• 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
14. 14
14
15. 15
15
16. 16
16
• Level: GCSE
• Subject: Maths
• Word count: 2123

# Investigate the shapes that could be used to fence in the maximum area using exactly 1000m of fencing each time.

Extracts from this document...

Introduction

Maths Coursework

 For this investigation, I will be looking at different shapes, and the areas the different shapes give. The exact question is:

A farmer has exactly 1000m of fencing, and wants to fence off a plot of level land. She is not concerned about the shape of the plot, but it must have a perimeter of 1000m. She wishes to fence of the area of land, which contains the maximum area.

Investigate the shapes that could be used to fence in the maximum area using exactly 1000m of fencing each time.

I will show the working in the form of formulas, putting results in tables, and then transfer the tables into graphs. Once this is completed, I will draw up a conclusion.

Prediction

My prediction is that as the number of sides increase, as will the area. I think this as the area of a rectangle, or any other quadrilateral, will have a bigger area than a triangle when using the same perimeter. I have no reason not to believe that this pattern of increasing sides/increasing area will continue.

I am going to start investigating different shape rectangles, all which have a perimeter of 1000m. Below are 4 rectangles (not to scale) showing how different shapes with the same perimeter can have different areas.

In a rectangle, any 2 different length sides will add up to 500, because each side has an opposite with the same length.

Middle

Because the regular rectangle was the largest before, I added 333.3 as a base length. This is the length of the base of a regular triangle. It is also an equilateral triangle.

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. This would also be a lot easier as many of the other shapes have millions of different variables.

 Because there are 5 sides, I can divide it up into 5 segments. Each segment is an isosceles triangle, with the top angle being 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 each. Because every isosceles triangle can be split into 2 equal right-angled triangles, I can work out the area of the triangle, using trigonometry. I also know that each side is 200m long, so the base of the triangle is 100m.Length of side =   1000    =  200                             5

Conclusion

20, 50, 100, 200, 500, 1000

On the following page is a table showing the results that I got.

 No. of sides Area (m2) 20 78921.894 50 79472.724 100 79551.290 200 79570.926 500 79576.424 1000 79577.210

The pattern remains the same as the number of sides goes up. You can clearly see that as the number of sides increases, the area does to. This is the same from a three- sided shape, all the way up to 1000. If this is the case, then what would the area be of a shape that at an infinite number of sides?

The final shape I am going to look at is the circle. Circles have an infinite amount of sides, so I cannot find the area by using the equation that I have used to find the other amount of sides out. I can find out the area of a circle by using π r2. But we don’t know the radius yet, so we have to work out the diameter. The circumference of a circle can be found using the equation is π d. I can rearrange this so that the diameter is circumference/ π. From that I can work out the area using the π r2 equation.

We know that the circumference has to be 1000, as that’s the amount of fencing given.

1000/ π = 318.310

Now we have the diameter of the circle. To find the radius, we simply divide by two, as the radius is half the diameter.

318.310/2 = 159.155

So we know have the radius. To find the area of a circle, you multiply π, by the radius squared.

π X 159.1552 = 79577.472m2

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 of gutters.

To find y I will use trigonometry: sinC= y/x y= xsinC To find h I will use trigonometry: cosC= h/x h=xcosC Now that I have found h and y I can find the area for the general case. Area= 1/2 (a+b)*h Area= 1/2 [(L-2x)+(L-2x+2xsinC)*xcosC Area= (L-2x+xsinC)xcosC I am unable to

2. ## A farmer has 1000 metres of fencing. She wants to use it to fence ...

both are the same an in theory it will work for every other shape. The only variable in the formula should be the number of sides and I will call it "n". I will use a pentagon to help me work out the formula, instead of righting the numbers I will substitute its formulas.

1. ## Investigate different shapes of guttering for newly built houses.

Because rectangles 3 and 4 give the same area this indicates that the best lengths to use must be halfway between them. I will now try to prove this: 7) a = 7.5 b = 15 Area = a?b = 7.5?15 = 112.5cm� This rectangle has proved my theory.

2. ## When the area of the base is the same as the area of the ...

10x4x4=160cm 11cm 2cm 44cm 4cm 11x2x4=88cm The maximum volume was when the height of the tray was 7cm. The area of the base though did not equal to the area of the 4 sides. 8x8=64cm =64x2=128cm The formula I used to calculate this was: lxbxh Height: Base length: Volume: Base

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

investigate, the same because I have found out from my investigation on regular rectangles and regular triangles that shapes with equal sides have the biggest area. So now all of my next shapes will have equal sides. Polygons: I am now investigating shapes with more than 4, and equal, sides.

2. ## GCSE Maths Coursework Growing Shapes

33 -3 5 42 -3 Formula for the perimeter of a triangular pattern when the added squares are pointing down = 9n-3. Check Formula = 9n-3 = 9�3 - 3 = 24 Number of Triangles/Area The yellow triangles represent the area Pattern no.

1. ## To investigate the isoperimetric quotient (IQ) of plane shapes using the calculation shown below.

Algebraic formula for pentagon: Area=0.5 =0.5 =5 =2.5 Perimeter = = = 5 IQ = = = =0.864806 Algebraic formula for hexagon: Area = 0.5sin60 =0.5sin60 =5 sin60 =3sin60 Perimeter = = = 6 IQ = = = = 0.906899682 Algebraic formula for octagon: Area = 0.5sin45 =0.5sin45 =5 sin45

2. ## To investigate the effects of a parachutes shape and surface area, on it time ...

Results: Experiment 1- Different Shaped Parachutes Shape of Parachute Drop 1 (seconds) Drop 2 (seconds) Drop 3 (seconds) Square 0.84 0.89 0.77 Triangle 0.8 0.78 0.82 Circle 0.98 1.04 1.01 Experiment 2 - Different Surface Area Radius of Parachute (cm)

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