The Fencing Problem
The problem:
A farmer has exactly 1000 metres of fencing, with it she wishes to fence off a plot of level land. She is not concerned about the shape of the plot, but it must have a perimeter of 1000 metres. What she does wish to do is fence of f the plot of land, which contains the maximum area.
The investigation:
To solve this problem I will investigate two things, different shapes and the same shape but with different figures for its sides.
I have decided to begin my investigation by looking at a four sided shapes, rectangles.
Rectangles:
A rectangle is a shape with four sides and four right angles. It has two parallel vertical sides and two parallel horizontal sides. I will construct four different rectangles. These shapes will all have a perimeter of 1000 metres.
Square:
As I have only 1000m of fence to use for its perimeter each side will need to be 250m in length. This is because a square has four sides equal in length.
Perimeter = 1000m
1000m ( 4 = 250m
The area of a square is found using the equation Base ( Height.
Base = 250m
Height = 250m
250 ( 250 = 62,500m²
Rectangle 1:
The height and base of a rectangle is of different height but each set of parallel lines is the same length. For rectangles there are many combinations I could make but I will start by looking at a rectangle with 300m and 200m sides.
Perimeter = 1000m
2 ( 300 = 600m
2 ( 200 = 400m
600m + 400m = 1000m
I will now find the area of the rectangle using these figures. To find the area of a rectangle the same formula is used as with the square, Base ( Height.
Base = 200m
Height = 300m
200 ( 300 = 60,000m²
Rectangle 2:
This rectangle will have a base of 100m and a height of. 400m
Perimeter = 1000m
2 ( 400 = 800m
2 ( 100 = 200m
800m + 200m = 1000m
I will now find the area of the rectangle using the same equation as before, Base ( Height.
Base = 100m
Height = 400m
100 ( 400 = 40,000m²
Rectangle 3:
This rectangle will have a base of 50m and a height of 450m.
Perimeter = 1000m
2 ( 450 = 900m
2 ( 50 = 100m
900m + 100m = 1000m
I will now find the area of the rectangle.
Base = 50m
Height = 450m
50 ( 450 = 22,500m²
Shape
Height (metres)
Base (metres)
Area (Metres squared)
Square
250m
250m
62,500m²
Rectangle 1
300m
200m
60,000m²
Rectangle 2
400m
00m
40,000m²
Rectangle 3
450m
50m
22,500m²
The above table displays the area of four different quadrilaterals and their dimensions. It shows that the wider the gap between the height and base in length then the smaller its area is. The shape with the largest area is the square its sides are all the same length, i.e. its base = its height. This type of shape is called a regular polygon. A polygon is a many sided shape and a regular polygon is a many sided shape with all angles and sides the same. The square is a regular polygon because its four sides are the same lengths and they all meet at 90(.
I have now decided to see whether it makes any difference to the area if you switch the height with the base. I will try this out on the first and second rectangles. I will not try it on the square, as all of its sides are the same any way.
First rectangle:
The dimensions of the first rectangle are:
Base = 200m
Height = 300m
Its area is:
200 ( 300 = 60,000m²
The dimensions of the new rectangle are the same but switched around:
Base = 300m
Height = 200m
The area of the new rectangle is
200 ( 300 = 60,000m²
Second rectangle:
This rectangle has dimensions of:
Base = 100m
Height = 400m.
Its area is:
100 ( 400 = 40,000m²
The dimensions of the new rectangle are the same but switched around:
Base = 400m
Height = 100m
The area of the new rectangle is
400 ( 100 = 40,000m²
This information shows that it makes no difference which way the figures are arranged on a rectangle or square. For example the base could be 100m and the height 400m or the base could be 400m and the height 100m, either way the area would always be the same. Evidence of this is shown above using the first and second rectangles as examples.
I have decided to work out a formula for quadrilaterals using the perimeter and the base length.
The area of a quadrilateral is found using the equation Base ( Height
Area = Base ( Height
As I am trying to only use the base length and perimeter I will need to find the height length using the base and perimeter length. The equation for this is:
Perimeter = 1000m
000 = 2H + 2B All sides must add up to 1000m
500 = H + B Equation for area uses one horizontal and one vertical side height and base.
