GCSE: Fencing Problem
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The Fencing Problem
I will investigate 2 different families of shapes and try to work out a pattern between them. These are the results from my investigation a b Area = 500(500a) (500b) (500c) c Area = 500(500333 1/3) (500333 1/3) (500333 1/3) =83334.248 Area = 500(500300) (500350) (500350) =47434.164 Area = 500(500250) (500450) (500300) =35355.339 Area = 500(500250) (500500) (500250) =6123.724 Area = 500(500150)
 Word count: 321

Maths Coursework  Beyond Pythagoras
52 + 122 = 132 52 = 5 x 5 = 25 (Shortest Side) 122 = 12 x 12 = 144 (Middle Side) 132 = 13 x 13 = 169 (Hypotenuse) 52 + 122 = 25 + 144 = 169 = 132 Perimeter 5 + 12 + 13 = 30 units Area 1/2 ( 5 x 12 ) = 30 square units The next triangle in the sequence is the 7, 24, 25 triangle. This triangle also satisfies the condition...... a2 + b2 = c2 Because....... 72 + 242 = 252 72 = 7 x 7 = 49 (Shortest Side)
 Word count: 4002

Maths Fence Length Investigation
56100 180 320 57600 190 310 58900 200 300 60000 210 290 60900 220 280 61600 230 270 62100 240 260 62400 250 250 62500 260 240 62400 270 230 62100 280 220 61600 290 210 60900 300 200 60000 310 190 58900 320 180 57600 330 170 56100 340 160 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 500 0 0 Using this formula I can draw a graph of base length against area.
 Word count: 2220

Fencing problem
This is so that I can show what the corresponding measurements are for each point on my graph. When I narrow my search down to rises and falls of 12.5, I obtained these results: This shows that the greatest area, which can be covered by a RECTANGLE, is 62343.75sq2m. A width of 262.5 and a length of 237.5 obtain this. The greatest area covered by a foursided shape however, is a square, 250X250, which covers 62,500 sq2m. Width (m) 300 287.5 275 262.5 250 237.5 225 212.5 200 Length (m)
 Word count: 978

Counting Stomata
This is a technique. Method: The leaf was placed onto graph paper and its area calculated. This was recorded. A small section of a given leaf was painted with nail polish and then let to dry. A section of clean selo tape was then placed carefully onto the painted section.
 Word count: 292

Maths Dots Investigation
Jonathan parsonage Investigation one. 0 dots inside shape. Number of dots joined Area(cm) 4 1 5 1.5 6 2 7 2.5 8 3 9 3.5 10 4 prediction 11 4.5 rule no. Of dots joined 1 2 Prediction I predict that when 11 dots are joined, the area will be 4.5cm. My prediction was correct. Formula D= no. Of dots joined A= D 1 A= area 2 Jonathan Parsonage investigation two.
 Word count: 550

Advantages and Disadvantages of To Proposed Sites of a Call Centre
in the NE England area are preserved to be reasonably successful and thus provide a reason for a possible call centre The Bolton site is situated in an area that may not be regarded as such a successful area to locate a call centre, the survey and results were taken from the views of call operators and so may not be a fair reflection on where is successful to locate a call centre. The results do offer some value in that the people on the 'shop floor' may have a good knowledge of call centres and their success; if the results are of any use then Longbenton is the best option.
 Word count: 1134

Maths Coursework  The Fencing Problem
170 330 56100 180 320 57600 190 310 58900 200 300 60000 210 290 60900 220 280 61600 230 270 62100 240 260 62400 250 250 62500 260 240 62400 270 230 62100 280 220 61600 290 210 60900 300 200 60000 310 190 58900 320 180 57600 330 170 56100 340 160 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 The
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The fencing problem
x width Area = 450 x 50 = 22,500m� I have experimented with quadrilaterals and have decided to create a table with lengths varying from 0 to 500, going up in 10's. I have created this in a table below using the formula: Length x Width = Area Length (m) Width (m) Area (m2) 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
 Word count: 1293

Maths Coursework: The Fencing Problem.
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 62100 240 260 62400 250 250 62500 From this table I can draw a graph to show the width against the area of a rectangle.
 Word count: 2252

The Fencing Problem
This means that you can work out the area if you only have the length of one side. To work out the area of the rectangle, I subtract 150 from 500, giving 350 and then times 150 by 350. I have made an equation to show this. x(500  x) = 1000 eg: 200(500  200) 200 x 300 = 60000m� As you can see, the graph has formed a parabola. According to the table and the graph, the rectangle with a length of 250m has the greatest area.
 Word count: 1609

History Coursework: Local Study, Stanton Drew Stone Circles
The main circle consists of 28 stones (30 originally), 2 stones have been located using probes. Only 4 stones in this circle remain standing the rest are recumbent. The circumference of the circle is 351.85m this was calculated by using half of the diameter and substituting it into the formula 2?r. The diameter was 112m or 368ft this gave me a radius of the circle being 56m. The average distance between the stones (along a straight line) was 7.4m and the average height of the stones (the four remaining standing)
 Word count: 3629

