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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. Medicine and mathematics

    Therefore we can substitute the ^x to be 15. This will then be: "0.95�5 x 10 = 4.63" "5 when rounded up". Though the following answers shows us that the formula given above is not 100% accurate as the answers should equal 5; as that is the half life of the insulin. As we know the half-life of Insulin, we can use the the function y=10(0.95)x to show how long it takes to reach its half-life. We substitute y=s - Where y is representing the half-life.

    • Word count: 2260
  3. Fencing Problem

    The graph shows a curve which represents the area of a given rectangle. The graph starts with the lowest value and reaches a peak which is achieved when the side lengths are the same, the graphs ends at the same value as it had started with. To make sure that my results are as accurate as they can be, I'm going to draw another table but in decimals and I will be looking between 249.8 and 250.2. Length Width Area (m) 249.8 250.2 62499.96 249.9 250.1 62499.99 250 250 62500 250.1 249.9 62499.99 250.2 249.8 62499.96 This table shows the same result that the maximum area is achieved when both the length and width is 250m.

    • Word count: 2796
  4. t shape t toal

    I will now find the formula for a 5 by 5 grid. Here is the 5 by 5 grid. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 The total of all the numbers inside the T-shape is 1+2+3+7+12=25 As you can see, the middle column is going down in 5's because of the grid size. 1 2 3 7 12 T-Total=25 T-11 T-10 T-9 T-5 T Here is a converted version of the original.

    • Word count: 2267
  5. Investigating different shapes to see which gives the biggest perimeter

    The area of this rectangle is the same as the area of Rectangle 2. This is because a x b is the same as b x a and therefore length (being a) x height (being b) is the same as height (being b) x length (being a). The height of this rectangle is the same as the length of rectangle 2 and the length of this rectangle is the same as the height of rectangle 2. So no matter which way round you put in values, in a multiplication the value will be the same.

    • Word count: 2911
  6. Math Coursework Fencing

    52500 400 100 40000 450 50 22500 500 0 N/A 249.99 250.01 62499.9999 250.99 249.01 62499.0199 The table shows that the rectangle with the maximum area, 62500m2 has lengths of 250 and widths of 250. Therefore this regular rectangle is a square. The values in this table can be used to create a quadratic graph which proves that the maximum area occurs when the length is regular. This quadratic graph has a maximum which occurs when the length is 250 metres, therefore this has to be the length which produces the maximum area.

    • Word count: 2540
  7. GCSE Maths Coursework Growing Shapes

    Length 1 1 2 3 3 5 4 7 5 9 D1 As there are all 2's in the D1 column, the formula contains 2n Pattern no. (n) Length Length - 2n 1 1 -1 2 3 -1 3 5 -1 4 7 -1 5 9 -1 Formula for length of squares - formula = 2n - 1 Check When n = 3 Length = 2n - 1 = 2 � 3 - 1 = 5 Number of Squares/Area Pattern no.

    • Word count: 2572
  8. the fencing problem

    Width (m) Lengths (m) Perimeter (m) Area (m�) 249.5 250.5 1000 62499.75 249.6 250.4 1000 62499.84 249.7 250.3 1000 62499.91 249.8 250.2 1000 62499.96 249.9 250.1 1000 62499.99 250 250 1000 625000 After refining my search into a decimal search this table can tell me that the maximum area stayed the same at 62500m(. I went up by 0.1 for the width and I went down 0.1 for the length so that width and length add up to 1000 meters. The maximum area from this search does not represent a rectangle, instead it represents a square.

    • Word count: 2167
  9. Maths Fencing Coursework

    This I an example to show the methods of how to work out a rectangle. The perimter is 1000m so that means all four sides Will be a fraction of 1000m. 275 cm two sides that are opposite will have the same Area this will make these sides as 275 m 25 cm these sides will then have to be 25 m. To work out the area I will have to do 25 X 275= 11875 The answer that I have worked out ensures me that the graph is right. So now I will have to work out more rectangles to ensure my graph is right.

    • Word count: 2727
  10. 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
  11. 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.

    • Word count: 2972
  12. 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
  13. 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
  14. 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!

    • Word count: 2944
  15. 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.

    • Word count: 2983
  16. 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
  17. The Fencing Problem

    In my investigation, I will prove to you that this is true. Method 1. Firstly, I am going to start with rectangular shape because they are the easiest to calculate seeing that to find the area is a simple multiplication of any two adjacent sides. 2. I will then look at other four-sided shapes like trapeziums and parallelograms and see if they have a larger area than the rectangle with the largest area. 3. Secondly, I will look at the different types of triangles and see which one gives me the biggest area.

    • Word count: 2141
  18. Difference in Japans Two Biggest Regions: Kanto and Kansai

    Osaka is the third-largest city in Japan. It is located on the island of Honshu, at the mouth of the Yodo river on Osaka Bay. The city is one of Japan's major industrial centers and ports, as well as the capital of Osaka prefecture. Historically, Osaka was the center of Japanese commerce. Osaka is known for bunraku (a type of puppetry) and kabuki theatre. Nowadays, most major companies have moved their main offices to Tokyo, but several major companies are still based in Osaka, including Sharp Electronics, and the West Japan Railway Company ("Osaka Prefectural", n.d.).

