area
A rectangular swimming pool with dimensions of 11m and 8m is built in a rectangular backyard. The area of the backyard is 1120m^2. If the strip of yard surrounding the pool is of uniform width, how wide is the strip?
So I tried to find a diagram but it doesn't seem to make sense because the backyard could have any dimensions…so i made the backyard 28 by 40 and then the answer would be 8.5 and 16 which isn't correct.
Note you are asked for the dimensions of the room, not the area of the room. Draw a picture of the scenario. I cannot emphasize how important this is. Now assign the unknown additional room lengths to the variable $x$.
Here is one way: Put the rug in the center (does not really matter), and let $x$ be the additional floor lengths needed on all 4 sides of the rug. Now you have dimensions $16+2x$, and $21+2x$. Now take that corresponding area in terms of $x$, set it equal to two times 336, and solve for $x$. Doing it this way, your new room dimensions will be $16+2x$, by $21+2x$. Don't respond until you have drawn a picture.
We are apparently in a 2D world here, so a “box” is in fact a rectangle, which has an area and a perimeter. The most trivial answer to “why aren't the two always proportional” would be a counter-question of “why should they”.
But this is all about intuition, so imagine you start with a cardboard square of $10\times10$ units. It has a perimeter of $4\times 10=40$ units. Cut it in half and you obtain two rectangles of $10\times 5$ units, each of which has $30$ units perimeter. That's because your newly introduced cut created new perimeter in what used to be the inside, namely twice the length of the cut, i.e. $2\times10=20$ units. OK, two rectangles are not one rectangle, but if you glue them together along their short edge, you loose $2\times5=10$ units of perimeter which now are glued and therefore contained in the inside, but you still have $2\times(5+20)=50$ units of perimeter; more than what you started with. So in this example, the cardboard is the “stuff” whose amount stays fixed, and the perimeter depends on the shape, can be gained by cutting and lost by gluing.
But there is the opposite case as well. Instead of a cardboard square, take a piece of string, $40$ units of length. If you form a square, you will get the same square as above, covering an area of $100$ square units. But if you create other figures with the string, you get other areas. So here the perimeter is what stays fixed, and the area varies.
There are a lot of topics related to this relation between perimeter and area:
Actually, on a side note, how IS the perimeter measured, is it thought of as part of the content? Though if not, how is it measured, what is measured?
The perimeter of a planar figure is a simple case of a topological boundary and as such not part of the “content” (i.e. interior) of the shape. As I just mentioned, measuring a boundaries in real life using a ruler can lead to results which depend on the ruler. So in most branches of mathematics, the problem is more one of computing the perimeter. One common approach starts by finding a piecewise parametric description of the perimeter, then integrating over that. But it depends a lot on what kind of shape you are talking about, how it is described.
What's also fascinating is that this doesn't seem possible with circles, yet with any other object it seems that it is.
A circle has a fixed ratio between area and perimeter. So does the square. The class of all rectangles has no such ratio, but if you e.g. fix the aspect ratio of the rectangles, e.g. only consider 16:9 TV screens, then you again have a fixed ratio between area and perimeter. If you want a more flexible class containing circles, you could consider all ellipses. There again the ratio between area and perimeter depends on the actual shape (and computing the perimeter can be really hard). So it all depends on what family of shapes you consider, whether knowing the area fixes the perimeter (or vice versa) or not.
Best Answer
Let the width of the strip of yard around the pool be $w$ metres. Then the whole yard is a rectangle that is $11 + 2w$ metres long and $8 + 2w$ metres wide - you can see this because there's a strip of width $w$ on the left of the pool, and a strip of equal size on the right, and similarly for the top and bottom.
So what's the area of the yard + pool? How can you relate that to stuff you know?