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# GCSE: Fencing Problem

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1. ## Fencing Problem

Base=200m Sides=400m Area: 1/2 x b x h 1/2 x 200 x 387 =38700m� 3. Base=300m Sides=350m Area: 1/2 x b x h 1/2 x 300 x 316 =47400m� 4. Base=400m Sides=300m Area: 1/2 x b x h 1/2 x 400 x 224 =44800m� Conclusion I have found that the three sided shape that had the biggest area when using 1000 meters of fencing, was a triangle with the measurements: Length=300m Sides=350m Height=316m Area=47400m� Equilateral Triangle I found that the biggest area of a triangle when the base goes up in hundreds each time was a triangle with the measurements: Length=300m Sides=350m Height=316m.

• Word count: 3047
2. ## Regeneration of the London Docklands

This is because the Docklands were not designed for the size of the more modern ships, not been wide enough or deep enough to allow the ships in. This meant that competition was starting to arise from other ports. These newer ports offered a facility to handle containers. With the competition a problem and after years of decline, the docks became too expensive to run, with the lack of trade and inefficiency of loading and unloading. By 1981, all the docks along the Thames were closed.

• Word count: 1100
3. ## Maths GCSE Courswork

triangle = 333.33/ 2 = 166.225m To find length A, I will now need to use the Pythagoras Theorem which is a2 + b2 = c2. I need to rearrange the formula so that I can determine length A: c2 - b2 = a2. Length A = c2 - b2 = a2 = 333.332 - 166.6652 = 83331.66m2 = V83331.66 = 288.67m If length A is 288.672m then that means that the height is also 288.67m. So now, I can work out the area of the whole triangle using the base and height: Area = base x height / 2 = (333.33 x 288.672)

• Word count: 2985
4. ## The Koch Snowflake

However, a closer look reveals another geometric progression! Let me show you how: 0.33333333/1=0.111111111/0.333333333=0.037037037/0.11111111=0.33333333:1 0.33333333:1 is the constant ratio. The length of the sides changes by this ratio with the progression of stages. Thirdly, we come to the perimeter. It is seen to increase with each passing stage. On closer observation, it is revealed that there is a geometric progression. This is seen again in: 4/3=5.33333333/4=7.111111111/5.333333333=1.33333333333:1 1.33333333333:1 is a constant ratio. The perimeter increases in this ratio with each passing stage. Lastly, the area of the snowflake is put under consideration. By now it seems clear that all the three parameters follow a geometric pattern and hence I am tempted to test the last one as well.

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5. ## Fencing Problem

gives the biggest area in terms of the rectangle. I shall plot my results onto a graph to see the correlation between them. The line of symmetry shows the biggest area when you have 1800m of fencing in the shape of a rectangle. I found the ideal dimensions were if the length and width of the rectangle were 450m. This gives you the biggest area of 202500cm�. Now that I have found the largest area in rectangles I shall move on to find the largest area in an isosceles triangle. I shall look at an isosceles triangle simply because they are easiest to calculate out of all the triangles.

• Word count: 1343
6. ## Fencing Problem

Perimeter = 275 + 225 + 275 +225 = 1000m Area = 275 x 225 = 61, 875m 8) Perimeter = 265 +235 + 265+ 235 = 1000m Area = 265 x 235 = 62,275m 9) Perimeter = 250 +250 +250+250 = 1000m Area = 250 x 250 = 62,500m To make the process easier and quicker, I am now going to put the areas I have already found into a table. Table to show areas of rectangles Length (m)

• Word count: 1890
7. ## Fencing Problem

We used the isosceles triangle because it gives a large range of outcomes. It did this because we can change the base and sides, this means we were able to work out the area of the triangle. It is also the easiest triangle to find the area of as the equilateral triangle only has one outcome and the scalene triangle has too many options to investigate systematically. However, with isosceles we can start with a base of 1m, 2m, 3m etc.

• Word count: 732
8. ## HL type 1 portfolio on the koch snowflake

* For , To confirm the value of 'r' we use a new pair of terms, Therefore, we conclude that the ratio between the successive terms of the geometric progression is 4. * For , Therefore we come to the conclusion that the ratio between the successive terms of the geometric regression is . * For , * For , Now substituting by their conjectures, we get, Therefore we come to the conclusion that the formula involves the calculation of a geometric series and that difference between successive terms, say .

