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• Level: GCSE
• Subject: Maths
• Word count: 1715

# The Fencing Problem

Extracts from this document...

Introduction

THE FENCING PROBLEM

The problem:

A farmer has 1000m of fencing and wants to fence a plot of land.  It can be any shape but must have a perimeter of no more than 1000m.  I am going to investigate the effects of using different shapes to get the largest possible area using the 1000m of fence.

Triangles:

I will now investigate different types of triangles as they have the least number of sides.  Isosceles and scalene triangles will be my starting point.

Hypothesis:

The triangle’s area will increase as the triangle’s height increases.

Below are examples of different triangles.

Here is an equilateral triangle, a triangle with all angles and sides the same:

Here is a right-angled triangle, a triangle with a right angle:

Middle

48112 .52 m2.

Next, I will investigate simple four sided shapes called quadrilaterals.  I will investigate different rectangles and squares.

## Hypothesis

I predict that the closer the length and width values get to equalling each other, the larger the area will be.

Below are a square and a rectangle:

100                                              150

50             RECTANGLE                50

100                SQUARE        100

150

100

The formula to work out the area is:

Area = Length x Height

E.g.

100 x 100 = 10000

50 x 150 = 7500

This shows just how different areas of shapes can be, even with the same perimeter.

 QUADRILATERALS Height Width Area 20 480 9600 40 460 18400 60 440 26400 80 420 33600 100 400 40000 120 380 45600 140 360 50400 160 340 54400 180 320 57600 200 300 60000 220 280 61600 240 260 62400 260 240 62400 280 220 61600 300 200 60000 320 180 57600 340 160 54400 360 140 50400 380 120 45600 400 100 40000 420 80 33600 440 60 26400 460 40 18400 480 20 9600

Conclusion:

I have chosen different values of length and width and found that the biggest area is created when the lengths and widths of the sides are equal, at 250m.  This quadrilateral, with all sides the same, is a square.
Therefore, the largest area for a four-sided shape is
62500m2.

Polygons:

The square is another regular shape, like the equilateral triangle, so from now on I will use regular polygons whilst finding out how the number of sides affects the area inside.

Hypothesis:

Conclusion

 POLYGONS: n Side Angle Radiens 1/2 side 1/2 angle Height Triangle area Total area 5 200 72 0.628319 100 36 137.6382 13763.82 68819.1 6 166.6667 60 0.523599 83.33333 30 144.3376 12028.13 72168.78 7 142.8571 51.42857 0.448799 71.42857 25.71429 148.323 10594.5 74161.48 8 125 45 0.392699 62.5 22.5 150.8883 9430.522 75444.17 9 111.1111 40 0.349066 55.55556 20 152.6376 8479.869 76318.82 10 100 36 0.314159 50 18 153.8842 7694.209 76942.09 20 50 18 0.15708 25 9 157.8438 3946.095 78921.89 30 33.33333 12 0.10472 16.66667 6 158.5727 2642.879 79286.37 40 25 9 0.07854 12.5 4.5 158.8276 1985.344 79413.78 50 20 7.2 0.062832 10 3.6 158.9454 1589.454 79472.72 60 16.66667 6 0.05236 8.333333 3 159.0095 1325.079 79504.74 70 14.28571 5.142857 0.04488 7.142857 2.571429 159.0481 1136.058 79524.04 80 12.5 4.5 0.03927 6.25 2.25 159.0731 994.207 79536.56 90 11.11111 4 0.034907 5.555556 2 159.0903 883.835 79545.15 100 10 3.6 0.031416 5 1.8 159.1026 795.5129 79551.29 200 5 1.8 0.015708 2.5 0.9 159.1419 397.8546 79570.93 300 3.333333 1.2 0.010472 1.666667 0.6 159.1491 265.2485 79574.56 400 2.5 0.9 0.007854 1.25 0.45 159.1517 198.9396 79575.84 500 2 0.72 0.006283 1 0.36 159.1528 159.1528 79576.42 600 1.666667 0.6 0.005236 0.833333 0.3 159.1535 132.6279 79576.74 700 1.428571 0.514286 0.004488 0.714286 0.257143 159.1539 113.6813 79576.94 800 1.25 0.45 0.003927 0.625 0.225 159.1541 99.47133 79577.06 900 1.111111 0.4 0.003491 0.555556 0.2 159.1543 88.41905 79577.15 1000 1 0.36 0.003142 0.5 0.18 159.1544 79.57721 79577.21

As you can see, the area increases as the number of sides go up.

Finding the area of the circle:

The circumference is 1000 (as this is the length of fencing available)

From this we can find the diameter of a circle with a 1000m circumference

Circumference = π x diameter

Diameter = circumference / π

= 1000 / π

= 318.3

Radius = ½ diameter

= 159.15                                                        r = 159.15 m               c = 1000 m

Area = π x radius2

= π x 25328.7

= 79522.5 m2

Conclusion:

This proves my theory, as you can

see the circle has the largest area out

of all the shapes I have experimented with.  This is because a circle has an infinite number of sides and if the area size goes up as the number of sides goes up, an infinite number of sides will obviously have the largest area.

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

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