average dining room table size
charles built arectangular table that has a perimeter of 20 feetand an area of 24 square feet. the table is longerthan it is wide. what are the lengthand width of the table? the length and widthare whole numbers. so it's longer than it is wide. so let's draw this table here. so the table mightlook something like this-- where this dimensionright over here is the length.
so this distance rightover here is the length. we could also writelength here if we want to show thatthis is the same, these two sides havethe same length. and then we could call thisdimension right over here-- this is the width. and this is also, ofcourse, the width, as well. this is a rectangle. so these two sides aregoing to be the same.
now, they tell usthat the perimeter is 20 feet, which isanother way of saying that the width plus the widthplus the length plus the length is equal to 20. and they tell us that thearea is 24 square feet. that's another way of sayingthat the width times the length is going to be 24. so we could write that down. width times the length isgoing to be equal to 24.
now, there's a lot of waysto solve this problem. and later on when youlearn more algebra, there's fancy algebraicways to do that. but we won't haveto resort to this. they tell us that the lengthand width are whole numbers. so we really should just beable to try out some numbers, because we know that thewidth times the length is 24. so we just have try outall of the whole numbers that when i were to taketheir product, i get to 24.
essentially, thefactors of 24, and then figure out which of thosesatisfy the perimeter up here. where if i take the width plusthe width-- essentially 2 times the width plus 2times 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 goingto call it a width column. another column i'mgoing to call it length.
and then i'm going towrite down the perimeter. i'll write perimeter. i'll just shorten itwith per-- maybe peri. i'll just write out thewhole word, perimeter. and then let's write out area. actually, let's do all that-- icould 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 makesure that everything we try out has an area of 24 square feet. so let's just thinkabout the factors of 24. well, it could be 1 and 24. so this literally could be 1, awidth of 1, and a length of 24. 1 times 24 is 24. and they tell us that thelength is longer than the width, that the table islonger than it is wide. so we want the largernumber under length.
so let's see, 1 times 24 is 24. but what is 1 plus1 plus 24 plus 24? well, that's going to be2 plus 48, which is 50. so this doesn'tmeet our condition that the perimeter is 20. so let's cross that out. so this one right overhere does not work out. let's try the otherfactors 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 meetour perimeter constraint. so we can-- that'snot going to be right. well, what about-- let's see,3 times 8 is also equal to 24. and what's 3 plus 3-- is6, plus 8 plus 8 is 16.
6 plus 16 is 22. well, we're gettingclose, 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 plus4 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 be4 feet, and our length is going to be 6 feet.