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• Level: GCSE
• Subject: Maths
• Word count: 1152

# Fencing Problem

Extracts from this document...

Introduction

Simran Singh Ghatore

11B

Mathematics

Fencing Problem

Aim-: A farmer has 100m of wire. She wishes to fence of an area of land into a shape, which will give her the maximum area.

• I have to measure and experiment with different shapes of 1000m circumference of perimeter to achieve the maximum area.
• I have to consider working from these shapes-:

- Triangles

- Pentagon

- Hexagon

- Circle

Prediction-: I predict that if the sides of a regular shape increase so will the area, coming to the conclusion of the circle holding the most area.

Square(l x w)*NOT TO SCALE

Rectangles (l x w)

In a rectangle, any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m X 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long.

 Length (m) Width (m) Area (m2) 480 20 9600 425 50 22500 400 100 40000 350 150 52500 325 175 56875 300 200 60000 250 250 62500

Middle

 Base (m) Side (m) Height (m) Area (m2) 480 260 10000 24000 400 300 223.607 44721.4 333.33 333.33 288.675 48112 300 350 316.228 47434.2 250 375 353.553 44194.125 180 410 400 36000 150 425 418.330 31374.75 50 475 474.342 11858.55

Graph

Conclusion

The regular triangle seems to have the largest area out of all the areas but to make sure I am going to find out the area for values just around 333.

 Base (m) Side (m) Height (m) Area (m2) 333 333.5 288.964 48112.450 333.25 333.4 288.747 48112.518 333.3 333.4 288.704 48112.522 333.5 333.3 288.531 48112.504 333.75 333.1 288.314 48112.410 334 333.0 288.097 48112.233

This has proved that once again, the regular shape has the largest area.

Because the last 2 shapes have had the largest areas when they are regular, I am going to use regular shapes from now on.

Regular Pentagon

Because there are 5 sides, I can divide it into 5 segments. Each segment is an isosceles triangle, with a top angle of 720. This is because it is a fifth of 360. This means I can work out both the other angles by subtracting 72 from 180 and dividing the answer by 2. This gives 540

Conclusion

I am going to apply the same method as before to solve the areas of these two regular shapes.

Hexagon:

1000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3

360 ÷ 6 = 60 ÷ 2 = 30

Adjacent= 831/3 m andHypotenuse= h

= 831/3 / h

=83.33333335/COS60

=166.6666667 = 166.66666672 – 83.33333352

= √20833.33332

h = 144.338

Area = ½ X b X H = ½ x 83 1/3 X 144.338 = 6014.065

6014.065 X 12 = 72168.784m2

Heptagon:

1000 ÷ 7 = 142.857 ÷ 2 = 71.429

360 ÷ 7 = 51.429 ÷ 2 = 25.714

= x/71.4285 = h= 148.323

Area = ½ X b X H = ½ X 71.429 X 148.323 = 5297.260

5297.260 X 14

= 74161.644m2

Octagon

TAN 67.5o = h/62.5

h= 62.5xTan67.5 o

=Area= ½ x 125 x h

COS= Adjancent/Hypotenuse = 62.5/x

= 62.5/COS67.5=163.320m

TAN=62.5 x Tan67.5= 150.888m

Area= ½ x 150.888 x 125= 9430.3 x 8

=73444m2

My predictions were correct and as the number of side’s increases, the area increases. Below is a table of the number of sides against area:

 No. of sides Area (m2) 3 48112.522 4 62500.000 5 68819.096 6 72168.784 7 74161.644 8 73444

Circle

I have already come to the conclusion that: as the number of side’s increases, the area increases. Circles have no sides and are infinite. This should prove my prediction correct.

C= 2 π x r

1000= 2 x 3.14 x r

1000= 6.28 x r

1000

6.28    = r

r= 159.23

Area = πr2

= 3.14 x 159.23 x 139.23

=79612.2m2

From this I conclude that a circle has the largest area when using a similar circumference.

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