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

Perfect Shapes

Extracts from this document...

Introduction

Perfect Shapes A perfect shape is said to be one that has the same area and perimeter. (1) Find the area and perimeter of the rectangle below. (2) Is the rectangle perfect? (3) Investigate perfect rectangles (4) Extend your investigation into other shapes. Perfect Shapes A perfect shape is defined to be one in which its perimeter is equal to its area. Rectangles I will begin by assuming that two of the sides of the rectangle are of length 5: A 5 by 1 rectangle is therefore not perfect and we need to investigate others: A 5 by 2 rectangle is not perfect either. I will tabulate further results in my quest to find a perfect 5 by something rectangle: Rectangle Length Width Area Perimeter Perimeter - Area 5 1 5 12 7 5 2 10 14 4 5 3 15 16 1 5 4 20 18 -2 5 5 25 20 -5 5 6 30 22 -8 5 7 35 24 -11 5 8 40 26 -14 5 9 45 28 -17 5 10 50 30 -20 Table 1 Looking at Table 1 it can be seen that no perfect shape has been found. ...read more.

Middle

7 28 22 -6 4 8 32 24 -8 4 9 36 26 -10 4 10 40 28 -12 This is interesting in that it shows that a 4 by something perfect rectangle is a square. Let us consider a general square to see if any other perfect square exists: A graph to illustrate this relationship might be useful: Graph 1 We can see from Graph 1 that in two places: and . The zero solution makes no sense but this confirms that if a square is 4 by 4 then it is perfect. As there is only one non-zero crossing point it also illustrates that there is only one perfect square. Let us return to the algebra briefly: Which confirms the result from the graph and proves that a 4 by 4 square is the only one that exits. It might be interesting to see if this pattern is true for other regular polygons, is there only one perfect equilateral triangle? Is there only one perfect equilateral pentagon? These questions I will attempt to address later. So far in this investigation the length of a rectangle has been fixed, it will be useful to relax this condition and consider the more general case of a perfect rectangle: Suppose that the length and ...read more.

Conclusion

#NUM! 25 #NUM! B C D E F G 2 y x h A P 3 Length of the equal sides Length of the base Perpendicular height Area of Triangle Perimeter of triangle Perimeter -Area 4 7 1 6.98 3.49 15 11.51 5 7 2 6.93 6.93 16 9.07 6 7 3 6.84 10.26 17 6.74 7 7 4 6.71 13.42 18 4.58 8 7 5 6.54 16.35 19 2.65 9 7 6 6.32 18.97 20 1.03 10 7 7 6.06 21.22 21 -0.22 11 7 8 5.74 22.98 22 -0.98 12 7 9 5.36 24.13 23 -1.13 13 7 10 4.90 24.49 24 -0.49 14 7 11 4.33 23.82 25 1.18 15 7 12 3.61 21.63 26 4.37 16 7 13 2.60 16.89 27 -10.11 Note the lack of sign change in the Area - Perimeter column suggesting that perfect 4, 4, x and 6, 6, x isosceles triangles do not exist. There are two sign changes in the 7, 7, x table and so it seems that two perfect isosceles triangles exist with these dimensions. Let me investigate this further: To be completed... let me know if you would like a copy of the completed version.......... Page 1 of 1 ...read more.

The above preview is unformatted text

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

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Investigating different shapes of gutters.

    9 You can see that for the multiples of three the angle, which gives the best area, is 30? and the best value for x is 1/3 of the Length (L). I notice that the maximum value on the two-way table is surrounded by symmetry as you can see more clearly on the following sheets.

  2. Investigate different shapes of guttering for newly built houses.

    The formula to find the circumference is: Circumference = ??diameter 60 = ??d 60?? = d 19.098593 = d Diameter ? 2 = radius 19.098593? 2 = 9.5492965 So the area of the semi-circle is: A =1/2??r� =0.5???9.5492965� =143.2394462cm� =143.2cm� Conclusion As nothing can be varied for the semi-circle the

  1. Geography Investigation: Residential Areas

    Cyprus Road should have a higher average area rating than anywhere I have surveyed in Basingstoke This hypothesis was not intended to be a long one and does not involve masses of writing and describing results etc. I wanted to put in this hypothesis as a test to the data on how the residents rated their own areas.

  2. Investigating different shapes to see which gives the biggest perimeter

    I will do this by picking length of 249.9m and height 250.1m. I will make another rectangle with length and height the other way round from the previous example, length of 250.1m and height of 249.9m. Verifying rectangle number 1: Area = 249.9 x 250.1 = 62499.99m� Verifying rectangle number

  1. GCSE Maths Coursework Growing Shapes

    No. of lines No. of lines - 9n2 1 1 3 -6 3 2 21 -15 5 3 57 -24 7 4 111 -33 9 5 183 -42 D1 As there are all 9's in the D1 column, the formula contains9n2 - 9n.

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

    The following gives reasons on why I asked each question and what I hoped it would tell me. Question 1- "Are you local to the area of Plymouth?" From this I could distinguish how popular the area was with both local people and also how many tourists were in the area.

  1. Impacts of tourism Positive and Negative effects in Castleton

    There is little vegetation because it was mainly a road and path. There was not much visible wildlife in this area but some birds were seen by the river. there were no pedestrians in the area. Point 2 Point 2 was situated near a traffic island in the centre of Castleton opposite the main car park entrance.

  2. Testing the Performance of an Ion Chromatography Column

    Buffer Day Two: Peak Number Area % RT (min) Area 1 6.325 3.34 42,716 2 22.765 3.8 153,743 3 3.339 4.19 22,548 4 34.748 5.18 234,669 5 14.112 5.66 95,304 6 18.71 8.67 126,359 Quinn 4 Soln. #1: Peak Number Area % RT (min)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work