Hi
I want to work out the surface area of my lawn. Problem is, is that it is a very convoluted shape but there are no breaks in the perimeter. Now I'm not great at maths, I'm ok with length x width m2, and LxWxH m3, but I dont know how to find the surface area of circle or the volume of a cylinder.
I'll tell you how I think I could find the area of my lawn:
Using a meter wheel I walk around the complete perimeter of the lawn until I arrive back where I started.
Say it measures 100 meters.
If I get a piece of rope exactly 100 meters long, and then join the ends together and make a big circle, could I then just work out the area of that circle using pi r squared or whatever it is to get my answer? (I would have to learn how to do the Pi thing though)
Would this be possible do you think? or am I missing something?
I would have thought that the area of this rope circle couldn't change as it is bound by the 100 meter perimeter length of the rope. So even if I made a big letter 'C' on the floor with rope the area of that C would equal the circle or any other irregular shape I made.
I'd greatly appreciate anyones help..
Thanks
Thanks guys for the replies.
Ron: I wish I had refreshed my browser 10 mins ago, I have just spent that time scratching my head as my circle area didn't match my hypothetical L shaped area doh!!
If the circle's perimeter = 100 mts and my L shape perimeter also = 100mts Why is the area of both not the same, when both their perimeters are = ?
In areas, even if the perimeter remains the same, a change in the shape of the enclosed area will result in a new area different from the original area....whether the figure is bounded by straight lines or curves.
Example, perimeter is 12m always:
area 1 ...a square whose sides are 3m each ...area = 3*3 = 9 sq.m.
area 2 ...a rectangle 4m long by 2m wide ...area = 4*2 = 8 sq.m.
area 3 ...a rectangle 5m long by 1m wide ...area = 5*1 = 5 sq. m.
It is only in volumes where even if the shape of the closed container is changed, the volume inside remains the same ....as long as the surface area of the closed container is not changed.
Umm, let us see. I was made to believe about the volumes by one of my teachers in high school and I carry that belief until now.
Sphere, radius = 1 unit:
volume = (4/3)pi(r^3) = (4/3)pi(1^3) = 4.18879 cu. units
surface area = 4pi(r^2) = 4pi(1^2) = 12.56637 sq. units
Cube of the same surface area as the sphere above:
surface area = 6(s^2) = 12.56637
s = sqrt[12.56637 /6] = 1.4472 units
volume = s^3 = (1.4472)^3 = 3.03101 cu. units
3.03101 cu. units is not the same as 4.18879 cu. units
Umm....
Meaning, the surface did not change but the volume changed.
Meaning my statement about the case in volumes is wrong.
Oh, well, we were wrong then.