GCSE: Fencing Problem
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 Level: GCSE
 Questions: 75

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 halflife of Insulin, we can use the the function y=10(0.95)x to show how long it takes to reach its halflife. We substitute y=s  Where y is representing the halflife.
 Word count: 2260

The Fencing Problem
x h]} x 5 * Hexagon => A = {1/2[(1000 � 6) x h]} x 6 * Octagon => A = {1/2[(1000 � 8) x h]} x 8 * Decagon => A = {1/2[(1000 � 10) x h]} x 10 * A polygon with 15 sides (Pentadecagon) => A = {1/2[(1000 � 15) x h]} x 15 * A polygon with 20 sides (Icosagon) => A = {1/2[(1000 � 20) x h]} x 20 * A polygon with n number of sides => Remains to be seen...mWaHaHa! * Circle => A = ?r� I will conclude this investigation in the form of a graph depicting the maximum areas for each shape examined, and I will review the shape which holds the most area whilst keeping with the specified perimeter (1000m).
 Word count: 6538

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

Fencing Problem
I would do the calculation (Sc), which would be 500m  400m = 100m * Following that I would multiply the product of (Sa) , (Sb) and (Sc) by S (500) to give me the result which will : 500 x 200 x 200 x 100 = 2000000000 m� * Then I would square root () the answer of S(Sa)(Sb)(Sc) to give me the area of the triangle which will be : 2000000000 m� = 44721.35955 m� PERIMETER = 1000 m AREA = 44721.35955 m� Equilateral I will start investigating triangles with Equilateral triangles.
 Word count: 5382

Maths:Fencing Problem
I will go up by a pitch of 1m. Base Height Area 245 255 62475 246 254 62484 247 253 62491 248 252 62496 249 251 62499 250 250 62500 251 249 62499 252 248 62496 I will double check again to be certain that I still get maximum at 250m length by 250m.I will use a pitch of 0.1m between 249.0 and 250.2. Base Height Area= l*w 249 251 62499 249.1 250.9 62499.19 249.2 250.8 62499.36 249.3 250.7 62499.51 249.4 250.6 62499.64 249.5 250.5 62499.75 249.6 250.4 62499.84 249.7 250.3 62499.91 249.8 250.2 62499.96 249.9 250.1 62499.99 250 250 62500 250.1 249.9 62499.99 250.2 249.8 62499.96 Using Microsoft
 Word count: 1483

The Fencing Problem
To save time by going through all the possible calculations by calculator, I am going to create a spreadsheet. I will change the length of the base of the triangle in steps of 10m. The formulas I will use in the spreadsheet are the same as above. They are: s = (1000b)/2 h = ( (((1000b)/2)�(b/2)�) a = ( (((1000b)/2)�(b/2)�) * (b/2) The results are as followed: base/m side/m height/m area/m� 10 495 494.975 2474.874 20 490 489.898 4898.979 30 485 484.768 7271.520 40 480 479.583 9591.663 50 475 474.342 11858.541 60 470 469.042 14071.247 70 465 463.681 16228.832 80 460 458.258 18330.303 90 455 452.769 20374.617 100 450 447.214 22360.680 110 445 441.588 24287.342 120 440 435.890 26153.394
 Word count: 3635

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

Perimeter Investigation
I decided to use regular shapes throughout my investigation because only regular shapes, i.e. shapes with equal sides, give the maximum area. So this is why I decided to use equilateral triangles. In my triangle investigation, I decided to start with a base of 50 metres and investigate the area using that base. I did not investigate triangles with a base above 450 metres because the area kept on decreasing. For example, the triangle with base 500 metres and sides 250 metres gave an area of 0 metres.
 Word count: 1274

Fencing Prblem
to the nearest whole number. Triangle 2 I chose a different size isosceles triangle to continue my investigation 425m 425m 150m I divide the Isosceles triangle 425 m2  75 m2 = Height m2 140625  5625 = 175000 Height = square root of 175000 Height = 418.33 to two decimal places To calculate the area of this Isosceles triangle we calculate: Area = 1/2 x base x height Area = 1/2 x 150 x 418.33 Area = 31,375 m2 (meter squared)
 Word count: 1818

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

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

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

t shape t toal
+ (T19) = Ttotal 5T  63 = Ttotal I will see if this new formula still works. 5 ? 20  63 = T total 100  63 =T  total 37 = 37 5 ? 21  63 = T total 105  63 = T  total 42 = 42 Part 2 I will now as part of my investigation use different grid sizes, transformations of the tshape and investigate the relationship between both. Then I will see how the tnumber and the ttotal relate to the new factors. The smallest grid size can only be a 3 by 3 grid because that is the smallest grid size a tshape can fit on.
 Word count: 1785

t shape t toal
From this we are able to work out some parts of the formula for a 9 by 9 grid. Taking the Tnumber of 20 as an example, we can say that the Ttotal is gained by: t = (20  19) + (20  18) + (20  17) + (20  9) + (20  0) = 37 As there are 5 numbers in each Tshape, we have to use five lots of twenty. The number above the 20 is 11, which is 9 less than 20; the other numbers in the Tshape are 1,2 & 3, which are 19,18, & 17 less than 20.
 Word count: 4517

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 Tshape 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 TTotal=25 T11 T10 T9 T5 T Here is a converted version of the original.
 Word count: 2267

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

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

Geography Investigation: Residential Areas
Figure 2 Source: Google Images Both these models are based on stereotypical towns and cities, thus they are not going to fit exactly for Basingstoke. However, when my investigation is over I will be able to see if the transition goes in the same order and come up with my own model for Basingstoke. Before I can investigate anything I must come up with what I want to find out, i.e. hypothesis that will be the basis of my investigation throughout.
 Word count: 10454

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

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

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

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

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

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

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