Maximum area of a fenced-in half a circle

  • #1
. Homework Statement

ujdAu5z
Let imagine you have gras field formed as a semi circle and you want to fence in that area.

The fence is connected to a wall, so you only have to fence in the area formed by the semi-circle.

You have to use 60 meters of fence bought at a hardware store.

Homework Equations


Area of semi-circle: A = 1/2*Pi*radius^2

Length of the perimeter of a semi-circle. L = Pi*radius + 2*radius

The Attempt at a Solution



And here is where it become tricky for me.

I know that part of the area which needs to be fenced in is the part formed by the semi-circle and excluding the walled area.

However if the length of the perimeter is L = Pi*radius + 2*radius

Am I allowed to ignore the 2*radius ? In my expression for the perimeter of the semi-circle ? (Because L = Pi*radius + 2*radius how you define length of the perimeter of semi circle according to my textbook. )

Yes or no? Because If I am allowed to ignore it then I get an area of the gras field which is 572 m^2, but if I include the 2*radius in the expression for the Length of the perimeter fence then the area of field is only 212 m^2.

Hope someone here can bring light to my question :)
 
Last edited by a moderator:

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
11,667
4,675
Can you post a picture showing the wall and the grass semicircle? If the wall is along the diameter of the semicircle, then you don't need the 2*radius.
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
. Homework Statement

ujdAu5z
Let imagine you have gras field formed as a semi circle and you want to fence in that area.

The fence is connected to a wall, so you only have to fence in the area formed by the semi-circle.

You have to use 60 meters of fence bought at a hardware store.

Homework Equations


Area of semi-circle: A = 1/2*Pi*radius^2

Length of the perimeter of a semi-circle. L = Pi*radius + 2*radius

The Attempt at a Solution



And here is where it become tricky for me.

I know that part of the area which needs to be fenced in is the part formed by the semi-circle and excluding the walled area.

However if the length of the perimeter is L = Pi*radius + 2*radius

Am I allowed to ignore the 2*radius ? In my expression for the perimeter of the semi-circle ? (Because L = Pi*radius + 2*radius how you define length of the perimeter of semi circle according to my textbook. )

Yes or no? Because If I am allowed to ignore it then I get an area of the gras field which is 572 m^2, but if I include the 2*radius in the expression for the Length of the perimeter fence then the area of field is only 212 m^2.

Hope someone here can bring light to my question :)

You do not need anybody here to tell you: the answer is given right in the question itself.
 
Last edited by a moderator:

Related Threads on Maximum area of a fenced-in half a circle

Replies
10
Views
18K
  • Last Post
Replies
1
Views
3K
Replies
1
Views
2K
Replies
3
Views
1K
Replies
4
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
11
Views
6K
Replies
22
Views
19K
Top