500 - B = H The height equals the perimeter minus the base.
To find area the formula is Base ( Height, the formula I will use is:
Area = Base ( 500 - Base
This formula is basically the same except that height is replaced with a formula that equals height, 500 - Base.
The formula for finding the area can be simplified
Area = Base ( 500 - Base
Area = Base (500 - Base)
Area = 500Base - Base²
Area = 500B - B²
This can also be used for height:
Area = Height ( 500 - Height
Area = Height (500 - Height)
Area = 500 Height - Height²
Area = 500H - H²
These formulas can work out the area of any quadrilateral with a perimeter of 1000m using either the base or the height.
I will test my formulae on three shapes to make sure that they are correct.
Test one:
The first test will be on a square, the square sides are all 250m in length and it has an area of 62,500m².
Area using base formula:
Area = 500B - B²
Area = 500 ( 250 - 250²
= 62,500m²
Area using height formula:
Area = 500H - H²
Area = 500 ( 250 - 250²
= 62,500m²
Test two:
This next test will be on a rectangle, the rectangle base is 200m and its height 300m in length and it has an area of 60,000².
Area using base formula:
Area = 500B - B²
Area = 500 ( 200 - 200²
= 60,000m²
Area using height formula:
Area = 500H - H²
Area = 500 ( 300 - 300²
= 60,000m²
Test three:
The last test is on a rectangle the base is 100m and 400m in length and it has an area of 40,000 m².
Area using base formula:
Area = 500B - B²
Area = 500 ( 100 - 100²
= 40,000m²
Area using height formula:
Area = 500H - H²
Area = 500 ( 400 - 400²
= 40,000m²
These results prove that my formulae are correct and that you can work out the area of a rectangle using just perimeter and either height ...
This is a preview of the whole essay
Test three:
The last test is on a rectangle the base is 100m and 400m in length and it has an area of 40,000 m².
Area using base formula:
Area = 500B - B²
Area = 500 ( 100 - 100²
= 40,000m²
Area using height formula:
Area = 500H - H²
Area = 500 ( 400 - 400²
= 40,000m²
These results prove that my formulae are correct and that you can work out the area of a rectangle using just perimeter and either height or base.
Scale Drawings:
I will now also try using scale drawings in my investigation to see if they help. I will do scale drawings of two further rectangles. The scale that I will use for my diagrams will be 1cm = 30m or 1cm = 3,000cm
Rectangle 4:
The first rectangle that I will look at will have a height twice the length of the base.
To work this out I will write down the ratio that they are to each other, two to one, 2:1. This means that the set of horizontal sides is twice as long as the vertical sides. Now I will find the number with which to multiply these numbers by to find the length of the sides. To do this I will divide the perimeter by six.
Perimeter = 1000m ( 6
= 166.666m
Now I times this number by the number of sides to find its length in metres.
Height = 166.666 ( 2
= 333.333m
Base = 166.666 ( 1
= 166.666m
Height = 333.333m
Base = 166.666m
To find the length of the sides I am going to draw according to my scale I will divide the lengths by 30, as this is my scale.
Height = 333.333 ( 30
= 11.111cm
Base = 166.666 ( 30
= 5.555cm
This means that on my diagram the height will be 11.111cm long but represent 333.333m. The base will be 5.555cm but represent 166.666m.
The area of the rectangle is 55,555.555m²
333.333 ( 166.666 = 55,555.555m²
Rectangle 5:
The second rectangle that I will look at will have a base three times the length of the height, 3:1. I will work this out the same way as before.
Perimeter = 1000m ( 8
= 125m
Now I times this number by the number of sides to find its length in metres.
Base = 125 ( 3
= 375m
Height = 125 ( 1
= 125m
Base = 375m
Height = 125m
This means that to find the actual length I am going to draw for my scale I will need to divide these lengths by 30.
