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GCSE: Fencing Problem

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  1. Maths Investigation: Drainage Channels.

    1/2A = 573cm� So the area of the semicircle is 573cm� Triangles Example 1 30cm 30cm As there are two sides to a triangle the 60cm length must be split in half, so each side is now 30cm long. But there can also be a variety of angles, ranging from 0� to 90� Here is the area for 30� Sin15� = x/30 X = 0.26 x 30 X = 7.76 Cos15� = h/30 H = 28.89 Area = 7.76 x 28.89 =

    • Word count: 520
  2. Tangled Triangles

    Then I went on to work out the combinations for 60 o as the angle and then for 80 o. This left me with the first nine combinations. The next task was to work out all the combinations using two angles. I started by using 40 and 60 which gave three combinations, then 60 and 80 and then finally 80 and 40. Now I had explored all the possible combinations of two angles and was left with nine more combinations, three for each.

    • Word count: 589
  3. Mathematics Gcse Coursework Tubes Investigation

    So this tells us that using the paper landscape holds a greater volume than using the paper portrait. The diagrams below show the volume of some shapes using both methods: Paper: 24cm by 32cm First method: using the paper landscape Cube Width � Length � Height = Area of Cube Width: 8cm Length: 8cm Height: 24cm 8 � 8 � 24 = 1536cm3 Triangular Prism Base � Height = Cross Section � 2 � height of tube. 32 � 3 = 10.7 length of one side 10.7 � 2 = 5.3 half of one side 5.35square + b square =

    • Word count: 1974
  4. Tubes Maths Investigation

    Below are two tables showing the volumes of different cuboid tubes. 32cm base Height (cm) Volume (cm3) 1x15 24 360 2x14 24 672 3x13 24 936 4x12 24 1152 5x11 24 1320 6x10 24 1440 7x9 24 1512 8x8 24 1536 24cm base Height (cm) Volume (cm3) 1x11 32 352 2x10 32 646 3x9 32 864 4x8 32 1024 5x7 32 1120 6x6 32 1152 From these results I can see that having a larger base and smaller height gives a larger volume.

    • Word count: 1375
  5. Mathematics Coursework - Beyond Pythagoras

    And also if you can see in the shortest side column, it goes up by 2. I have also noticed that the area is 1/2 (shortest side) x (middle side). 3) In this section I will be working out and finding out the formulas for: * Shortest side * Middle side * Longest side In finding out the formula for the shortest side I predict that the formula will be something to do with the differences between the lengths (which is 2).

    • Word count: 1924
  6. The Fencing Problem

    Shape Equation Total area Perimeter Equilateral: 333.3+333.3+333.3 24,052� 1000�3= 333.3 288.64 x (333.3�2) =48,103� x 2 To find the area of this regular triangle I must: * Divide 1000 which is the perimeter by 3 which is the number of the polygon's sides to give me 333.3 * Find the vertical height. To do this I use Pythagoras theorem a�+b�=c� but because I have to find the shorter area I use an invert equation and I could say a�= c�- b�. So I half the base and I draw a line of symmetry (making two triangles)

    • Word count: 679
  7. Environment Questionnaire

    Do you think that there is a lot of noise pollution in your neighbourhood? ? Yes ? No 4. If so; on which day, and is there a particular time? ? Monday- Thursday ? Friday- Sunday ?

    • Word count: 210
  8. The Distribution of Daises, Plantains, Moss and Grass.

    These readings would include the plants that I have chosen to measure, i.e. daises, plantains, moss and grass. In these areas it is possible to get more than 100% coverage because it is possible that the area is 90% covered by moss and yet growing up through it is a 40% covering of grass. I must take the same amount of quadrates in each area to make sure there is a fair test. It is essential to have a fair test other wise the results I take would be inaccurate.

