# Investigating different shapes to see which gives the biggest perimeter

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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:

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:

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:

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:

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

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.

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:

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

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:

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

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:

Conclusion

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.

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.

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