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Area & perimeter word problem: table

Sal solves for the dimension of a table. Created by Sal Khan.

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  • primosaur tree style avatar for user obeaver
    What if the length and width are decimals?
    (3 votes)
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  • starky sapling style avatar for user Jaelynn Hopkins
    I dont get any of this and ive been working on it for weeks
    (8 votes)
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  • blobby green style avatar for user Maryam Yusuf
    how did u get 20 for the width and 24 for the length
    (4 votes)
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    • male robot hal style avatar for user Jesse
      There is nowhere in the video that those assignments are made. In the problem statement, the perimeter is 20 feet, and the area is 24 square feet. Also, at roughly , Sal has the equations you will need to find length and width.
      (2 votes)
  • blobby green style avatar for user Phish255
    A rectangular field is enclosed by 320 feet of fencing. If the length of the field is 6 feet more than its width, what is its length, in feet?
    (4 votes)
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    • duskpin ultimate style avatar for user Polina Vitić
      First, pick variables for each side and the perimeter:
      l = length
      w = width
      p = perimeter

      We know that the outside of the rectangle (or perimeter) is w + l + w + l, and we also know that it is 320 feet:
      p = w + l + w + l
      p = 2w + 2l
      320 = 2w + 2l

      We also know that the length is 6 more than the width:
      l = 6 + w

      So now we can plus this value for l into the equation we got above to solve for w:

      320 = 2w + 2l
      320 = 2w + 2(6 + w)
      320 = 2w + 12 + 2w
      320 = 4w + 12
      etc.

      Once you have solved for w, you can plug that value back into this equation to find l:
      l = 6 + w

      Finally, double check that the values you found for l and w are valid by plugging them into this equation:
      320 = 2w + 2l
      (Double w plus double l should equal 320.)

      Hope this helps!
      (11 votes)
  • aqualine seedling style avatar for user Graham
    why does this cost me energy points
    (3 votes)
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  • blobby green style avatar for user rkshpathak01
    if i took a wire of 44m and bent it into a shape of a circle.how i find its radius and area of circle.if i bent same wire in the shape of square what is the measure of each side
    (3 votes)
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    • stelly blue style avatar for user Kim Seidel
      If your wire is 44m and it has been bent into a circle, then the 44m is the circumference of the circle.
      Use the formula for circumference to find the radius.
      Circumference = 2 Pi r; or 44 = 2 Pi r
      Divide 44 by 2 Pi, and you get r.

      To find the area, use your value of r in the area formula for a circle: Area = Pi r^2

      If your wire is bent into a square, then the 44 becomes the perimeter of the square. Use the formula for perimeter to find the length of one side: Perimeter = 4x where "x" is the length of 1 side. Or, with your info: 44 = 4x

      See if you can finish the calculations. If you get stuck, comment back.
      (3 votes)
  • marcimus pink style avatar for user Sam
    Can't the width and the length be switched and you get the same answer?
    (2 votes)
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  • blobby green style avatar for user jvictor
    I don't get it
    (4 votes)
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  • duskpin seedling style avatar for user Ethan186
    how do you do this i undrestand the firts part but not the second
    (2 votes)
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    • duskpin ultimate style avatar for user Meriberry
      He was trying to find the length and width using a guess-and-check method. What I mean by that is he was trying to find two numbers that multiplied to make 24 square feet, and the same two numbers plus the same to numbers again to make 20 feet. 1 x 24 = 24, but 1+24+1+24 did not equal 20. Same with 2 and 12, and 3 and 8. However 4 and 6 met both qualifications, so 4 and 6 are the length and width. Does that make sense?
      (3 votes)
  • duskpin seed style avatar for user talamohammad
    what if the length and the width are the same? do we have to do something diffrent
    (3 votes)
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Video transcript

Charles built a rectangular table that has a perimeter of 20 feet and an area of 24 square feet. The table is longer than it is wide. What are the length and width of the table? The length and width are whole numbers. So it's longer than it is wide. So let's draw this table here. So the table might look something like this-- where this dimension right over here is the length. So this distance right over here is the length. We could also write length here if we want to show that this is the same, these two sides have the same length. And then we could call this dimension right over here-- this is the width. And this is also, of course, the width, as well. This is a rectangle. So these two sides are going to be the same. Now, they tell us that the perimeter is 20 feet, which is another way of saying that the width plus the width plus the length plus the length is equal to 20. And they tell us that the area is 24 square feet. That's another way of saying that the width times the length is going to be 24. So we could write that down. Width times the length is going to be equal to 24. Now, there's a lot of ways to solve this problem. And later on when you learn more algebra, there's fancy algebraic ways to do that. But we won't have to resort to this. They tell us that the length and width are whole numbers. So we really should just be able to try out some numbers, because we know that the width times the length is 24. So we just have try out all of the whole numbers that when I were to take their product, I get to 24. Essentially, the factors of 24, and then figure out which of those satisfy the perimeter up here. Where if I take the width plus the width-- essentially 2 times the width plus 2 times the length-- I am going to get to 20. So let's figure that out. So let me make two columns here. So one column, I'm going to call it a width column. Another column I'm going to call it length. And then I'm going to write down the perimeter. I'll write perimeter. I'll just shorten it with per-- maybe peri. I'll just write out the whole word, perimeter. And then let's write out area. Actually, let's do all that-- I could write it just like that. Let me try out-- make it a table here. So I have a table here. And then I can try things out. And what we can do is just make sure that everything we try out has an area of 24 square feet. So let's just think about the factors of 24. Well, it could be 1 and 24. So this literally could be 1, a width of 1, and a length of 24. 1 times 24 is 24. And they tell us that the length is longer than the width, that the table is longer than it is wide. So we want the larger number under length. So let's see, 1 times 24 is 24. But what is 1 plus 1 plus 24 plus 24? Well, that's going to be 2 plus 48, which is 50. So this doesn't meet our condition that the perimeter is 20. So let's cross that out. So this one right over here does not work out. Let's try the other factors of 24. It could be 2 and 12. Once again, 2 times 12 is 24. But what's 2 plus 2? It's 4, plus 12 plus 12. So it's 4 plus 24. That's going to be 28. Well, that doesn't meet our perimeter constraint. So we can-- that's not going to be right. Well, what about-- let's see, 3 times 8 is also equal to 24. And what's 3 plus 3-- is 6, plus 8 plus 8 is 16. 6 plus 16 is 22. Well, we're getting close, but it's still not a perimeter of 20. So that's not going to be right. Now, what about 4 and 6? Once again, 4 times 6 is 24. And what's 4 plus 4 plus 6 plus 6? Well, that's 8 plus 12, which is indeed equal to 20. So that works out. Our width is going to be 4 feet, and our length is going to be 6 feet.