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# Area & perimeter word problem: table

CCSS Math: 4.MD.A.3

## 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.