Base = 375 ( 30
= 12.5cm
Height = 125 ( 30
=4.166cm
This means that on my diagram the base will be 12.5 cm long but represent 375m. The height will be 4.166cm but represent 125m.
The area of the rectangle is 46,875m²
125 ( 375 = 46,875m²
After carrying out the investigation into scale I have scrapped and abandoned the idea as it is too difficult for me to do. This is because I do not have the resources or knowledge to continue to do this or even do it accurately. However I will add this data to my table showing area of quadrilaterals.
This data in the table below continues to prove that the closer the height length is to the base length then the larger the area is.
Shape
Height (metres)
Base (metres)
Area (Metres squared)
Square
250m
250m
62,500m²
Rectangle 1
300m
200m
60,000m²
Rectangle 2
400m
00m
40,000m²
Rectangle 3
450m
50m
22,500m²
Rectangle 4
66.666 m
333.333 m
55,555.555m²
Rectangle 5
25m
375m
46,875m²
I can also show this data as a graph, on the graph I will plot area against base.
This graph shows that the highest area is achieved for a rectangle when the base is 250m long.
As I have found that if a regular quadrilateral has a larger area than an irregular quadrilateral I will test this theory on other shapes. The shapes that I will look at are triangles and hexagons. In investigating these two shapes I hope to prove my theory that a shape achieves its largest area when it is regular. I will begin by looking first at the triangles and then looking at the hexagon I will then compare these two new lots of data with that of the quadrilaterals.
Triangles:
The triangle will be slightly harder to investigate than the quadrilateral. This is because to find the area of the triangle it is necessary to do working out in order to complete the equation to find the area of a triangle. This is because the equation for triangle area is height multiplied by half base. However this is not always possible by just using information from the perimeter. This is because the height of a triangle is not always part of the perimeter.
Triangle 1:
I will now look at an equilateral triangle. An equilateral triangle is a triangle with all sides and angles the same.
This means that an equilateral triangle perimeter will be 1000m divided by 3.
Perimeter = 1000m
1000m ( 3 = 333.333´m
This means that each side will be 333.333´m long.
I will now find the triangles area,
half base multiplied by height.
The base is 333.333´m long
To find the height I need to cut the triangle through the centre from top to bottom. This halves the length of the bottom and leaves me with a right-angled triangle. I can now work out the height of the triangle using Pythagorean theorem. This states that the square of the length of the hypotenuse of a right-angled triangle equals the sum of the squares of the lengths of the other two sides. By rearranging this equation you can work out the height of the triangle.
Hypotenuse ² - Base ² = Height ²
VHeight ² = Height
In this case the equation is:
333.333´² - 165.5² = 83333.333´²
V83333.333´² = 288.675metres
I can now enter the height of the equilateral triangle, as it is the height of the right-angled triangle that I have just worked out.
Base = 333.333´metres
Height = 288.675 metres
Area = .5 ( 333.333´ ( 288.675
= 48,112.5m²
Triangle 2:
The next triangle that I will look at is a right-angled triangle; a right-angled triangle is a triangle where one angle is 90(. The 1000m perimeter will needed to be shared between three sides. To work this out I am going to need to know the size of the ratio of one side to another. For example 7:3:1.
The ratio that I will use is 3:2:1. From this I can work out their individual lengths using the total length of 1000m.
Perimeter = 1000m
3 + 2 + 1 = 6
1000m ( 6 = 166 + 2/3
I can now multiply this number by the ratio to find their length as part of the triangle with a perimeter of 1000m. This process can be used on any sized shape.
166 2/3 ( 3 = 500m
166 2/3 ( 2 = 333 1/3m
166 2/3 ( 1 = 166 2/3m
The lengths of the sides are shown below
Side A = 166 2/3m
Side B = 333 1/3m
Side C = 500m
I will now find the area of the triangle. Because this is a right-angled triangle I will not need to find the height, as side A is the height of the triangle.
.5 ( 333.333 ( 166.666 = 27,777.777m²
Triangle 3:
This triangle will be a scalene triangle, all sides of different length but all angles less then 90(. The ratio of these sides will be 8:6:4.
