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

Four Sided Shapes Investigations

Extracts from this document...

Introduction

FOUR SIDEDED SHAPES RECTANGLES INTROUCTION: I am going to start my investigation with a simple shape, the rectangle. 400 m AREA = BASE x WIDTH = 400m x 100m 100 m 100 m = 40000m2 400 m AREA = BASE x WIDTH 300 m = 200m x 300m = 60000m2 200 m 200 m ...read more.

Middle

= BASE x WIDTH 375m = 375m x 125m = 46875m2 125m 125m 375m AREA = BASE x WIDTH 275m = 275m x 225m = 61875m2 225m 225m 275m AREA = BASE x WIDTH 251m = 251m x 249m = 62499m2 249 m 249m 251m AREA = BASE x WIDTH 475m = 475m x 25m = 11875m2 25m 25m 475m I made the chart below using a spreadsheet using a formula: (500 - Y) ...read more.

Conclusion

x Y BASE (m) WIDTH (m) AREA (m2) 1 499 499 25 475 11875 50 450 22500 75 425 31875 100 400 40000 125 375 46875 150 350 52500 175 325 56875 200 300 60000 225 275 61875 250 250 62500 275 225 61875 300 200 60000 325 175 56875 350 150 52500 375 125 46875 400 100 40000 425 75 31875 450 50 22500 475 25 11875 499 1 499 ...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. Geography Investigation: Residential Areas

    I have targeted the streets that have in my opinion serious issues and in some cases, like that of Bounty Road, I have pin pointed a good value to the area to explain why I have found results like I have.

  2. Maths Coursework: Equable Shapes

    -a (3a - 8 ) = 0 - Now I work out what "a =" -a = 0 ( a = 0 ) 3a - 8 = 0 ( a = 2 2/3 ) Therefore a = 2 2/3 Now after doing this same method for several more rectangles the following pattern emerged...

  1. Perfect Shapes

    It seems reasonable to assume therefore that a perfect 5 by (something between 3 and 4) rectangle exists. Notice also that as we progress down the Perimeter - Area column the difference between the area and the perimeter is getting larger, this seems to suggest that another perfect shape will not be found.

  2. An Investigation into the Varying Isoperimetric Quotients of Differing Shapes.

    I have labeled the length of the sides . To find the area of the square I am going to use trigonometry. To use trigonometry I need a right angled triangle with another angle. At the moment I don't have this.

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