math help, please show me how to do this?

Question

Can someone show me the steps that are involved in solving this problem??
A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 20ft by 28ft. The total area, garden plus walkway, is to be 1100ft^2. What must be the width of the walkway? (Use a quadratic equation)

Answer 1

if the width of the garden is 28ft then the width of the walkway is 27ft.
THis is because:
20 * 28 = 560
560sqft for the garden plus 560 sqrft for the walkway come out to 1120sqft. So we are over by 20 sqft. Therefore, you deduct 1 foot of width to get the answer of 1100sqft. The walkways has the dimensions of 27ft by 20ft.

Answer 2

Here are the basic steps to solving this. First, you have to visualize what is really happening. So on a sheet of paper draw a rectangle that is wider than high. Now around the outside of that draw another rectangle that has even spacing on both sides and top. That is the picture from 100 feet of what you are working with.

Lets see what we know about this so far.

1. We know that the area of a rectangle is Length * Width.

2. We know that we have the dimensions of the inner rectangle and we only know that the space around it is even. So using our algebra knowledge, we can call this distance "X".

3. We know that the total area is 1100 sqare feet.

So we can set up some basic statements about this.

1. The area of the inner rectangle is 28*20 or 560 square feet.

2. The length of the outer rectangle is 28 + 2x (one side is just "x", but remember the sides are even and there are two sides so we have to use "2x").

3. The height of the outer rectangle is 20 + 2x (same logic and since the width of the space around the inner rectangle is the same on all four sides we can call this amount "X" also.)

4. We can now set up a formula for the outer rectangle as follows.
length * width = area
(28 + 2x) * (20 + 2x) = 1100

5. That formula becomes
560 + 96x + 4x^2 = 1100 (using FOIL)

6. re-written into quadratic equation order you now have
4x^2 + 98x + 560 = 1100

7. Which becomes
4x^2 + 98x - 540 = 0

8. Use the quadratic formula to solve.

If you require additional help solving this you can go to the link below. Good luck.

Answer 3

(20 + 2x) * (28 + 2x) = width * length of 'total area'

560 + 40x + 56x + 4x^2 = 1100ft^2

560 + 96x + 4x^2 = 1100

& solve as a quadratic for x...

I hope you understand why the width is '20 + 2x'... it is the width of the garden plus two widths of the paths (which I've called x)...

Same goes for length, length of garden plus two widths of path..

Tidy up quadratic by making it 'equal zero' (get all values on one side of the 'equals' sign), once there you should know the rest...

answer is 'around' 4.5 ft for the width of path...

Whatever the guy who wrote the post below is doing, it's not Mathematics..

Steve (below again) is solving another problem entirely...

I've actually saved this page as of a moment ago... it will feature in a presentation on Mathematics & Culture that I'm giving next summer...

Answer 4

lp= total length of path in feet
wp = width of path in feet
lg = length of garden in feet
wg = width of garden in feet

(wp * lp) + (wg * lg) = 1100
wp * lp = 1100 - (wg * lg)
wp *(20+28+20+28) = 1100 - (20*28)
wp = (1100 - 20*28)/(20+28+20+28)
wp = 540/96 = 5.625 or (5 feet and 7 1/2 inches)