Perimeter = 1000m
8 + 6 + 4 = 18
1000m ( 18 = 55 5/9
I can now multiply this number by the ratio to find their length as part of the triangle with a perimeter of 1000m. This process can be used on any sized shape.
55 5/9 ( 8 = 444 4/9m
55 5/9 ( 6 = 333 1/3m
55 5/9 ( 4 = 222 2/9m
The lengths of the sides are shown below
Side A = 444.444m
Side B = 333.333m
Side C = 222.222m
To find the area I must first find the height of the triangle. To do this I will need to split the triangle into two right-angled triangles. I will use triangle A to find the height, but I do not have enough information on the right-angled triangle to do so yet. This means that I am going to have to use trigonometry to work out angle C. Looking at the original triangle I will work out angle C. To do this I will use the lengths of the hypotenuse and opposite side. Because I am using hypotenuse and opposite I will have to use sin to work out the angle.
Sin = Opp ( Hyp
Sin = 333.333 ( 444.444
= 0.75m
Sin0.75 = 48.59(
This angle means that I now have enough data to find the height of the right-angled triangle A.
Sin = Opp ( Hyp
If I rearrange this formula I can work out the length of the opposite side.
Opp = Sin ( Hyp
Opp = Sin 48.59 ( 444.444
= 166.665m
This means that I now have the necessary data to find the area of the original triangle.
Base = 444.444m
Height = 166.665m
I will now find the area of the triangle using the following formula, half base multiplied by height.
.5 ( 444.444 ( 166.665 = 37,036m²
I have compiled a table using the data that I have found for the triangles to show the relation between area, base and height.
Shape
Height (metres)
Base (metres)
Area (Metres squared)
Triangle 1
288.675 m
333.333m
48,112.5m²
Triangle 2
66.666 m
333.333m
27,777.777m²
Triangle 3
66.665 m
444.444 m
37,036m²
This table shows again that the closer the height length is to the base height the larger the area. It shows that wider the gap between the height and base in length then the smaller its area is. It shows the same trend as that of the quadrilateral table, that is that a regular-sided shape has the largest area.
Hexagons:
Hexagons are six sided shapes. I will look at three hexagons to find which has the highest area.
Hexagon 1:
This first hexagon will be a regular hexagon, have all sides and angles the same.
Perimeter = 1000m
1000 ( 6 = 166.666m
I will now find the area of the hexagon. To find the area of the hexagon I will split it into triangles. To find the area of these triangles the equation is half Base ( Height. As the hexagon is regular all of the triangles will be equilateral and will be identical this means that I only need to find one which can be multiplied to find the rest of the triangles. It also means that all of the triangle sides equal the same as the base, 166.66m. However to find the height of the triangles it is necessary to split them in half to find their height.
To find the height of the triangle I must now use Pythagorean theorem to find its height.
This states that the square of the length of the hypotenuse of a right-angled triangle equals the sum of the squares of the lengths of the other two sides. By rearranging this equation you can work out the height of the triangle.
Hypotenuse ² - Base ² = Height ²
VHeight ² = Height
In this case the equation is:
166.666´² - 83.333´² = 20,833.33´²
V20,833.33² = 144.333m
I can now enter the height of the equilateral triangle, as it is the height of the right-angled triangle that I have just worked out.
Base = 166.666m
Height = 144.333m
Area = .5 ( 166.666 ( 144.333
= 12,027.777m²
This is the area of one of the six triangles, to find the area of the hexagon I need to multiply this area by six.
12,027.777 ( 6 = 72,166.662m²
Hexagon 2:
This hexagon will have sides of 133.333m, 133.333m, 133.333m, 200m, 200m and 200m.
Perimeter = 1000m
3 ( 133.333 = 400m
3 ( 200 = 600m
600m + 400m = 1000m
I will now find the area of the hexagon. To find the area of the hexagon I will split it into one square and two triangles.
The section that I will look at first is the square. The area of a square is found using the equation, Area = Base ( Height
Base = 133.333m
Height = 133.333m
133.333m ( 133.333m = 17,777.777m²
The area of triangles is found using the equation, half Base ( Height.