    • Word count: 2334
  9. Tubes Investigation

    x 24 = 1440cm( A = 9 x 7 = 63 V = 63 x 24 = 1512cm( A = 8 x 8 = 64 V = 64 x 24 = 1536cm( A = 7 x 9 = 63 V = 63 x 24 = 1512cm( This examination proved my theory again, that squares have the biggest volume. This proves that shapes that are regular have a bigger volume than those that are not regular. The biggest volume the rectangle ever reached was 1536cm�, to enforce this fact I have drawn a table: 24cm length Length x Width x Depth

    • Word count: 2978
  10. The Fencing Problem

    Triangle Where h= height, a = area, n = number of sides and l = length of each side. Equilateral Triangle = 48112.5243m2 Square length x width = Area 250 x 250 = 62500m2 Pentagon (Regular, 5 sides) Triangle is a section of the pentagon. Pentagon = 68819.0960235m2 Hexagon (Regular, 6 sides)

    • Word count: 533
  11. Beyond Pythagoras

    a right-angled triangle The Perimeter = 5+12+13= 30 units The Area x5x12= 30 units The units 7,24 and 25 can also be length in appropriate units-of a right-angled triangle The Perimeter = 7+24+25= 46 units And the Area

    • Word count: 291
  12. Beyond Pythagoras

    Smallest number + middle number + largest number = Perimeter Example: - Perimeter = 3 + 4 + 5 = 12 units To find the area of the right - angled triangle I need to know the height and the length of the triangle. These on the triangles are the smallest number and the middle number. Example: - Numbers 3,4 and 5 Area = 0.5 or 1/2 * 3 * 4 = 6 units. To make things easier you can change the sides to numbers instead of saying that one side is the smallest side or the largest side.

    • Word count: 2932
  13. Surface Area: Volume Ratio Investigation

    * Heat may escape from the water while I am pouring it into the glassware. * Human error e.g. misreading measurements I predict that the smaller the surface area compared to the volume the longer it will take for heat loss. I have decided on this prediction after researching homeostasis. Homeostasis is the maintenance of a constant internal environment, by balancing bodily inputs and outputs and removing waste products by an animal. In my research looked at a polar bear. A polar bear lives in a cold, harsh habitat. Polar bears have a number of ways of keeping warm in this weather but most important to this investigation is that polar bears have a smaller surface area compared to their body core.

    • Word count: 1191
  14. Koch’s Snowflake Investigation

    Shape Number 1 2 3 4 No. Of Edges 3 �4 12 �4 48 �4 192 Length of Each Edge 9 �3 3 �3 1 �3 ? Perimeter 27 36 48 64 I have now noticed that the perimeter is increasing in a geometric sequence, by ? of its value each time. It is also increasing by a larger amount each time, and so is a divergent series. For example: 27 � 1? = 36 36 � 1? = 48 48 � 1?

    • Word count: 1173
  15. Using Tai Po as a case study, is clustering move evident in an older of newer shopping area?

    Background The research area for this investigation will be Tai Po (see figure 1) of the many residential and industrial new towns in Hong Kong. Covering an area of 147.43 square kilometers and a population of over 320,000, Tai Po is one of the least densely populated settlements in Hong Kong. This makes Tai Po one of the least crowded areas for shopping. Tai Po has two main shopping districts, the old and new town center which are approximately 0.5 km apart and is divided by the Lam Tsuen River (figure 2).

    • Word count: 997
  16. To investigate the areas of different shapes when they are joined together on square dotted paper

    I will also be predicting shape areas and explaining patterns and formulas and trying to justify why the formulas work. I will also be explaining the decisions that I have made. I ultimately aim to find a formula that is able to give me the area of any shape drawn on dotted paper, then testing it to see if it works on any shape (the nth term). Equipment I will need: - Calculator, pens, pencils, rubber and square dotted paper.

    • Word count: 2315
  17. Calculate the Area of a Shape

    Now I know I am defiantly correct with my formula I can find area of a shape of any size as long as there are no dots inside. E.G N=68 A=68 - 1 = 33cm2 Now I have investigated shapes with 1 dot inside I will put my answers into a table. Number of dots N Area A 3 1 1/2 4 2 5 2 1/2 6 3 7 3 1/2 * After creating a table 2, I have noticed a pattern in the area column.

