Math homework help??

Question

I'm really confused on some problems I'm having a test on this week!! Any input on any problem would be great .. i just need them explained!!!!

1. a rectangle is three times as long as it is wide. If its length and width are both decreased by 2 cm, its area is decreased by 36cm(squared). Find its original dimensions. Make a sketch.

2. A small city park consists of a rectangular lawn surrounded on all sides by a 330 m(squared) border of flowers 2.5 m wide. Find the area of the lawn if the entire park is 5 m longer then it is wide.

Note: cm & m are centimeters and meters!

Also- if there are any sites you think could help me understand this sorta thing, please give it to me! I need a lot of help!

Thanks in advance!!!!

note:: the probelms aren't related at all they are two totally different ones.

Answer 1

ok, let's talk through them:

1. so mathematically speaking: the area of a rectangle is the length of the sides multiplied together. we'll use x and y. sketched, it would look something like

--------x---------
y| |y
---------x--------

area = x * y (* = multiply)

length is 3 times as the width means x = 3 * y

so area = (3 * y) * y = 3 y^2

so if the length (x) is decreased by 2, or [3y - 2]

and width (y) is decreased by 2, or [y - 2]

then the new area is decreased by 36, or
[newarea = area - 36]

remember that area = 3 y^2 so area - 36 = (3 y^2) - 36
also remember that newarea = new length * new width = (3y - 2)*(y - 2)

so all of that becomes
newarea = 3 y^2 - 36 = (3y - 2)*(y - 2)

so multiplying everything out on the right side gives you:
3y^2 - 36 = 3y^2 - 6y - 2y + 4

now solve for y and you get:
- 36 = -8y + 4
8y = 40
y = 5

remember x = 3y so x = 3*5 or 15

2. is very similar to 1.
rectangular lawn means area = x * y again

so the border trim of flowers (if i'm reading you right) is altogether 330m^2. the border trim is 2.5m wide.

your border trim area is is 4 little rectangles - see if you can draw them out based on what i write:

2 of them are the (width of rectangle) y * 2.5m (width of flower bed) or
= 2 * (y * 2.5) = y * 5

and the other 2 areas are (length of rectangle) x - 2.5m from both sides (or 5m because they were counted in the other two rectangles) times 2.5m width
= 2 * (x - 5) * (2.5) = 5x - 25

so all of this = 330m^2

but in order to only solve for one variable at a time, we have to look at the fact that the lawn is 5m longer than it is wide, or

x = y + 5

so put it all together and you get:

5y + 5x - 25 = 330
5y + 5(y+5) -25 = 330
10y + 25 - 25 = 330
10y = 330
y = 33

x = 33 + 5 = 38

Hope this all makes sense to you. Have fun.

Answer 2

hay there...
First of all, that homework is very hard and it took me more than 20 minutes to solve only the first question...
If you assume that the length is x one way to solve is by solving the following equation:
(x-2)(3x-2)=(x^2) – 36 but you won’t be able to get real numbers after solving it.
Therefore the only solution is to do the following:
Try to get the equation for the removed part from rectangle which will be an L shaped with thickness of 2 cm, length of 3x and width of x. Therefore the equation for the area for that shape is
2x + 6x = 40 and by solving the equation you get x = 5 which is the width and therefore the length is 3x which is 15..

try it: the area of 5 x 15 rectangle is 75
and if we decrease 2 frrom both it will be 3 x 13 which is 39
75 - 39 = 36

I did my best to try to explain it, but if you didn't get it, send me an e-mail and i'll write the solution on paper and scan it to send it to you by e-mail...

I'll try solving the second one

Answer 3

i'll do part 1:
a rectangle is three times as long as it is wide has equation:
3W = L (1)

length and width are decreased by 2 , its area is decreased by 36 has equation:
(W-2)(L-2) = LW - 36 (2)

So sub. 1 into 2 to find W
3W^2 - 2W - 6W + 4 = 3W^2 - 36
8W = 40

W = 5
therefore L = 15 from eq.(1)

Answer 4

1. think the width is x and length is 3x. and im not too good with algerbra but u mite be able to figure it out.
2. um, im confuzzled by the question's altitude to lawns and flowers.