The first triangle that I will look at is triangle A.
Triangle A has a base of 133.333 metres but its height must be found. I will find its height using Pythagorean theorem.
Hypotenuse ² - Base ² = Height ²
VHeight ² = Height
Hypotenuse = 200m
Base = 66.666m
I can now find the height of triangle A.
200 ² - 66.666² = 35,599.888m
V35,599.888 = 188.679m
Using the height I can now find the area of triangle A.
.5 ( 66.666 ( 188.679 = 6,257.864m²
Now I will find the area for triangle B. But again I must first find the height of the triangle.
Hypotenuse = 200m
Base = 188.679m
200 ² -188.679 = 4,400.234m
V4,400.234 = 66.334m
I can now find the area
.5 ( 188.679 ( 66.334 = 6,257.916m²
I can now add up all these areas to find the total area of the hexagon.
6,257.916 + 6,257.864 + 17,777.777
= 30,293.609m²
Hexagon 3:
This hexagon will have sides of 250m, 200m, 100m, 150m, 100m and 200m.
200 ( 2 = 400m
00 ( 2 = 200m
50 + 250 = 400
400 + 200 + 400 = 1000m
Perimeter = 1000m
To find the area I will split the hexagon into a rectangle and a trapezium. The first section that I will look at is the rectangle.
The equation for rectangle area is:
Area = Base ( Height
I can now use this equation to find the area
Base = 250m
Height = 200m
250 ( 200 = 50,000m²
I will now find the area of the trapezium.
The equation for trapezium area is:
Area = ((Base + Top) ( 2) ( Height
As I do not yet know the trapeziums height I will have to work it out.
I will use Pythagorean theorem to find it. I will cut a triangle of the end of the trapezium. Its base equals the base of the trapezium minus its top divided by two
Base = (Base - Top) ( 2
Base = (250 - 150) ( 2
= 50m
The hypotenuse of the triangle equals 100m
Hypotenuse = 100m
Base = 50m
To find the height the equation is:
Hypotenuse ² - Base ² = Height ²
VHeight ² = Height
I will now use this equation to find the height.
100² -50² = 7500
V7500 = 86.602m
Now that I have found the height I can work out the trapeziums area.
Area = ((250 + 150)( 2) ( 86.602
= 4,330.127m²
The total area of the hexagon is the trapezium area plus the rectangle area.
4,330.127 + 50,000 = 54,330.127m²
I will now input my information into a table showing the area of a hexagon and its relation to the side lengths.
Shape
Side 1
Side 2
Side 3
Side 4
Side 5
Side 6
Area
Hexagon 1:
66.666m
66.666m
66.666m
66.666m
66.666m
66.666m
72,166.662m²
Hexagon 2:
200m
200m
33.333m
33.333m
33.333m
200m
54,330.127m²
Hexagon 3:
250m
200m
00m
50m
00m
200m
54,330.127m²
This table also shows that the regular shape provides the biggest area. This shape is a regular polygon. I will now begin investigating regular polygon to find which has the largest.
Regular Polygons:
A regular polygon is a many sided shape with all angles and sides identical.
I hope to find the regular-sided polygon with the largest area I will look at regular polygons ranging from three sides to ten sides and a fifteen-sided shape. Although I have already found the area of the three, four and six sided polygons I will still include these in my investigation. I will only show workings and necessary information, as I will be repeating the processes that I have used previously to find lengths of sides and area. I will begin with the triangle.
As I have found earlier with the hexagon that to find the area of polygons they must be split into triangles I will repeat this for these polygons, the difference being that these are regular polygons which means that the triangles inside the polygons are all the same. This means that the area of just one triangle need be found as the area can be multiplied by the number of triangles in the polygon to find it's the total area. I will find the area by splitting the polygon in triangles. However I do not know the necessary information to find height from just this. I will need to split the triangles in half into equilateral triangles to find their height. From this I can now find the height as I know the base length and the angle of where all the triangles meet.
