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GCSE: Fencing Problem

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  1. GCSE Maths questions

    • Develop your confidence and skills in GCSE Maths using our free interactive questions with teacher feedback to guide you at every stage.
    • Level: GCSE
    • Questions: 75
  2. the fencing problem

    Width (m) Area (m) 50 450 22500 100 400 40000 150 350 52500 200 300 60000 250 250 62500 300 200 60000 350 150 52500 400 100 40000 450 50 22500 The table shows all the possible rectangles with the perimeter of 1000m and what the area would be. As you can see on the table the shape with the length and width of 250m gives the biggest area (62500) and as you can see on the previous page that shape is a square.

    • Word count: 914
  3. fencing problem part 2/8

    To work out the areas of quadrilaterals, I will use Bretschneider's formula. The formula is . In this formula a, b, c and d are the lengths of the four sides of the quadrilateral. The s stands for the semiperimeter and ? is half the sum of two opposite angles. As we want the maximum area for a quadrilateral, the number which has to be square rooted should be as high as it can be. This can only be achieved if the 'abcdCos2?

    • Word count: 763
  4. Fencing problem

    I only investigated six rectangles as after E the rectangles have the same numbers but on different sides. This means the area would be the same and so it was pointless to continue at this rate of increase so I looked into it further to see if I was able to find a bigger area. I also put these results into a table. Rectangle Length (M) Height (M) Area (M ) G 290 210 60900 H 280 220 61600 I 270 230 62100 J 260 240 62400 After investigating further I found no rectangle with a larger area. I decided not to investigate further as the closer the measurements get to becoming square the bigger the area becomes.

    • Word count: 662
  5. fencing problem

    Firstly I will be drawing 4 triangles with the perimeter of 1000 meter. Secondly I will draw the quadratral family for example squares and rectangles. Next I'll draw polygons up to 5sides to 10sides, and I will be calculating the nth side using the nth formula. To end with I will calculate the area of the circle. I predict that the circle will get the maximum area because its area is immeasurable and predict that area of a polygon will rise depending of the number of sides. Triangles I will begin my investigation by drawing 4 triangles to find the maximum area with the perimeter of 1000m� shape B I W A 1

    • Word count: 981
  6. Koch Snowflake

    The value of N, the first term is 3 and the common ratio is 4. In l the first term is 1 and the common ratio 1/3. The variable P has a first value 3 and a common ratio of 4/3. The value of P is also the product of N and l. To find An we can use the following formula, which had been accumulated with the findings from A1, A2, and A3: Relationships with n N against n: As the number of sides increase with the rise in, each stage will have a number of sides which will be four times higher then the prior stage.

    • Word count: 799
  7. The Koch Snowflake

    1/3, 1/3 x 1/3 = 1/9, 1/9 x 1/3 = 1/27 , etc; The perimeter was found by multiplying the Number of sides with the length of the sides and there is no evident relationship between the values of perimeter. The area was found by calculating the triangles added to the main triangle and then adding that to the area of the main triangle. As each value of n increases, the difference between each successive value of An decreases. 2.

    • Word count: 641
  8. Assignment 2 Pond area

    The calculator will then be used to find the exact answer to each and hence it will be possible to asses which mothod produces the best results. Finding the area under y =10SIN(0.08x)+25 by the trapezium rule A fairly good approximation could be gained by dividing the line into 3 sections each 20m wide.

    • Word count: 455
  9. The fencing problem - maths

    Area(cm2) 333.3 333.3 289 48162 I also found it's the last triangle with a 1000cm diameter that isn't a nonsense triangle, where the base is larger than the hypotenuse. So far in my investigation I have found the largest areas for rectangle and triangles. From the results of both of them I have found that the ones with the largest area are regular shapes, such as a square or equilateral triangle. So I believe from now on in my investigation, I am only going to investigate regular shapes.