• Word count: 1871
9. ## Fencing - maths coursework

x 50m 22500m� The formula I have used to find out the area of the rectangles is =A3*B3 My table and my graph show that the square has the greatest area: 62,500m� I have realised that the best area is a square so I am now going to investigate in between the sides of 245-255 Base (x) Height (y) Base x height Area 245m 255m 245m x 255m 62475m� 246m 254m 246m x 254m 62484m� 247m 253m 247m x 253m 62491m� 248m 252m 248m x 252m 62496 249m 251m 249m x 251m 62499m� 250m 250m 250m x 250m 62500m� 251m

• Word count: 2883
10. ## topshell populations

Suckering to the rock stops water from escaping so this prevents dehydration. Also sticking to the rock will lower the chances of being eaten by a predator. Monodonta lineata are more tolerant of raised temperatures and are often abundant in the middle and upper parts of the shore and are only found on the western parts of the British Isles where the sea temperature is raised by the influence of the Gulf Stream. At saltern cove there is a partially sheltered area where a rocky outcrop protects the area behind from the largest waves meaning that the area behind is very calm.

• Word count: 3341
11. ## Koch Snowflake Math Portfolio

The perimeter is directly proportional to the stage. Pn ? n iv. An - The area: The relationship between the successive values of the area is that the values increase by 3 � 4n-1 times with each stage. The area is 9n directly proportional to the stage. An ? n 2. i. Nn v/s n ii. ln v/s n iii. Pn v/s n iv. An v/s n 3. n (stage) Nn (no. of sides) 0 3 1 12 2 48 3 192 i.

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12. ## Fencing Problem

Therefore, to get the sides it would be: A = 1000 x 3 = 250 250 12 B = 1000 x 4 = 1000/3 12 C = 1000 x 5 = 1250/3 1000/3 , 12 I will move on to look at scalene triangles. In this case, it is safe to assume the sides and put any lengths so long as they add up to 1000m. Scalene Triangles To find the area of these triangles I can use Hero's Formula since I have all the sides given: A = V[s(s-a)

• Word count: 4001
13. ## A farmer has 1000 metres of fencing. She wants to use it to fence off a field. The fencing has to enclose the maximum area possible; my task is to find the shape and dimension that will give the largest area

400 200 400 38729.83 450 150 400 29580.4 500 100 400 0 S= 500 Largest Area 44721.36 Graph: Analyse I noticed that the isosceles triangle (400 by 300 by 300) made the largest area of the 9 shapes above; I also noticed that the isosceles also had the largest perpendicular height; this is a valuable analyses point, if I use another formula to work out the area Of a triangle: Base x Perpendicular Height 2 If the base is fixed and the Perpendicular height is at its upper bound the highest result will be at its highest.

• Word count: 1399
14. ## A length of guttering is made from a rectangular sheet of plastic, 20cm wide. What is the best position for the folds so that the guttering carries the maximum amount of water?

10 20 10 50 10 10 5 0.868241 4.924038765 11.73648 21.73648 10.86824 53.51564 20 10 5 1.710101 4.698463104 13.4202 23.4202 11.7101 55.01948 30 10 5 2.5 4.330127019 15 25 12.5 54.12659 40 10 5 3.213938 3.830222216 16.42788 26.42788 13.21394 50.61232 50 10 5 3.830222 3.213938048 17.66044 27.66044 13.83022 44.44948 60 10 5 4.330127 2.5 18.66025 28.66025 14.33013 35.82532 70 10 5 4.698463 1.710100717 19.39693 29.39693 14.69846 25.13585 80 10 5 4.924039 0.868240888 19.84808 29.84808 14.92404 12.95766 90 10 5 5 3.06287E-16 20 30 15 4.59E-15 21 10 5 1.79184 4.667902132 13.58368 23.58368 11.79184 55.04315 22 10 5 1.873033 4.635919273 13.74607

• Word count: 2919
15. ## The Koch Snowflake

5 1/3 � 4 = 1.333... 7 1/9 � 5 1/3 = 1.333... Therefore each successive term is the result of multiplying the previous term by 1.333... which is equal to 4/3. The perimeter is increasing in a geometric sequence, by 1/3 of its value each time. It is also increasing by a larger amount each time, and so is a divergent series. This is connected to how I first calculated perimeter, as it is shown on the table that I multiplied the last perimeter by 4 then divided it by 3 to find the next perimeter.

• Word count: 1686
16. ## 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. She wishes to fence off the plot of land which contains the maximum area. I will be investigating the shape

Now here is an exert from the central part of the table which contains the maximum value for the area, I've highlighted this value in yellow. 400 600 240000 410 590 241900 420 580 243600 430 570 245100 440 560 246400 450 550 247500 460 540 248400 470 530 249100 480 520 249600 490 510 249900 500 500 250000 510 490 249900 520 480 249600 530 470 249100 540 460 248400 550 450 247500 560 440 246400 570 430 245100 580 420 243600 590 410 241900 600 400 240000 Now notice that the maximum values for rectangles when drawn out like so is a square!

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17. ## To investigate the ratio of Area:Perimeter for triangles (2) To investigate the suggestion made by students that a 40,60,80 triangle would maximise the above mentioned ratio.