To find the length of the base of the right angle triangle I need to divide the perimeter by the number of sides, this gives me the length of the triangles. From here I half the number to find the base of the equilateral triangle as it is half of one of the triangles. I can simplify this to half the perimeter divided by the number of sides.
Where n = the number of sides
Base = (1000 ( n) ( 2
Base = 500 ( n
The angle is found in much the same way. The total angle of where all the triangles meet in the polygon is 360 degrees. The angle of the right-angled triangle is half that of one of the original triangles. If I divide 360 degrees by the number of sides it gives me the angle of one original triangle. I now half this angle to find the angle of the right angled trangle. I can again simplify this so that I do not need to divide by two. To do this I simply start with 180 degrees and divide by the number of sides.
This is because the angle inside the right angled triangle is half that of the original
Angle = (360 ( n) ( 2
Angle = 180 ( n
With this information I can use trigonometry to find the height. As I know the angle T the side that I know the length of is the opposite and the side that I am trying to find is the adjacent. This means that I can use the equation tangent = opposite over adjacent. If I rearrange this equation I can use it to find the length of the adjacent. This equation is opposite over tangent = adjacent.
By adding the workings to find the angle of tangent and to find the opposite I am left with.
Opp = 500 ( n
Tan = 180 ( n
Height = Opp ( Tan
Height = 500 ( n ( tan (180 ( n)
This means that I can find the height of any triangle that is part of a regular polygon split into equal triangles with a perimeter of 1000m. This can be used on any regular polygon with any number of sides.
To find the area of a triangle the equation is half base multiplied by height. From the information above I can find the height and base. This means that I can use this information to find the height of the triangle.
Area = .5 ( Base ( Height
The equation for the area of a triangle using my formulas above is:
Area = (.5 ( (1000 ( n)) ( (500 ( n) ( (tan 180 ( n)
This means that I can work out the area of a triangle that is a regular polygon with a perimeter of 1000m, or that is part of a regular polygon that is split into equal triangles with a perimeter of 1000m.
To find the total area of the polygon I need now only to multiply the area of the triangle by the number of sides, n.
The equation for the complete area of a regular polygon is:
Area = n (.5 ( (1000 ( n)) (((500 ( n) ( (tan (180 ( n))
This formula means that I can find the area of any regular polygon with a perimeter of 1000m.
I will now test this formula on the previous regular polygons whose area I found, the equilateral triangle, square and the regular hexagon and two new regular polygons using the same method, regular pentagon and the regular heptagon.
Equilateral triangle:
The area of an equilateral triangle with a perimeter of 1000m is
Area = .5 ( 333.333´ ( 288.675 = 48,112.5m²
I will now test my formula to see if I come up with the correct area.
Area = 3 (.5 ( (1000 ( 3)) ( ((500 ( 3) ( (tan (180 ( 3)) = 48,112.5m²
This is correct.
Square:
The area of an equilateral triangle with a perimeter of 1000m is
Area = 250 ( 250 = 62,500m²
I will now test my formula to see if I come up with the correct area.
Area = 4 (.5 ( (1000 ( 4)) ( ((500 ( 4) ( (tan (180 ( 4)) = 62,500m²
This is correct.
Regular Pentagon:
The area of a regular pentagon with a perimeter of 1000m is
Area = (.5 ( 200 ( 137.638) ( 5 = 68,819.096m²
I will now test my formula to see if I come up with the correct area.
Area = 5 (.5 ( (1000 ( 5)) (((500 ( 5) ( (tan (180 ( 5)) = 68,819.096m²
This is correct
Regular Hexagon:
The area of a regular hexagon with a perimeter of 1000m is
Area = (.5 ( 166.666 ( 144.333) ( 6 = 72,166.662m²
I will now test my formula to see if I come up with the correct area.
Area = 6 (.5 ( (1000 ( 6)) ( ((500 ( 6) ( (tan (180 ( 6)) = 72,166.662m²
This is correct.
Regular Heptagon:
The area of a regular heptagon with a perimeter of 1000m is
Area = (.5 ( 142.857 ( 148.322) ( 7 = 74,161.478m²
I will now test my formula to see if I come up with the correct area.
