# Maximum area of a fenced-in half a circle

• Mathman2013
In summary: You are allowed to ignore the 2*radius in the expression for the perimeter of the semi-circle because the fence only needs to surround the semi-circle and not the entire enclosed area. Therefore, in summary, you can ignore the 2*radius in the expression for the perimeter of the semi-circle.
Mathman2013
. Homework Statement

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:
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.

Mathman2013 said:
. Homework Statement

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:

## 1. How is the maximum area of a fenced-in half circle calculated?

The maximum area of a fenced-in half circle is calculated by finding the radius of the circle, dividing it by two to get the radius of the half circle, and then using the formula A = πr^2/2 to find the area.

## 2. Why is the maximum area of a fenced-in half circle important in scientific research?

The maximum area of a fenced-in half circle is important in scientific research as it can be used to optimize space usage and efficiency in experiments or simulations involving circular areas.

## 3. Can the maximum area of a fenced-in half circle be calculated for any size of circle?

Yes, the maximum area of a fenced-in half circle can be calculated for any size of circle as long as the radius is known.

## 4. Are there any real-world applications of the maximum area of a fenced-in half circle?

Yes, the concept of the maximum area of a fenced-in half circle can be applied in urban planning to determine the most efficient use of land for circular structures such as parks or roundabouts.

## 5. How does the maximum area of a fenced-in half circle relate to other geometric concepts?

The maximum area of a fenced-in half circle is related to other geometric concepts such as optimization and the Pythagorean theorem, which can be used to find the radius of a half circle. It is also a special case of the more general concept of maximum area for any given shape.

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