Calculation As no side lengths have been give then we could assume any one side is any length we want. By defining the length of one side the others are obviously defined. We can calculate the other length by use of the sine rule. We will make a decision: Let the length of side 'a' be 1m. The perimeter is = a + b + c: By the sine rule a = b = c Sin 40 sin 80 sin 60 Therefore b = (a sin40)/(sin80)

• Word count: 1229
18. ## To investigate the effects of a parachutes shape and surface area, on it time of decent.

The length of string tied form the parachute to the washer - (30cm) The weight tied to the parachutes - (10+-1g) o Uncontrolled Variables: The wind o Independent Variables: Experiment 1 - the shape of the parachute Experiment 2 - the surface area of the parachute (cm2) o Calculated Variables: The surface area of each parachute (cm2) o Measured Variables: Method: Experiment 1 - Different Shaped Parachutes > first you need to get the same surface area for each of the three shapes, this is done by using the formulas, A=l2 for the square parachute, A=0.5*b8h for the triangular parachute

• Word count: 1504
19. ## Proving a2 + b2 = c2 Using Odd Numbers

Perimeter Odds Perimeter = a + b + c a= 2n + 1 b= 2n2 + 2n c= 2n2 + 2n + 1 = 4n2 + 6n + 2 Proving that this formula is correct: If I take n=3 = 4x9 + 18 + 2 = 36 + 18 + 2 = 56 So if you look along the odds table on n=3 the perimeter is 56 this proves my formulae is correct.

• Word count: 412
20. ## Biological Individual Investigation What Effects Have Management Had On Grasses In Rushey Plain, Epping Forest? Abstract

Grasslands are important in maintaining biodiversity. Many grasses and shrubs, as well as wildlife only populate areas of grassland. Grazing is the most important arresting factor in stopping grasslands from developing into woodland. In 1953, myxomatosis was accidentally introduced into Britain. Myxomatosis is a viral infection, which is spread by fleas. As a flea bites a rabbit, a small amount of live virus is injected into the blood. The virus is then transmitted to the lymph nodes, where it can spread to multiple sites. The virus causes swelling around the eyes, nose, ears, anus and genitalia of the rabbit.

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21. ## A farmer has brought 1000 metres of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the biggest size. I will investigate

The best shape is a regular quadrilateral (square). Length Width Area 450 50 22,500m� 400 100 40,000m� 350 150 52,500m� 300 200 60,000m� 250 250 62,500m� 200 300 60,000m� 150 350 52,500m� 100 400 40,000m� 50 450 22,500m� Area of a rectangle in terms of n: n n Area= n x n = nm Triangles Now I am going to work out the area of triangles, I am going to use Isosceles triangles because I know the base length and so I can work out the other two as they are the same.

• Word count: 1348
22. ## A Settlement Enquiry

I'm also going to find other geographical theories to see if they agree with the hypothesis. Method I collected my data by visiting three areas around terminal 5, these were Staines, Ashford and Shepperton. Firstly I went to Ashford, then Shepperton and finally Staines. We went as a class. When we arrived at each destination we split into groups of two's and three's and walked around asking people in the area questions. We asked each individual ten questions mainly about what they thought the effects of Terminal 5.

• Word count: 1660
23. ## My investigation is about a farmer who has exactly 1000 metres of fencing and wants to use all of it to fence off the maximum area of land possible. I want to find the regular polygon that will have the maximum area with a perimeter

So, instead, I am going to increase the width to 50m so the length will be 450m: A = L x W A = 450 x 50 A = 22500m2 From this I can see that the area gets bigger as the rectangle gets shorter and fatter, therefore I will carry on looking at rectangles that are shorter and fatter rather than taller and thinner. I am going to increase the width again, this time to 100m, the length is now 400m, I am doing this to see if I am correct in saying that as the rectangle gets shorter

• Word count: 3055
24. ## The fencing problemThere is a need to make a fence that is 1000m long. The area inside the fence has to have the maximum area. I am investigating which shape would give this.100m

I can put this into an equation form. 1000 = x(500 - x) Below is a table of results, worked out using the formula. I have gone down by taking 10m off the base every time. Height (m) x Area (m2) 0 500 0 10 490 4900 20 480 9600 30 470 14100 40 460 18400 50 450 22500 60 440 26400 70 430 30100 80 420 33600 90 410 36900 100 400 40000 110 390 42900 120 380 45600 130 370 48100 140 360 50400 150 350 52500 160 340 54400 170 330 56100 180 320 57600 190 310 58900 200 300 60000 210 290 60900 220

• Word count: 2223
25. ## Maths Coursework: The Fencing Problem

48083.26 6 340 340 320 48000.00 From the table I've noticed that the closer the sides are to each other then the area is bigger. I used Heron's formula to work out the area using the 3 sides: VS(S-A) (S-B) (S-C) where S = (A+B+C) / 2 Since the equilateral triangle has all equal sides it should have the biggest area out of the triangles. In this case it's true with 48112.52 m2 (2 d.p.). This graph is used to show that as the difference between the 3 sides' increases, the area decreases.

• Word count: 1496