic by finding factors at positive 32 at sum to negative 12. Those factors are negative 8 and negative 4. And finally, we set each factor equal to 0, so w can equal positive 8 or w can equal positive length would be 8. Our unit of measurement is in inches on all sides and we can quickly check to verify the perimeter and area for our problem. If we add up all the sides around the object, we'll get 24 inches. So yes, that checks with our original problem. And if we multiply the length times the width, 8 times 4 equals 32 inches squared, which was our original area. Now, it's okay if you didn't get all of this, but hopefully, you made your way at least to the factoring. Great job if so.

Geometry Applications with Quadratics 2

Here's your second problem using area and perimeter. This time, you want to construct an enclosed dog run using 46 feet of fencing. If you only have enough concrete to cover 120 square feet, what should be the dimensions of your dog run? Now, you might not even know what a dog run is, but it's just like a patio or even a rectangular garden. Try doing some research to find out what the diagram looks like. Then, see if you can create a diagram to help you problem solve. When you think you have the length and width, enter them here.

Geometry Applications with Quadratics 2

The length should be 15 feet and the width should be 8 feet. Nice problem solving for getting that one correct. To start solving, I would draw a rectangle. We'll have a width here and a length here. A dog run is just an enclosed rectangular area where a dog can play. This would be a bird's eye view or looking down from up top, into a dog run. We want to start by working on an equation for the perimeter and the area. Since we enclosed the dog run using 46 feet of fencing, we know that all of these or these sides added together would be the perimeter or 46 feet. We have enough concrete to cover 120 square feet. So the area of this rectangle, in the gray, is 120 feet squared. That's what we can use for A. Now that we have these two equations, we can start to solve our system of equations. We want to solve this equation for the length, and then substitute that expression in here. I've taken this equation and rewritten it here. We'll divide each of the terms by 2, and then we'll subtract w from both sides to get 23 minus w equals l. We don't change the equation involving the area, and now, we're ready to substitute the value of l, here, for 23 minus w. So after we substitute, we'll have the equation 120 equals the quantity 23 minus w, times w. We distribute the w to each of the terms to get this new equation. And now, we want to solve this quadratic. We add w squared to both sides and subtract 23w from both sides to get our new equation. We factor this quadratic by finding factors of a 120 that sum to negative 23, those factors are negative 15 and negative 8. This means we set each factor equal to 0, so w can equal positive 15 or w can equal positive 8. We usually consider the width to be the shorter side, so the width will be 8 feet, and the length will be 15 feet. And, as a quick check, if we add up all the sides around the rectangle, we'll get 46 feet, which was our original perimeter. And for the area, if we multiply the length times the width, we'll have 15 times 8, which equals 120 feet squared. This also checks. This means the dimensions are 15 feet by 8 feet, the length by the width.

Geometry Applications with Quadratics 3

Here's our third practice problem for geometry applications using quadratics. If it takes 42 feet of screening to enclose a sunroom at a house, and if the area of that room is 108 square feet, what are the dimensions of that room? Enter the length in feet here, and the width in feet here.

Geometry Applications with Quadratics 3

Here the length is 12 feet and the width is 9 feet. Great work if you found those. As always, we want to start by writing the two equations for perimeter and area. We know we're going to surround or enclose our sunroom with 42 feet of screening. This means that 42 must be the perimeter. The area of the room is equation to get 21 minus w equals l. We get that by dividing each of the terms by two first, and then subtracting w from both sides. Now we're ready to substitute. We replace l with it's equivalent expression, 21 minus w here. Then we distribute w to each of the terms to get 108 equals 21 w minus w squared. Next we want to set this equation equal to 0, so we'll move these two terms to the left side. So, our w squared term will end up being positive. So, adding w squared to both sides, and subtracting 21 w from both sides we'll get this equation. Next we want to factor this quadratic by finding factors of 108 that sum to negative 21. Those factors are negative 12 and negative 9. Then we set this factor equal to 0, and this factor equal to 0. So, w can equal 12, or w can equal 9. We know the width should be the shorter side, so we'll have this width as 9 feet and this length as 12 feet. And as a quick check we can add up all the sides to make sure that the sum is 42 feet, that's our perimeter. And finally we check the area by multiplying the width times the length. Or 9 times