    • Word count: 1754
  18. Perfect Shapes

    An extra column has been added: Perimeter - Area, this allows area and perimeter to be quickly contrasted. A negative value here indicates that the area is larger than the perimeter and a positive one that perimeter is larger than area. The smaller (magnitude) this value, the closer the rectangle is to being perfect. Notice that a 5 by 3 rectangle is the closest to being perfect. A value of zero in this column would indicate a perfect shape and it is interesting to note that a sign change takes place between 5 by 3 and 5 by 4.

    • Word count: 2357
  19. The Fencing Problem

    The first column in my table determined which triangle I was finding the area of. I numbered these to save confusion from 1-50. The Base column had the possible value of the base, which ranged from 10 to 500 going up in 10. The other two sides were also done in two separate columns, which went up in 5. The three columns, which had the base, side one and side, two all added up to 1000. There was a formal involved even in these simply sequence of numbers. For the base the formula was the previous cell plus 10.

    • Word count: 1764
  20. Acoustics Assignment

    These are the points which undergo the maximum displacement during each vibrational cycle of the standing wave. In a sense, these points are the opposite of nodes, and so they are called antinodes. A standing wave pattern always consists of an alternating pattern of nodes and antinodes. Fundamentals. The first possible standing wave is called the fundamental, alternatively called the first harmonic, the second possible standing wave is called the 2nd harmonic, the third possible standing wave is called the 3rd harmonic, and so on. The fundamental is the lowest frequency produced by the instrument or source.

    • Word count: 3003
  21. Beyond Pythagoras

    When this happens as a general rule the formulae must have 2n then I found the difference between 2n and the actual length in each case, this happened to be 1. Example of a=2n+1: The answer to a when n=6 is 13.The answer to the formula: 2x6+1=a 2x6=12 12+1=13 so the formula is correct. b=2n2 +2n This is because there is a second difference between each number. The first difference is not the same every time but the difference between this first difference was a constant of 4.

    • Word count: 2107
  22. Borders Investigation

    For our 2D cross, I will define the first cross as a single square, but give it a value of n equal to zero. This means that the value of n for any 2D cross shape can be seen as the number of borders that surround the centre square, or the number of squares which the shape extends away from the centre. The diameter can now be conveniently expressed as : Perimeter If we measure the perimeter of the first four crosses and put them on a table together with their respective values of n, we can derive a formula that expresses the perimeter in terms of n for any cross.

    • Word count: 2343
  23. Investigating the Relationship Between Heat Loss and Surface Area

    I can back this up with evidence such as; a baby is smaller than an adult, and loses heat more quickly than an adult. This is because of the surface area:volume of the baby is higher than the adult's. The adult holds more volume than the baby, but has less surface area in comparison to the baby.

    • Word count: 388
  24. Take ivy leaves from the North and South side of a hedge to come to the conclusion of what natural factors will effect the ivy leaves and why.

    I also think that the ivy leaves that are in the shade most of or all the day will be larger because they do not get as much energy from the sun as the ones on the other side so it they do ever get sun light they will have to be bigger so they can take in as much energy as possible. Other factors which might affect my results is the kind of tree the ivy has grown up because of how much nutrients and minerals it is getting from the ground.

    • Word count: 830
  25. Fencing Problem - Maths Coursework

    54400 350 150 52500 360 140 50400 370 130 48100 380 120 45600 390 110 42900 400 100 40000 410 90 36900 420 80 33600 430 70 30100 440 60 26400 450 50 22500 460 40 18400 470 30 14100 480 20 9600 490 10 4900 Using this table I can draw a graph of height against area. This is on the next sheet. As you can see, the graph has formed a parabola. According to the table and the graph, the rectangle with a base of 250m has the greatest area.

    • Word count: 1412

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