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  • Level: GCSE
  • Subject: Maths
  • Word count: 2911

Investigating different shapes to see which gives the biggest perimeter

Extracts from this document...

Introduction

M Suhaib Vaiyani

Fencing Coursework

Introduction:

I am going to investigate different types and sizes of shapes with a perimeter of 1000m in each one. This is because a farmer needs the biggest possible area available using 1000m of fence. I am going to find out which shape with the same perimeter, has the biggest area. I will start my investigation using rectangles because it is the easiest shape to work out to start of with.

Rectangles:

Rectangle 1:

image05.png

image28.pngimage33.pngimage22.png

image40.pngimage41.png

I will calculate the area of the rectangle using the formula:

Area = Length x Height

So for the above rectangle, the area is:

Area = 450 x 50 =22500m²

Therefore, the area of the above rectangle is 22500m². I will now carry on the same procedure for the next 3 rectangles which I going to investigate. I will also increase the height by 100m and decrease the Length by 100m each time.

Rectangle 2:

image00.png

image01.png

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Area = 350 x 150 = 52500m²

So the area of the above rectangle is 52500m². This is already more than Rectangle 1 and therefore it proves that yes the area of a rectangle or any other shape changes as the length and Length changes.

Rectangle 3:

image03.png

image03.png

Area = 250 x 250 = 62500m²

Above is a square with both height and length of 250m and perimeter of 1000m. The area of this square is more than the area of the other rectangles investigated above.

Rectangle 4:

image04.png

image02.png

Area = 150 x 350 = 52500m²

This is the last rectangle which I have investigated. The area of this rectangle is the same as the area of Rectangle 2. This is because a x b is the same as

...read more.

Middle

Above is my first isosceles triangle which I am going to investigate. The length of the side is 425m and base is 150m. But to work out the area of the triangle, I need to know the height. I am going to do this by cutting the triangle in two halves to get a right-angled triangle.

image18.png

image16.png

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Note, the base is 75m because the isosceles has been halved so the base of the isosceles is 150m while the base of the right-angled triangle is 75m. I can now use Pythagoras’s theorem to work out the height of the triangle (a).

a² = c² – b²

= 4252 – 752

= 180625 – 5625

= 175000

a = √175000

= 418.3m

So the height of the isosceles triangle is 418.3m because the height of the isosceles is the same as the height of the right-angled triangle. Now using the formula,     Area = ½ base x height, I can now work out the area of the isosceles triangle.

Area = ½ base x height

= (½ x 150) x 418.3

= 75 x 418.3

= 31374.8m²

Triangle 2:

image15.png

image19.png

image20.png

Cut it in half to get a right-angled triangle so I can apply the Pythagoras’s theorem:

image21.pngimage19.png

        a² = 375² - 125²

        = 140625 – 15625

        = 125000

        a = √125000

        = 353.6m

So the height of the isosceles is 353.6m. I can now work out its area:

        Area = ½ base x height

        = (½ x 250) x 353.6

        = 125 x 353.6

        = 44194.2m²

Triangle 3:

image15.png

image23.png

image24.png

Cut it in half to get a right-angled triangle so I can apply the Pythagoras’s theorem:

image18.png

image23.pngimage25.png

image26.png

        a² = 325² - 175²

        = 105625 – 30625

        = 75000

        a = √75000

        = 273.9m

So the height of the isosceles is 273.9m. I can now work out its area:

        Area = ½ base x height

        = (½ x 350) x 273.9

        = 175 x 273.9

        = 47925.7m²

Spreadsheet:

...read more.

Conclusion

image48.png

Each side equals 125m (1000/8) and each base of right-angled triangle equals 62.5m (½ x 125) and each angle within the right angled triangle is equal to 22.5° (360/16). So I can now work out the height of the right-angled triangle:

        Adjacent = Opposite/Tan 22.5

        Height = 62.5/0.41

        = 150.89m

So area equals:

        Area = ½ base x height

        = (1/2 x 125) x 150.89

        = 62.5 x 150.89

        = 9430.52.m²

Area of Octagon = 9430.52 x 8 = 75444.174m²

Decagon:

The final polygon I am going to investigate is the Decagon, a 10 sided shape. The sides’ lengths are going to be equal yet again and I believe this final polygon should give me the biggest area out of all the shapes I have done so far. The perimeter will also be the same, 1000m.

image49.png

The above regular decagon has 10 sides. Each side is 100m long. The base of each right-angled triangle found when the isosceles is cut is 50m long. The angle within the right-angled triangle is equal to 18°. Using trigonometry, I can now work out the height of the triangles.

        Adjacent = Opposite/Tan 18

        Height = 50/0.32

                                        = 153.88

So area equals:

        Area = ½ base x height

        = (1/2 x 100) x 153.88

        = 50 x 153.88

        = 7694.21m²

Area of Decagon = 7694.21 x 8 = 76942.1m²

The area of the Decagon is yet again the largest of all shapes investigated so far. I have now proven that the more the number of sides in a regular shaper, the bigger its area. I am now going to find out how big the area can be in any regular shape. To make it easier and quicker, I am going to put this in spreadsheet software and produce a chart to present the data better.

        - Page  -

...read more.

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