SOLUTION: I'm having trouble with this one! A rectangular garden has dimensions 3m by 4m. A path is built around the garden. The area of the garden and path is 6 times as great as the area o

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I'm having trouble with this one! A rectangular garden has dimensions 3m by 4m. A path is built around the garden. The area of the garden and path is 6 times as great as the area o      Log On


   

Question 712636: I'm having trouble with this one! A rectangular garden has dimensions 3m by 4m. A path is built around the garden. The area of the garden and path is 6 times as great as the area of the garden. What is the width of the path?
Found 3 solutions by jim_thompson5910, josgarithmetic, nerdybill:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The area of the garden is 3*4 = 12 square meters.

Multiply this by 6 to get 12*6 = 72. We do this because "The area of the garden and path is 6 times as great as the area of the garden"

So the area of the garden+path = 72 square meters

So if x = width of the path, then the width goes from 3 to 3+2x meters. This is because you're adding x to each direction on the width. Similarly, the length jumps from 4 to 4+2x meters

The area of the garden+path in algebraic form is

%283%2B2x%29%284%2B2x%29

but this is 72 square meters which is what we found earlier, so

%283%2B2x%29%284%2B2x%29+=+72

Solve for x

%283%2B2x%29%284%2B2x%29+=+72

12+%2B+6x+%2B+8x+%2B+4x%5E2+=+72

12+%2B+6x+%2B+8x+%2B+4x%5E2+-+72+=+0

4x%5E2+%2B+14x+-+60+=+0

2%282x%5E2+%2B+7x+-+30%29+=+0

2x%5E2+%2B+7x+-+30+=+0%2F2

2x%5E2+%2B+7x+-+30+=+0

Now use the quadratic formula to solve for x

x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%287%29%2B-sqrt%28%287%29%5E2-4%282%29%28-30%29%29%29%2F%282%282%29%29 Plug in a+=+2, b+=+7, c+=+-30

x+=+%28-7%2B-sqrt%2849-%28-240%29%29%29%2F%284%29

x+=+%28-7%2B-sqrt%2849%2B240%29%29%2F%284%29

x+=+%28-7%2B-sqrt%28289%29%29%2F4

x+=+%28-7%2Bsqrt%28289%29%29%2F4 or x+=+%28-7-sqrt%28289%29%29%2F4

x+=+%28-7%2B17%29%2F4 or x+=+%28-7-17%29%2F4

x+=+10%2F4 or x+=+-24%2F4

x+=+5%2F2 or x+=+-6

Toss out the negative solution since a negative width of the path makes no sense at all.

So the only reasonable answer is 5/2 = 2.5 meters

Therefore, in the end, the width of the path is 2.5 meters

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Enough information about garden, path around, how they relate; and we assume that except for the corners, width of the path around garden is uniform.
In case you have trouble imagining some of this, try drawing a picture.


Area of garden: 3%2A4 m^2.
Area of garden plus path: %283%2B2w%29%284%2B2w%29 m^2
Area of garden plus path is 6 times area of garden alone:
%283%2B2w%29%284%2B2w%29=6%2A3%2A4+


That should be enough to understand the solution process for this example. w is the width of the path. Solve for w.

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangular garden has dimensions 3m by 4m. A path is built around the garden. The area of the garden and path is 6 times as great as the area of the garden. What is the width of the path?
.
Let x = width (m) of path
then
area of "path+garden" = (2x+3)(2x+4)
area of "garden" = 3*4 = 12
.
From:"The area of the garden and path is 6 times as great as the area of the garden." we get our equation:
(2x+3)(2x+4) = 6*12
4x^2+8x+6x+12 = 72
4x^2+14x+12 = 72
4x^2+14x-60 = 0
2x^2+7x-30 = 0
rewrite middle term:
2x^2+12x-5x-30 = 0
group:
(2x^2+12x)-(5x+30) = 0
2x(x+6)- 5(x+6) = 0
(x+6)(2x-5) = 0
x = {-6, 5/2}
we can throw out the negative solution (extraneous) leaving:
x = 5/2
or
x = 2.5 m