Equable shapes Maths Investigation
3 9 27 24 3 10 30 26 4 1 4 10 4 2 8 12 4 3 12 14 4 4 16 16 4 5 20 18 4 6 24 20 4 7 28 22 4 8 32 24 4 9 36 26 4 10 40 28 5 1 5 12 5 2 10 14 5 3 15 16 5 4 20 18 5 5 25 20 5 6 30 22 5 7 35 24 5 8 40 26 5 9 45 28 5 10 50 30 6 1 6 14 6 2 12 16 6 3 18 18
 Word count: 1713

Four Sided Shapes Investigations
300 m AREA = BASE x WIDTH 350 m = 150m x 350m = 525000m2 150 m 150m 350 m 250m AREA = BASE x WIDTH = 250m x 250m = 62500m2 250m 250m 250m AREA
 Word count: 211

The Fencing Problem
400m Perimeter = 1000m 2 ( 400 = 800m 2 ( 100 = 200m 800m + 200m = 1000m I will now find the area of the rectangle using the same equation as before, Base ( Height. Base = 100m Height = 400m 100 ( 400 = 40,000m� Rectangle 3: This rectangle will have a base of 50m and a height of 450m. Perimeter = 1000m 2 ( 450 = 900m 2 ( 50 = 100m 900m + 100m = 1000m I will now find the area of the rectangle.
 Word count: 5550

The Fencing Problem
What she does wish to do is fence off the plot of land, which contains the maximum area. The answer is a circle but in this course work I will try to explain why. All numbers will be rounded to 2 decimal places. Regular polygons have the largest areas This shows that the closer the values of the height and width the larger the area.
 Word count: 300

The multiplier effect with reference to Newport and Setubal
If it is a manufacturing industry, then other industries may set up nearby to keep transport costs down. A good example of the multiplier effect is in the region of Portugal called Setubal. The carmakers Ford and Volkswagen decided to build a �1.7billion car manufacturing plant in the area, which created three thousand jobs.
 Word count: 287

A Hydraulic Device.
This should be greater than one for the device to be worth making. Examples of hydraulic devices are quite common. Car brakes depend on a small force applied at the brake exerting a large force on the wheel to stop the car. I selected two different sized glass syringes and connected them with a rubber tube filled with water.
 Word count: 220

The fencing problem
Area = (500333.333) * V [333.333�  (500333.333) �] Area = 166.667 * V [83333.333] Area = 48112.52243 From this result I can see that my formula works. Here is a table of results for the areas of isosceles triangles As the length of the base of an Isosceles triangle is increased (and so the lengths of the equal sides decrease) the area decreases. The same thing happens when the length of the base is decreased and the length of the equal sides is increased.
 Word count: 1438

The Fencing Problem
3Sided Shapes For an equilateral triangle I know that all sides are equal and therefore the length of each side must be 1000/3, if the perimeter is 1000m. To find the area of a simple 3sided shape, you do: 1/2 base length x height. To find the perpendicular height (h), we can use Pythagoras Theorem, or trigonometry. I would rather use Pythagoras for the investigation of the 3sided shapes, because we always know two of the side's lengths, whereas we don't always know the angles.
 Word count: 3346

Tube Maths Investigation
1 2A = 10.66 x v(10.662  5.332) x V = 1182cm3 This is the result for a triangle with height of 32cm, and perimeter of 24cm: 1 2A = 8 x v(82  42) x V = 887cm3 From this a formula could be established: 1 2V = p x v p 2  p 2 x x h 3 3 6 V = p x v p2 xh 6 12 The next shape I used as a base was a hexagon.
 Word count: 1428

Estimation of ð
Fig. 2 To find the area of the inscribed nsided polygon, we first have to find the area of one isosceles triangle, which has one angle that is formed by subtending lines from the centre to 2 consecutive vertices, e.g. A and B (fig. 2). The angle ? at the centre As the area of a triangle is given by the formula, , substituting values from Fig. 2 into the formula gives: Area of one triangle, So area of inscribed polygon 3.
 Word count: 2482

Measuring Stomatal Numbers on Abaxial and Adaxial Surfaces
Leave to dry. 4) When complete, remove in turn, a large an area as is possible with the fine forceps of the clear coating. 5) Put on slides respectively with the clear slips situated on top, marking their reference names next to each. 6) Setup the microscope and place Upper1 underneath. 7) Measure out using the 0.3m ruler and a ball point marker pen a square of 0.3cm by 0.3cm to give an area of 0.09cm�.
 Word count: 400

The Fencing Problem
Triangles Next, I will look at the maximum area of a triangle with a perimeter of 1000m. I will use the cosine rule to work out one of the angles: Cos A = (b2 + c2  a2) / (2bc) Cos A = (2502 + 3502  4002) / (2 x 250 x 350) Cos A = 0.14 so A = 81.787o Now that I have one angle, I can work out the area: Area = 1/2 x b x c x Sin A Area = 1/2 x 250 x 350 x Sin (81.787)
 Word count: 2247