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Introduction  ## Introduction

The fencing problem involves finding a solution for a farmer who wants to fence a levelled piece of land. The farmer has exactly 1000m of fencing therefore he wants to achieve the maximum possible area. This will require me to test all shapes and see which shape gives the biggest area. The main shapes that I will look at are the following;

1. Rectangles
2. Trapeziums
3. Parallelograms
4. Kites
5. Triangles
6. Polygons greater than 4 sides
7. Circles

Rectangles

In geometry, a rectangle is defined as a quadrilateral where all four of its angles are right angles.

From this definition, it follows that a rectangle has two pairs of parallel sides of equal length; that is, a rectangle is a parallelogram. A square is a special kind of rectangle where all four sides have equal length; that is, a square is both a rectangle and a rhombus.

The area of a rectangle is the product of its length and its width; in symbols,  . For example, the area of a rectangle with a length of 5 and a width of 4 would be 20, because 5 × 4 = 20.
I first started by drawing two rectangles and finding out their area using the formula: A table will be best to record my results

 Length=L (m) Width=500-L (m) A=LxW (m)2 0 500 N/A 50 450 22500 100 400 40000 150 350 52500 200 300 60000 250 250 62500 300 200 60000 350 150 52500 400 100 40000 450 50 22500 500 0 N/A 249.99 250.01 62499.9999 250.99 249.01 62499.0199

Middle  =h

4  Apply parallelogram formula for area   It is clear that a table I necessary as it will allow me to work out the best parallelogram in the quickest manner. Because I am dealing with angles here, I will need to create many tables to test out all possible combinations of lengths, widths and angles.

 Angle 30° 60° 90° Length m 450 450 450 Width m 50 50 50 Height m 25 43.3 50 Area m² 11250 19485 22500 Angle 30° 60° 90° Length m 400 400 400 Width m 100 100 100 Height m 50 86.6 100 Area m² 20000 34640 40000 Angle 30° 60° 90° Length m 350 350 350 Width m 150 150 150 Height m 75 129.9 150 Area m² 26250 45465 52500
 Angle 30° 60° 90° Length m 300 300 300 Width m 200 200 200 Height m 100 173.2 200 Area m² 30000 51960 60000
 Angle 30° 60° 90° Length m 250 250 250 Width m 250 250 250 Height m 125 216 250 Area m² 31250 54125 62500

The highlighted cells again show the shape that gives the highest. This shape is a square thus again proving my prediction that the square gives the biggest area so far.

Shown Geometrically  Transition Giving biggest area

Kites

In geometry, a kite, or deltoid, is a quadrilateral with two pairs of equal adjacent sides. Technically, the pairs of sides are disjointcongruent and adjacent. This is in contrast to a parallelogram, where the equal sides are opposite.

I predict that the closer the shape gets to a square, the bigger area it will provide. The general formula for calculating the area for a kite is:

• If d1 and d2 are the lengths of the diagonals, then the area is Conclusion

The area of a circle is calculated using this formula: The circumference of a circle is calculated using this formula: Since there is only 1 circumference possible we will only be able to get one area.

We don’t know the radius but because the perimeter, The formula can be rearranged to give us the radius:   Plug r into area of a circle formula: Simplify:    The area of a circle with a circumference of a 1000m is approximately, 79577.47155m². So far the circle gave us the largest area.

This is because the circle has infinite number of sides and comparing the two formulas shows that if n was to be infinity it would have the biggest area.    Proving Geometrically  Transaction Giving us a bigger area

The proof above shows that as number of sides tends to infinity the area of the polygon will tend to that of a circle’s. Therefore, when n is an infinite number of sides, its area will be equivalent to that of a circles and this the largest possible area for the given perimeter of a 1000m.

In conclusion; if the farmer wanted the shape with the maximum area and a perimeter of 1000m, he ought to fence his land in the shape of a circle. | The Fencing Problem

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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Opp = 500 ( n Tan = 180 ( n Height = Opp ( Tan Height = 500 ( n ( tan (180 ( n) This means that I can find the height of any triangle that is part of a regular polygon split into equal triangles with a perimeter of 1000m. • Over 160,000 pieces
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