• 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
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
• Level: GCSE
• Subject: Maths
• Word count: 1438

# The fencing problem

Extracts from this document...

Introduction

The fencing problem I am going to investigate the shapes of a plot of land. The plot has a fixed perimeter of 1000m and I will investigate what shape gives the maximum area with a fixed perimeter. I will only investigate regular shapes, as there are an unlimited number of irregular shapes. Triangle First I will investigate triangles as they are the first shape I though of and are very simple. The formula to work out the area of a triangle is - 1/2 x base x height This formula is simple but I will make a formula that gives you the area but you don't need to know the height to save time calculating the areas. Isosceles triangle- Let the equal sides be x. X X < 1000- 2 x > Total perimeter = 1000m First I need to work out the height using Pythagoras. H = x � - 1/2 (1000 -2 x)� H = x � - (500 - x) � H = V[x� - (500 - x) �] So the area is - Area = 1/2 (1000 -2 x) * V[x� - (500 - x) �] Area = (500-x) * V[x� - (500 - x) �] This formula will only work for Isosceles triangles. I will now test it out on an equilateral triangle. 333.3� 333.3� hh 333.3� h� = 333.333� - 166.67� h� = 83333.333 h = 288.6751346 Area = 166.667 * height Area = 48112.52243 Now I will work out the area using my formula. ...read more.

Middle

a graph of this equation is - Gradient A = 500 - 2x The power of both of the terms is reduced by 1 and the terms are multiplied by the old power. The maximum point on the graph is where the gradient is equal to zero. Gradient A = 500 - 2x 0 = 500 - 2x 2x = 500 x = 250 Using this method the result is the same as my trial and improvement method. Next I am going to investigate regular polygons. They are the next shape I thought of. The area of a polygon can be worked out by splitting it into isosceles triangles. If the polygon has a fixed perimeter of 1000m then the base will be 1000/n (n = number of sides in the polygon). Half the base will equal 500/n. S o the angle at the top of the triangle will be 360/n. So the angle at the top of the right angled triangle will be 180/n (half). The remaining angle in the right angled triangle will be 90 - (180/n). Because if the right angle is 90� than the sum of the other angles must be 90�. To work out the area formula for an n sided polygon I must first work out the height using trigonometry. N = number of sides P = perimeter (1000m) ...read more.

Conclusion

As the number of sides in the shape increase it looks more and more like a circle. The formula for the area of a circle is area = ?r� As the radius is unknown I will work it out using a different formula, my previous formula only works for shapes that can be divided into triangles. Perimeter = 2?r, so Radius = perimeter/2? Radius = 1000/2? Radius = 159.1549431 Area = ? * 159.1549431 * 159.1549431 Area = 79577.47155 This is the largest area possible from a given perimeter of 1000m. Earlier I stated that as the number of sides is increased the area increased also, but that does not work for a circle as it only has 1 side. As the number of lines of symmetry increases the total area increases. This statement is true because an equilateral triangle has 3 lines of symmetry, a square has 4 lines of symmetry, a pentagon has 5 lines of symmetry and a circle has infinite lines of symmetry. This graph shows that as the number of sides is increased the total area increase but will never be as great as the circles total area even if it is a thousand sided shape. Total area = 1000000/(4*1000) * tan (90 - (180/1000)) Total area = 250 * tan 89.82 Total area = 79577.20975 This area is very close to the area of the circle but is still smaller. The largest possible area with a fixed perimeter of 1000m is 79577.47155m which is a circle. ...read more.

The above preview is unformatted text

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. ## Investigating different shapes to see which gives the biggest perimeter

This is because I don't know the hypotenuse of the triangle. For this matter, the solution is to use trigonometry. If I am able to work out one side and one angle within a right-angled triangle, I will be able to use trigonometry to find out the height.

2. ## Fencing problem.

This can be found by excluding 1800 from the exterior angle. Interior angle = 1800 - Exterior angle Interior angle = 1800 - 51.4 = 128.60 Now by using the help of the interior angle I have calculated I shall find of the angles within one of the five triangles that can be made in a regular heptagon.

1. ## The Fencing Problem

and the resulting value is the area. Area = {1/2[(1000 � 15) x h]} x 15 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 66.67 h 66.66666667 33.33333334 Regular Polygons - Icosagon (20-sided-shape) As shown, I have divided the polygon into triangles, and found the area of one of the triangles; using trigonometry (tan)

2. ## Fencing Problem

(I never investigated the area of trapeziums as it shared the same properties of parallelograms so I would of ended up with the same area). After investigating Quadrilaterals and looking at my table of results for Quadrilateral shapes I realised that my hypothesis was true and that the shape with

1. ## t shape t toal

This proves my equation correct as the correct translated T-total is generated. t = (5x - 7g) - b(5g) + (a5) Only if you split the diagonal translation into two parts the horizontal movement and a vertical movement you can use this formula to generate you new T total It

2. ## Geography Investigation: Residential Areas

There are a high proportion of houses privately owned in Portacre Rise and Cumberland Avenue. These houses were bigger, more land and often semi detached. I spoke about this point earlier on in the enquiry. I believe that Portacre Rise and Cumberland Avenue will have a higher percentage of residents of an older generation.

1. ## Fencing Problem

= 500 - 1/2 B We can therefore replace L to the above formula: A = 1/2 x B x V [(500 - 1/2 B) 2- (B2/4)] I will start the Base from 10 m and move upwards. I am hoping to reach to a point where I would obtain a maximum area after which the area starts to decrease.

2. ## Regeneration has had a positive impact on the Sutton Harbour area - its environment, ...

This would give us an indication as to how the local people feel about it and whether they have been affected in a positive or negative way. Question 7- "Do you think that there are a suitable amount of shopping stores and services in the area?"

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