GCSE: Fencing Problem
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The fencing problem.
This means that you can work out an area even if you have the length of one side. To work out the area of a rectangle with a base length of 200m, I subtracted 200 from 500, which is 300 and then multiplied 200 by 300 I can put this into an equation.. =1000= n (500n) n = length of side. n 500n Area 25 475 11875m 50 450 22500m 75 425 31875m 100 400 40000m 125 375 46875m 150 350 52500m 175 325 56875m 200 300 60000m 225 275 61875m 250 250 625000m As you can see from the above table, the value that produces the highest area is 250.
 Word count: 1273

Beyond Pythagoras.
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 4, 2, 4 * the last digit of 'c' is in a pattern 5, 3, 5 * the square root of ('b' + 'c') = 'a' * 'c' is always +1 to 'b' * 'b' increases by +4 each term * ('a' x 'n')
 Word count: 1235

Find out the relationship of the dots inside a shape of different sizes.
DIAGRAM 4 AREA DOTS PERIMETER 4 cm2 25 16 DIAGRAM 5 AREA DOTS PERIMETER 5 cm2 41 20 I have realised that the comment I made before for the dots and area is incorrect. What I said was that after starting with four the number will double itself on each stage i.e. 4, 8, 16... etc. PREDICTION: I think that the shape with the area of 6 cm2 will have 61 dots and its perimeter will have 24 dots. RESULTS TABLE SIZE (cm2)
 Word count: 1235

A group of researchers set up a series of observation sites in the Arctic on a circumference of a circle of radius 50km.They need to set up a base camp so they can visit a different site each day.
A 50 km B  Plane = 100 mph Wind speed = 0 = 100*1.6 = 160 kmh1 Time = Distance/Speed = 50/160 = 18.75 minutes For the whole journey A to B then back to A would be: Time from A to B multiplied by 2. 2*18.75 = 37.5 minutes Equation Time (A to B to A) = 2(Distance/Speed) OR Time (A to B to A) = Total Distance/Speed Model 2 1. Base at centre of the 360 degree circle 2.
 Word count: 1010

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(sa)(sb)(sc), s=perimeter/2 = 1/2 abSinC (for triangles "u" and "aa" only)
 Word count: 2221

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� 225625625 = 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

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

Evalaute the Success of the Regeneration of London Docklands. Which Groups Have Benefited, Which Have Not From the Investement Made By the LDDC?
They sold the derelict land very quickly by clearing it and leaving it ready for developers to expand on. They also achieved this as they offered very good packages on the land for new companies to set up on; no fees or rent for 10 years was a main factor in this attraction scheme. New companies saw this as a perfect stepping stone or base to build on, and this also gave the Docklands a new revived image with modern firms based there. Lack of planning permission and cheap construction made establishment very easy. The LDDC benefited from selling this land, not only did it improve the image but it provided them with more money, as well as the grant to build around the new firms to complete the package of the revamped area.
 Word count: 1272

The fencing problem.
= r Area = Pi x r� Area = Pi x (Circumference / (Pi x 2)) � Pi x (1000m / (pi x 2)) � = 79577.45m� I predict that for regular shapes the more sides the shape has the higher the area is. A circle has infinite sides in theory so I will expect this to be of the highest area. The above only tells us about regular shapes I still haven't worked out what the ideal shape is.
 Word count: 1263

Investigating different shapes of gutters.
* (height) I have drawn a semicircular prism to illustrate the cross sectional area (yellow) and the height (red). The height isn't needed because it will be the same in all the shapes I use. The maximum volume is determined by the maximum cross sectionalarea. First I will find the area of the circle then divide that by two to get the area of the semicircle. Area of a circle= =? * radius I don't know the radius but I do know the circumference is 2L, which is equal to 2?r Circumference= = Radius ( r )
 Word count: 1504

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

Fencing problem.
Finally, I will experiment with the area of the circle, which is: 2?r (2*3.124*radius). For each shape, I am going to arrange the data into a results table, so that it can be easier for me to analyse the results. To help me explain the trend or pattern of results that I will be working out on spreadsheets (in Microsoft Excel), it is necessary that I translate the results onto graphs to back up the conclusions that I state for each shape.
 Word count: 3279

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 panelblock housing, which cannot be destroyed, "they need to consider the further development and improvement of these regions."
 Word count: 2274

Fencing problem.
290/2=145 area = 145x290=42050 290m c This in an isosceles triangle. Angles a and b = 75o so angle c = 30o 400m I will now use trigonometry to find the height. I will use the "tan" rule tan76= opp/100 = so opp= tan76 x 100 opp=373m area= 100 x 373 = 37300 a b 200m As you can see from the results an equilateral triangle has the largest surface area of any triangle. This is interesting because an equilateral triangle has all equal sides. I believe that this will prove significant in later shapes as well.
 Word count: 1348

