# Math Coursework - The Fencing Problem

# Math Coursework - The Fencing Problem

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More ↓A farmer has 1000m of fencing and wants to fence off a plot of level

land.

She is not concerned about the shape of plot, but it must have a

perimeter of 1000m. So it could be:

[IMAGE]

Or anything else with a perimeter (or circumference) of 1000m.

She wishes to fence of the plot of land with the polygon with the

biggest area.

To find this I will find whether irregular shapes are larger than

regular ones or visa versa. To do this I will find the area of

irregular triangles and a regular triangle, irregular quadrilaterals

and a regular square, this will prove whether irregular polygons are

larger that regular polygons.

Area of an isosceles irregular triangle:

========================================

(Note: I found there is not a right angle triangle with the perimeter

of exactly 1000m, the closest I got to it is on the results table

below.)

To find the area of an isosceles triangle I will need to use the

formula 1/2base*height. But I will first need to find the height. To

do this I will use Pythagoras theorem which is a2 + b2 = h2.

[IMAGE]

[IMAGE]

First I will half the triangle so I get a right angle triangle with

the base as 100m and the hypotenuse as 400m. Now I will find the

height:

a2 + b2= h2

a2 + 1002 = 4002

a2 = 4002 - 1002

a2 = 160000 - 10000

a2 = 150000

a = 387.298m

Now I will find the area:

100*387.298 = 3872.983m2

My table shows the areas of other irregular triangles, but to prove

that regular shapes have a larger area I will show the area of a

regular triangle:

Area of a regular triangle:

Tan30= 166.6666667/x

X= 166.666667/Tan30

X= 288.675m

288.675*166.6666667

= 48112.5224m2

This shows clearly that the regular triangle's area is larger than the

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Check your paper »## Math Coursework - The Fencing Problem Essay

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### Related Searches

irregular quadrilateral's area and a regular square's area.

The formula to find a four-sided shape's area is b*h.

Irregular quadrilateral:

[IMAGE]

[IMAGE][IMAGE]Base= 450m

Height= 50m

50*450= 22500m2

Regular quadrilateral:

[IMAGE]

All sides= 250m

250*250= 62500m2

To further back up my prediction that regular polygons have a larger

area I have made a table (below) that clearly shows that the polygons

I have researched have a smaller area. So I will see what regular

polygon has the largest area. I also found that there are the same

amount of isosceles triangles in the regular polygon as there are

sides:

[IMAGE]

[IMAGE]

When finding out whether regular polygons have a larger area than

irregular polygons I found that shapes with a larger amount of sides

have a larger area. E.g. regular triangle area= 48112.5224m2, regular

quadrilateral area= 62500m2. So I decided to find the polygon with the

largest area. To do this more efficiently I devised a formula in excel

and by hand which make the process faster. There are two separate

formulas because excel works with radians so I needed to adapt the

formula so it could work in excel. Here are the two formulae:

[IMAGE]

In excel: (aaa= this will change for each shape, it is a cell

reference and refers to the amount of sides in the polygon)

=(500/A3)/(TAN(PI()/A3))*500

This is what the equation means:

[IMAGE]The 500/n calculates the height of the right angle triangle

within a polygon.

[IMAGE] Tan 180/n

[IMAGE]

[IMAGE][IMAGE]The x 500 is the simplified version of my first

equation: 1000 x n 500 = x 500

[IMAGE][IMAGE]

N

[IMAGE][IMAGE] 2 1

This finds the overall area of the regular polygon and also simplifies

out to what is now x 500.

The excel formula is the same, but excel works with radians so I

needed to change the equation so that I works with radians. Therefore

instead of Tan 180/n, it is now tan p/n (cell reference). This is

because p= 180°.

So if I wanted to find the area of a decagon I would do the following

by hand:

500/10 x 500

[IMAGE]

Tan 180/10

= 50/ Tan 18 x 500

=76942.088m2

Because I can put my formula in excel I can see the area from a

variety of different polygons. Here are my results:

Sides in polygon

Area of polygon (m2)

Sides in polygon

Area of polygon (m2)

3

48112.5224

28

79243.2632

4

62500.0000

29

79265.9324

5

68819.0960

30

79286.3705

6

72168.7836

31

79304.8609

7

74161.4784

32

79321.6437

8

75444.1738

33

79336.9227

9

76318.8172

34

79350.8725

10

76942.0884

35

79363.6429

11

77401.9827

36

79375.3632

12

77751.0585

100

79551.2899

13

78022.2978

200

79570.9265

14

78237.2548

300

79574.5626

15

78410.5018

400

79575.8353

16

78552.1796

500

79576.4243

17

78669.5221

1000

79577.2097

18

78767.8031

2000

79577.4061

19

78850.9402

3000

79577.4425

20

78921.8939

4000

79577.4552

21

78982.9345

5000

79577.4611

22

79035.8270

23

79081.9600

24

79122.4387

25

79158.1509

26

79189.8169

27

79218.0258

From these results I plotted a graph (separate sheet). From the graph

you can see that the more sides there are in a shape the larger the

area. I decided to test one more shape and that is a circle. A circle

with a circumference of 1000m has the largest area. From this I can

say that the circle is the shape with the most amount of sides and is

also a regular polygon. But also the circle has a infinite amount of

sides which also makes it the shape with the largest area.

Therefore according to all my calculations I can safely say that the

fence should be made into a regular circle shape with a perimeter of

1000m.