• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6

# The Fencing Problem

Extracts from this document...

Introduction

Mike Kounoupias

I am investigating a problem to do with a plot of land with either a perimeter or a circumference of 1000 m. There are many possibilities for the she farmer to plot her fence round the area of land. She can plot in triangles, hexagons, decagons, rectangles, squares etc.

First, I am going to try to find out the greatest area for a rectangle. If I try using a rectangle with length of 400 m, and a width of 100 m, I am going to find out its area. Here, I have a rectangle of width w, which must have a length of 500 – w, if the perimeter is to 1000 m.

CHECK:

W + 500 – w + 500 – w + 500 – w.

= 500 w – w2

A = 500 w – w 2

Its area is w (500 – w)

A = 500w - w2

Middle

20000

Area

22500

40000

52500

60000

62500

60000

52500

40000

I shall draw a graph and find the value of w for which A is greatest.

The graph below shows me that a rectangle with a perimeter of 1000 m has a maximum area when it becomes a square, having sides of 250 m. The area is then 62 500 m2. I am now going to do the same with a triangle. I will find out the greatest area for a triangle.

Area of Triangle = ½ Base * Height

a= ½ 200 * 300 = 30 000 m

This is much less than the area of a square. Perhaps if I make the base and the height equal, the result will improve.

Pythagoras gives y = x + x.

y =       2x =    2x

P = x + x +  2x

2x +   2x = 1000

x ( 2 +  2 ) = 1000

x = 1000 / 2 +  2~ 292.9 M

A = 1/2 * 292.9 * 292.9

= 42, 893.2 m

Would  an equilateral triangle be better than a right angled triangle, since it has 3 equal sides?

Conclusion

Reg Polygon of “n” sides

1000/h = length of side

180 / ntan (180/n) = 500/n/h

h = 500 /n / tan(180/n)

Area of reg. polygon with n sides

= n * 500 / n * 500 / n / tan (180 / n ) = 250,000/ n TAN (180 / n )

I shall use a table to test my  formula for various values of n.

 n Area 3 250 000/ 3 tan 60 = 48112.523 5 250 000/ 5 tan 36 = 68819.096 10 350 000/10 tan18 = 76942. 088

These three results confirm that my general formula is true, because the values agree with the previous areas.

 100 250 000/ 100tan 1.8 = 79551.289 1000 250 000/ 1000 tan .18 = 79577.209 10000 350 000/10000 tan.018 = 79577.468

A circle is very close to a regular polygon having a very large number of sides. What is the area of a circle with perimeter of 1000 m ? I will calculate this value.

P = 1000 = 2πr

R = 1000/2π = 500/π

= 159.1549431 m

Area = πr2.

Π(159.1549431)2

= 79577.47155 m2

Conclusion

For a given length of fencing (such as 1000 m ) the greatest possibleb area which the farmer could enclose  would be given by a circle pf radius 159.1549431 m,; the biggest shape it could be is a circle.

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

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Fencing Problem essays

1. ## Fencing problem.

The trigonometry function TAN shall be used to find the height, which can also be known as length of opposite. TAN � = Opposite � Adjacent TAN 64.30 = Opposite � 71.5m Opposite = TAN 64.30 � 71.5m Opposite = 148.6m = Height.

2. ## The Fencing Problem

and the resulting value is the area. Area = {1/2[(1000 � 10) x h]} x 10 100 100 100 100 100 100 100 100 100 100 h 100 50 Regular Polygons - Pentadecagon (15-sided-shape) As shown, I have divided the polygon into triangles, and found the area of one of the triangles; using trigonometry (tan)

1. ## Fencing Problem

Rectangle 62499.99 1000 metres 249.1 x 250.1 2nd largest quadrilateral. The more narrowed down the closer dimension were to a square. Parallelogram (Trapezium) 39998.48 1000 metres On table of parallelograms. The smallest quadrilateral amongst the other 2. I stopped testing the values, as the area was nowhere near the area of a square.

2. ## Fencing Problem

170 34527.16 333.44 48112.515 190 37401.54 334 48112.233 210 39982.81 335 48110.712 230 42253.70 336 48107.879 250 44194.17 337 48103.725 270 45780.73 350 47925.724 Observation From the results of the table, the area increases as the base increase. However, there reaches a point where the areas stops increasing and in fact starts to decrease.

1. ## The Fencing Problem

I will start off with the base of the triangle increasing by 50m each time. Then I will zoom in until I find the right triangle. Below is a table showing the isosceles triangles' base, side, height and area in metres and metres�.

2. ## Beyond Pythagoras

Difference 1 8-4 4 4 2 24-16 8 4 3 48-36 12 4 4 80-64 16 4 5 120-100 20 4 Rule = 4n So the rule that the adjacent line column follows is = 4n�+4n This is how I found out the rule the third column follows (hypotenuse line)

1. ## The Fencing Problem

1000 60000 310 190 120 1000 58900 320 180 140 1000 57600 330 170 160 1000 56100 340 160 180 1000 54400 350 150 200 1000 52500 360 140 220 wwee eew steeeeud eee eent cee eneetral eecoee uk! 1000 50400 370 130 240 1000 48100 380 120 260 1000

2. ## Graphs of Sin x, Cos x; and Tan x

Question 1 Find the size of angle R. The Answer Did you get 25.4�? Well done! Teacher's Note If not, remember that Multiplying both sides by 4, we get = 0.4293... Therefore, R = 25.4� Question 2 Find the length of YZ. The Answer Did you get 4.40cm? Well done!

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to