I have a really simple math question, but one that I can’t seem to solve due to my pitiful math skills. It goes: There’s a rectangle that is 3m wide by 4m long. It is surrounded by a sidewalk of uniform width. Now if this sidewalk is 48m^2 what are the dimensions of the sidewalk?
Thanks in advance.
each sidewalk side is the square root of 60m, about 7.75m
oops, misread it, it’s 6m on each sidewalk side
Is there some more mathematical way you could help me solve this question?
If the sidewalk is of uniform width, the corners must be quarter circles. Four of them make up a full circle.
Calling the width, W, their area is:
PI*W^2
The other sections of sidewalk are rectangular. Two have length 4, two have length 3. Their area is:
(24 + 23)*W=14W
So, the equation you want to solve is:
PI*W^2 + 14W = 48
This is a quadratic whose positive root is 2.27.
So the width you are looking for is 2.27m
Guess I read it right the first time…
If the area of the sidewalk is 48m^2 and the area of the middle part is 12m^2 (3m x 4m = 12m^2), then the total area must be their sum, 60m^2.
Since the sidewalk is a square (“uniform width”), then the side length is the square root of 60m^2, or about 7.75m.
Area = length x width
A square’s Area = width x width (since length = width)
With a uniform sidewalk width, you can break the sidewalk into two 4xN rectangles; two 3xN rectangles; and 4 N^2 squares that total 48m^2.
Construct a equation from this and solve for N. The the outer dimention of the sidewalk will be 2N+4 by 2N+3.
The rectangle 3m by 4m has an area of 3m*4m = 12m^2.
The sidewalk has an area of 48m^2.
The entire thing has an area of 60m^2.
Since the sidewalk is of an uniform width, the dimensions must be in proportion with the original rectangle.
We can let the dimensions of the rectangle be 3x and 4x, where x is some variable.
3x*4x = 60
7x^2 = 60
x^2 = 8.57 (approx.)
x = 2.9 (approx.)
Substituting this back into the dimensions we set (3x and 4x), we get 3(2.9) = 8.7m and 4(2.9) = 11.6m.
So, the sidewalk will have dimensions of 8.7m by 11.6m.
I hope I did all this right.
We each read the problem differently. I read it from a birds’ eye view and he read it as a donut around the building.
I suppose I should’ve posted the answer, but it is 7m by 9m
I got a width of 2.68m
Interesting how daniel, puggyfish, and I all came out with different interpretations of what “uniform width” meant within the context of this problem.
While I tend to prefer so-called “word problems” where a problem must be extracted from the text, it’s frustrating when there is ambiguity in the wording. With my luck, I end up solving a completely different problem. Fortunately, my instructors were willing to accept correct solutions to different problems in such cases.
Puggyfish had the same idea of uniform width as me, but used 48 instead of 60 (you use 60 because the middle rectangle is 12)
Here is the equation:
PI*W^2 + 14W = 60
uh… never mind… ignore my responses.
Well thank you everyone for your help, I did manage to figure out the question with your help.
Puggyfish was right in the first place… and I really need to read my old math books again…
SSSS
SPPS
SSSS
The S represents the sidewalk and the R the rectangular plaza.
Break the sidewalk into a rectangular piece on each side of the plaza and one square in each corner. If the width of the sidewalk is w, then the corners each have area w[sup]2[/sup], the long pieces have area 4w, and the short pieces have area 3w. Adding up we get 4w[sup]2[/sup]+14w=48. Now plug into the quadratic formula and do the rest of the homework set your own damn self.