    • Word count: 899
  10. 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
  11. 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
  12. PE Coursework

    His movement after the pass is hit is always pretty good. Weaknesses: Harry cannot pass well with his weaker, right, foot. Harry's passing can sometimes be wayward over long distances. He often panics under pressure and hits poor passes. He sometimes under-hits short passes. Crossing Strengths: Harry is good at hanging high balls into the opposition's penalty area. He is good at picking out a player when there are a few players in the area to find. When beating the opposition right back, Harry is not afraid to put an early cross in.

    • Word count: 430
  13. GCSE Maths Coursework

    A square has 4 sides therefore to find the perimeter you times the sides to 4. And 4X4 is also how to work out the area therefore they both come to make 16 This all because a square has EQUAL SIDES My interpretation does not work with rectangles because the sides are NOT EQUAL 2.Triangle Working out the equation for the area of an equilateral triangle 1cm 1cm 1cm An equilateral triangle is one were all the sides have equal length ie.

    • Word count: 550
  14. Local area report on Rhondda Valley.

    The area has one rail link to Cardiff and other major towns. It has an approximate population of seventy two thousand five hundred people. The population is aging as younger people migrate from the area to seek work. The local employment is now mostly light industry and service industry as the coalmines have been closed for many years. Unemployment now runs at a high level locally, particularly for younger people. The leisure facilities in Rhondda are improving but are not satisfactory given the size of the population. Advantages and Disadvantages The advantages of living in the area are that: * There is a good community spirit. * Good availability of nursery education.

    • Word count: 784
  15. Development strategy to redevelop London Docklands

    The main project had various areas and themes for regeneration. Architectural regeneration The architectural value of Docklands had been ignored during the 1960's and 1970's. Many buildings were lost, notably at St Katharine Docks. The London Docks in Wapping and the West India Docks on the Isle of Dogs (buildings from the early and mid-19th century) were demolished without thinking about it. Older plans for development didn't even appreciate the important value of the architecture in this area Some plans were created to clear post-war housing instead of redeveloping it.

    • Word count: 828
  16. The Fencing Problem

    A = 500w - w2 If I multiply the length (400 m) and the width (100 m), I get the answer of 40,000 m2. However, if I increase the width, I get the following answers: Width 50 100 150 200 250 300 350 400 - width 2 2500 10000 22500 40000 62500 90000 122500 16000 500 width 25000 50000 75000 100000 12500 150000 175000 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.

    • Word count: 815
  17. Fencing problem.

    With a 1000 metres of fencing you could create a quadrilateral field with a maximum of 62500m�. Triangles I decided to look at triangles next. Right angle triangles don't work well with this type of investigation so I've decided not to do them. To find the area I used this method: e.g. H� = a� + b� 25 x 474.341 x2 A = 11,868.55 2 475� = 25� + b� 225625-625 = b� V 225000 = Vb� b = 474.341 Once I had done all the calculations for the triangles from 50 m base to 400 m base in 50 m increments I had enough information to create a graph.

    • Word count: 633
  18. Beyond Pythagoras

    72 + 242 = 252 72 = 7 x 7 = 49 242 = 24 x 24 = 576 252 = 25 x 25 = 625 Therefore, 72 + 242 = 49 + 576 = 625 = 252 7, 24, 25 Perimeter = 7 + 24 + 25 = 56 Area = 1/2 x 7 x 24 = 84 From the first three terms I have noticed the following: - * 'a' increases by +2 each term * 'a' is equal to the term number times 2 then add 1 * the last digit of 'b' is in a pattern

    • Word count: 748
  19. Fencing problem.

    Method; Draw three rectangles with a total perimeter of 1000m each, and then calculate their areas. Formula; Scale; Area =l x w, OR, l x b 1cm : 100m Rectangle 1 Rectangle 2 Rectangle 3 More rectangles Length/m Width/m Area/m2 50 450 22500 100 400 40000 150 350 52500 200 300 60000 250 250 62500 300 200 60000 350 150 52500 400 100 40000 450 50 22500 Stage 2; Triangles Now I plan to investigate other shapes to use to build the fence around the field so that I can find out if they give me a bigger area than the rectangles do. The three triangles will have the same base and a perimeter of 1000m.