    • Word count: 2545
  19. The Fencing problem.

    From this evidence, I can put this into an equation. The equation to work out a rectangle is, 1000 = width (500 -width). By using the equation, you can make a prediction table. Length (m) Width (m) 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 220 280 61600 230 270

    • Word count: 2049
  20. Beyond Pythagoras

    (Shortest side) � + (Middle side) � = (Longest side) � This I have down by multiplying the triples by 2 and 3. This is shown as below: TRIPLES 3, 4, 5 X 2 = 6, 8, 10 X 3 = 9, 12, 15 5, 12, 13 X 2 = 10, 24, 26 X 3 = 15, 36, 39 7, 24, 25 X 2 = 14, 48, 50 X 3 = 21, 72, 75 Therefore, 6� + 8� = 10� and 9� + 12� = 15� 36 + 64 = 100 81+144 = 225 100 = 100 225 = 225 They can also be called as Pythagorean triples.

    • Word count: 2599
  21. Mathematics Coursework - Fencing problem

    Right angle triangles: A triangle that has a right angle is called a right angle triangle. Formula: (Base x Height) ?2 Right angle triangle 1: Hypotenuse: 22 + 42 = AB2 4 + 16 = AB2 20 = AB2 AB = 4.47m (2 d.p.) Perimeter: (2 + 4 + 4.47) m = 10.47m Ratio: If 10.47 = 1000 2=X 4=Y 4.47=Z 1000 ? 10.47 = 95.5 Length: 2=X X = 2x 95.5 X = 191 m 4=Y Y = 4 X 95.5 Y = 382 m Area: (191 x 382 ? 2) m = (72962 ?

    • Word count: 2104
  22. Fencing problem

    I will investigate an equilateral triangle because it is in the range we are looking for. The equilateral triangle has a larger area than all the other triangles I have found. I will now find areas of triangle around 333.3 to prove that the equilateral triangle has the largest area. Here is a table to show the results I got from investigating triangles with a base around 333.3m: Base (m) Side (m) Height (m) Area (m�) 333 333.5 288.963 6655 48 112.450 333.25 333.375 288.747 2944 48 112.518 333.3 333.3 288.675 1288 48 112.522 333.4 333.3 288.617 3938 48 112.520

    • Word count: 2092
  23. Find the biggest value of the ratio area / perimeter for some triangles.

    due to lack of information or simply impossibility to draw: The triangles that are impossible to draw are; l, m, o, p, v, w, x, y, z, ab and ac. I have only managed to find one triangle that is that same as another. This is triangle q, which is the same as triangle k. To work out the areas, perimeters and ratios of the triangles, I will use the below formulas: Perimeter = a + b + c Area = square root of s(s-a)(s-b)(s-c), s=perimeter/2 = 1/2 abSinC (for triangles "u" and "aa" only)

    • Word count: 2221
  24. Geographical Inquiry into the proposed redevelopment plan of the Elephant and Castle.

    Regeneration can be seen inside every city within the developed world, as it has become a trend. The regeneration of areas will occur, where the inner city area becomes run down and derelict and it is felt that it needs rejuvenation. My work will become focused through the relevance of this topic. Regeneration can happen all over as it is a global concept which is a growing trend. An international example is East Berlin, which has benefited from huge capital investment. Berlin's urban planner Heinz Willumat says, "reform requires a strategic vision." 700,000 people live in panel-block housing, which cannot be destroyed, "they need to consider the further development and improvement of these regions."

    • Word count: 2274
  25. Maths Investigation on Trays.

    Table of Formulae x Area Base Area 4 sides Volume 1 (18-2)2 4(18-2) 1(18-2)2 2 (18-4)2 8(18-4) 2(18-4)2 3 (18-6)2 12(18-6) 3(18-6)2 4 (18-8)2 16(18-8) 4(18-8)2 5 (18-10)2 20(18-10) 5(18-10)2 6 (18-12)2 24(18-12) 6(18-12)2 7 (18-14)2 28(18-14) 7(18-14)2 8 (18-16)2 32(18-16) 8(18-16)2 9 (18-18)2 36(18-18) 9(18-18)2 Here is the algebra I am using to find out the different results for the three table headings. The algelbra here is easy to figure out, it has no tricky parts Table of results for Grid 18 by 18 x Area Base Area 4 sides Volume 1 256 64 256 2 196 112 392 3 144 144 432 4 100 160 400 5 64 160 320 6

    • Word count: 2100
  26. Investigate different shapes of guttering for newly built houses.

    The sides will be equal because if I made the lengths of the sides different the water would run out over the shorter side. So each side will be 15cm. To find the area of the triangle I will be using the formula: Area = 1/2 a?b?sin c C = 50? Area = 1/2 a?b?sin c = 0.5 ?15?15?sin 50? = 86.17cm� C = 60? Area = 1/2 a?b?sin c = 0.5 ?15?15?sin 60? = 97.43cm� C = 70? Area = 1/2 a?b?sin c = 0.5 ?15?15?sin 70?

    • Word count: 2073

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