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# Area & perimeter word problem: table

CCSS.Math:

## 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 might look something like this where this dimension 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 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 plus the length is equal to 20 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 with width times the length 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 to try all the whole numbers that when i were to take their product i get to 24. essentially the product the factors of 24 and then figure out which of those satisfied 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's make 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 perry i'll just write out the whole word perimeter and then let's write out area 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 that the length is longer 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 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 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