    • Word count: 970
  20. The rain forest of Amazonia.

    The amount of rubber liquid that he gets might be very little, but he does virtually no damage to the forest. One of the local rubber tree cutters tells "It has been really hard to compete with those big rubber tree companies who just cut down all the trees in the area and only plant rubber trees in the area which gives them such a big amount of rubber liquid, but it also damages the forest so much!" Also, the land in the area are free from the government which destroys even faster, another fact is that the big companies

    • Word count: 685
  21. In this investigation, we have been told to find out the largest possible volume for an open ended tube made from a piece of paper.

    can see that in order to get the optimum volume of the tube using a square or rectangular base, using the 24cm side as the base, it is best to use a square base. Next, I shall see if the same applies to using the 32cm side as the base. Here are the results for the above: 24cm Side y Side x Length of side x Length of side y Area of base (xXy) Volume of tube (area x 24)

    • Word count: 847
  22. The purpose of this experiment is to plan design and carry out an experiment to compare the stomatal density of the upper and lower side of a leaf of my choice.

    Repeat this on the lower surface of one of the other leaves * When varnish is dry peel layer off upper surface of the first leaf. (Use forceps) * Place this on a microscope slide with a tiny amount of water and label the slide U (upper surface) * Repeat this process with the lower surface of second leaf. (label this slide L) * Examine both slides under microscope using high and medium power. * Take an area on each leaf to count stomata.

    • Word count: 879
  23. The Fencing Problem.

    Results Analysis I am going to display my results firstly, in tables. The table in which I record these initial results may resemble this example for a rectangle; A table to show the area of a rectangle with a perimeter of 1000m made up of different combinations of lengths of its sides Number Length(m) Breadth(m) Area (m�) 1 2 3 4 5 At a glance of the table, I will be able to distinguish which combination gives the maximum area.

    • Word count: 864
  24. The Fencing Problem.

    However, I cannot take this for granted and I think using one more shape will be useful in order to back up my theory. Area = 52 500m This proves my theory regarding squares and I shall now put my results into a graph to show what I have found. Length (m) Width (m) Area (m) 400 100 40 000 300 200 60 000 250 250 62 500 150 350 52 500 I will now further my investigation by looking at shapes of a different nature: Regular Pentagon The regular pentagon has 5 sides, and as we get 1000m of fencing, this means each side will be 200m (1000?5=200).

    • Word count: 925
  25. Probability is the Likelihood, or chance, that an event will occur, often expressed as odds, or in mathematics, numerically as a fraction or decimal.

    The probability of a double (two numbers the same) is 6/36 or 1/6 since there are six doubles in the 36 events: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). Independent events are those which do not affect each other, for example rolling two dice are independent events, as the rolling of the first die does not effect the outcome of the rolling of the second die. If events are described as mutually exclusive it means that if one happens, then it prevents the other from happening. So tossing a coin is a mutually exclusive event as it can result in a head or a tail but not both.

    • Word count: 843
  26. Operation Cedar Falls.

    This procedure consisted of detailed maps of information on enemy activity obtained from a variety of sources over an extended period of time. As more data were plotted, patterns of activity and locations emerged. It then became possible to focus prime attention on those areas of intensive or unusual activity. Aerial observation and photography, sensors, patrol reports, infrared devices, sampan traffic counts, enemy probes of Regional and Popular Forces posts, agent reports, civilian movement reports, reports of increased antiaircraft fire, 177 separate enemy facilities found by the 11th Cavalry Regiment, 156, or 88.1 percent, were located within 500 meters of the locations previously reported.

    • Word count: 728

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