• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month   # Fencing problem

Extracts from this document...

Introduction

Page  of

Maths Coursework

## Fencing problem

I have been asked to investigate the maximum area enclosed by 1000 metres of fencing. To do this I need to use various shapes, to start with a square.   Secondly I shall look at rectangles.    Rectangle Length (M) Height (M) Area (M) A 450 50 22500 B 400 100 40000 C 350 150 52500 D 300 200 60000 E 250 250 62500 F 200 300 60000

The biggest quadrilateral in the table was E which is a square, meaning the square has a  larger area than rectangles. 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

Middle Triangle Equal sides  (m) Base Total area (m  ) A 300 400 44721.35955 B 350 300 47434.1649 C 400 200 38729.83346 D 450 100 4330.127019

The biggest area I found was with triangle B although as I increased by 50m each time I think that I could investigate this triangle in more depth with smaller intervals to find a larger area.

I continued with the same formula as before but just applied it to another group of triangles. These are my results.

 Triangle Equal sides (m) Base (m) Total area (m  ) A 305 390 45731.553658 B 310 380 46540.3051128808 C 315 370 47165.9305007334 D 320 360 47623.5235991628 E 325 350 47924.7237817015 F 330 340 48083.2611206854 G 335 330 48105.353132 H 340 320 48000 I 345 310 4774.208523

Although I could go further into depth with Triangle G I am

Conclusion                  Now that I know the perpendicular height I can use the same formula as before to get the area.

Base X height ÷ 2

125 X 150.8883476 ÷ 2 = 9430.521725

Area = 9430.521725m

This gives me the area of one of eight equal triangles in an octagon so now I must multiply this number by eight.

9430.521725 X 8 = 75444.1738

As expected the area has again increased in size. I think that for regular shapes the more sides the shape has the bigger the area is. A circle has infinite sides in theory so I think this will have the biggest area. This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Fencing Problem essays

1. ## The Fencing Problem

the shapes I have looked at so far all resemble a regular polygon. Consequently, this prompts me to explore regular polygons, as they evidently hold the highest areas. Below are some diagrams which elucidate why parallelograms are not ideal for achieving the maximum potential area.

2. ## Fencing problem.

above data in the graph below: The graph above obviously indicates that decagon has the most coverage of area, this in other words means that it covers the most area therefore the 1000m fencing shall be used to most out of all the polygons.

1. ## Fencing Problem

* Then I add the method to find the area of the entire polygon to the formula to figure out the area of the Isosceles triangle. * This formula can be cancelled down to this much simpler and easier version to use.

2. ## Maths Fencing Coursework

=44721.35955 Side a: 400 Side b: 400 Side c: 200 V(500*(500-400)* (500-400)* (500- 200)) =38729.83346 Side a: 450 Side b: 450 Side c: 100 V(500*(500-450)* (500-450)* (500- 100)) =22360.67977 Side a: 250.5 Side b: 250.5 Side c: 499 V(500*(500-250.5)* (500-250.5)* (500-499))

1. ## Regeneration has had a positive impact on the Sutton Harbour area - its environment, ...

Our interview took place with Ruby, who has been campaigning against some aspects of the regeneration. It meant that we could ask various questions and get a comprehensive answer to which we could ask other things, these are things which the questionnaire didn't allow us to do.

2. ## The Fencing Problem

Now that I have calculated the height of the triangle I can now find the area of it. The formula for the area of a triangle is: (h x b) � 2 (h x b) � 2 (223.607 x 400)

1. ## Fencing Problem - Math's Coursework.

Reproduction or retransmission in whole or in part expressly prohibited 435 430.116 27957.540 140 430 424.264 29698.480 150 425 418.330 31374.750 160 420 412.311 32984.880 170 415 406.202 wwdf dfw stdfdfud edf dfnt cdf endftral dfcodf uk. 34527.170 180 410 400.000 36000.000 190 405 393.700 37401.500 200 400 387.298 38729.800

2. ## The Fencing Problem.

all that needs to be done is the following: > With the rectangle, you simply multiply 'side A' with 'side B' > With the square, all you have to do is square (2) 'side A' So now that I knew how to calculate the areas of the square and the • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 