Maths Investigation on Trays.
Table of Formulae x Area Base Area 4 sides Volume 1 (182)2 4(182) 1(182)2 2 (184)2 8(184) 2(184)2 3 (186)2 12(186) 3(186)2 4 (188)2 16(188) 4(188)2 5 (1810)2 20(1810) 5(1810)2 6 (1812)2 24(1812) 6(1812)2 7 (1814)2 28(1814) 7(1814)2 8 (1816)2 32(1816) 8(1816)2 9 (1818)2 36(1818) 9(1818)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
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Fencing Problem.
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 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. This shape is also called a square.
 Word count: 1404

Investigating what affects the time taken for a paper helicopter to reach the ground.
4m allowed all the above to happen without discounting our results My prediction is that as the surface area decreases the time taken for the helicopter to fall 4m will increase. Theory: I think this will happen because if an object has a larger surface area it will have more weight, this creates more drag or air resistance and create a turning force for it to move through the air. I do know though that objects that have a larger surface area will create more air resistance or drag, than a smaller falling object and this increased air resistance might slow my helicopter down.
 Word count: 1164

Investigate the affects of the surface area: volume ratio on the cooling of an organism.
All the rest I will keep the same. Therefore my results should be more accurate. Prediction: I predict that the beaker with the smaller will lose heat faster than the larger beaker. This is because of the differences in surface area relative to volume. In a larger object, of the same shape, the amount of surface area per unit of volume is less than in an object of a smaller size. This can be more easily seen with cubes. Surface area to volume ratios of above cubes Small cube 1: 6 large cube 1: 3 The smaller cube has a
 Word count: 1350

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

Geography Coursework in North Kensington.
Most of the houses were built in the 1980's according to a typical Victorian housing pattern. We found from pictures of the houses that they gave the impression of being well looked after, many with fresh coats of paint and small back gardens out of sight. The second area was the Cheltenham Estate which consists of the Trellick Tower with a variety of flats going from 1 bedroom to 3 bedrooms and the building has a height of 322 feet. It is a very unique building in the way in which it has been designed. It's very tall and upright, however, the building does not recede as far back as we might expect it to.
 Word count: 3281

The Fencing Problem.
Using rectangles as the shape we can see the areas are always increasing until the length and the width equal each other; this is because if you go beyond the point where the shape is a square the areas would start to repeat itself. This brings us to a conclusion that the highest area that we can get is when we have a square where the sides equal each other. The point where we are using a square using a is where the length and the width is 250m each this gives us a maximum area fir rectangles being at 62500m2.
 Word count: 1771

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

Investigating heat loss in different sized animals.
From my preliminary experiments I found out that the larger the object the smaller the surface area to volume ratio. I can show this by: Surface area = 6cm2 Volume = 1 cm3 SA: VR = 6/1 = 6 Surface Area = 24cm2 Volume = 8 cm3 SA: VR = 24/8 = 3 Surface area = 54cm2 Volume = 27cm3 SA: VR = 54/27 = 2 These cubes show me that the cube with the most blocks has the smallest surface area to volume ratio, but the cube with the least amount of blocks has the largest surface area to volume ratio.
 Word count: 1639

The aim of this investigation is to find what is the maximum area you can obtain with the perimeter of 1000m.
I have put it in a table because it is easier to analyze and evaluate. The formula that I entered The formula used here is: =SQRT (500*(500A2)*(500B2)*(500C2)) After examining the triangles lengths and areas I have found out that isosceles triangles are the triangles with the larger areas this is confirmed by the result table. Also another way of checking that my theory is right is by trying out the formula 1/2(base x height). When either the height or the base is a large number the area of the triangle will be larger. In the diagram below you can visually tell that when the height is at its highest possible point the area is at its largest.
 Word count: 1786

The Fencing Problem.
I will change the value of the widths and go up in increments of 10m. I will not use negative numbers for they are realistically impossible. Mathematically negative lengths are possible but I know that investigating this won't give me the answers I want. I will put my results in a table now. Width Length Difference between W and L Perimeter =SUM(A:A+B:B)*2 Area=SUM(A:A*B:B) 0 500 500 1000 0 10 490 480 1000 4900 20 480 460 1000 9600 30 470 440 1000 14100 40 460 420 1000 18400 50 450 400 1000 22500 60 440 380 1000 26400 70 430 360 1000 30100 80 420 340 1000 33600 90 410 320 1000 36900 100 400
 Word count: 2015