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## 6th grade foundations (Eureka Math/EngageNY)

### Unit 5: Lesson 1

Topic A: Foundations- Area & perimeter word problem: table
- Area & perimeter word problem: dog pen
- Area & perimeter of rectangles word problems
- Find a missing side length when given perimeter
- Classify shapes by line and angle types
- Worked example: Classifying triangles
- Classify triangles by angles
- Classify triangles by side lengths

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# Area & perimeter word problem: table

CCSS.Math:

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

## Want to join the conversation?

- What if the length and width are decimals?(3 votes)
- If the length and width are decimals, you multiply them anyway to get the right answer. You could convert them into fractions, but that's a lot harder to do. Fractions are another story...(2 votes)

- I dont get any of this and ive been working on it for weeks(8 votes)
- how did u get 20 for the width and 24 for the length(4 votes)
- 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 roughly1:37, Sal has the equations you will need to find length and width.(2 votes)

- 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)
- 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)

- why does this cost me energy points(3 votes)
- watching videos gives you energy points and energy points gives you badges(2 votes)

- 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)
- 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)

- Can't the width and the length be switched and you get the same answer?(2 votes)
- Yes you will have the same result but the longer side is always the length and the width is always the shorter side. So no you can not switch the sides.(8 votes)

- how do you do this i undrestand the firts part but not the second(2 votes)
- 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)

- what if the length and the width are the same? do we have to do something diffrent(3 votes)
- nope u still do the same thing(1 vote)

## 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.