Area = 7 (.5 ( (1000 ( 7)) ( ((500 ( 7) ( (tan (180 ( 7)) = 74,161.478m²
This is correct.
From this testing I can see that my formula is correct, I will use it when I am trying to find the area of the regular polygons from three sides up to ten sides. I will proceed to investigate the regular polygon with the largest area.
Regular 3 Sided Polygon:
A regular-sided polygon is called an equilateral triangle.
Area = 3 (.5 ( (1000 ( 3)) ( ((500 ( 3) ( (tan (180 ( 3)) = 48,112.5m²
Diagram
Regular 4 Sided Polygon:
A regular four-sided polygon is called a square.
Area = 4 (.5 ( (1000 ( 4)) ( ((500 ( 4) ( (tan (180 ( 4)) = 62,500m²
Diagram
Regular 5 Sided Polygon:
Area = 5 (.5 ( (1000 ( 5)) ( ((500 ( 5) ( (tan (180 ( 5)) = 68,819.096m²
Diagram
Regular 6 Sided Polygon:
A regular six-sided polygon is called a regular hexagon
Area = 6 (.5 ( (1000 ( 6)) ( ((500 ( 6) ( (tan (180 ( 6)) = 72,166.662m²
Diagram
Regular 7 Sided Polygon:
A regular seven-sided polygon is called a regular heptagon.
Area = 7 (.5 ( (1000 ( 7)) ( ((500 ( 7) ( (tan (180 (7)) = 74,161.478m²
Diagram
Regular 8 Sided Polygon:
An eight-sided regular polygon is called a regular octagon.
Area = 8 (.5 ( (1000 ( 8)) ( ((500 ( 8) ( (tan (180 ( 8)) = 75,444.173m²
Diagram
Regular 9 Sided Polygon:
A nine-sided regular polygon is called a regular nonagon
Area = 9 (.5 ( (1000 ( 9)) ( ((500 ( 9) ( (tan (180 ( 9)) = 76,318.817m²
Diagram
Regular 10 Sided Polygon:
A ten-sided regular polygon is called a regular decagon.
Area = 10 (.5 ( (1000 ( 10)) ( ((500 ( 10) ( (tan (180 ( 10)) =76,942.088m²
Diagram
Regular 15 Sided Polygon
A fifteen-sided regular polygon is called a regular.
Area = 15 (.5 ( (1000 ( 15)) ( ((500 ( 15) ( (tan (180 ( 15)) = 78,410.501 m²
Diagram
I will now enter my results in the following table.
Shape
Number of Sides
Area (m²)
Equilateral Triangle
3
48,112.5
Square
4
62,500
Regular Pentagon
5
68,819.096
Regular Hexagon
6
72,166.662
Regular Heptagon
7
74,161.478
Regular Octagon
8
75,444.173
Regular Nonagon
9
76,318.817
Regular Decagon
0
76,942.088
Regular 15 sided polygon
5
78,410.501
From these results I can see that the more sides that a shape has the larger area it has. The largest regular polygon in my table has an area of 78,410.501 m².
As my results show that the area of a regular polygon increases as the number of sides increases, I will investigate the maximum number of sides. I know that a circle has an infinite number of sides so I believe that it will have the largest area. I will now investigate the area of a circle.
Circle:
The area of a circle is found using the equation pi multiplied by the radius squared equals area.
Area = ?r²
However the only information that I have for the circle is its circumference which is the 1000m perimeter. To find the radius of the circle I will rearrange the formula circumference = 2?r.
Circumference = 2?r
Radius = C ( 2?
I will now find the radius.
1000 ( 2? = 159.154m
Radius = 159.154m
With the information that I have just found I can now find the area of the circle.
? ( 159.154²
Area = 79,577.471m²
This area is the largest area that I have found out of all the shapes that I have investigated. It proves my earlier statement that a circle has a larger area.
From my investigation I conclude that any regular polygon has a larger area than an irregular polygon with the same number of sides. I have found that the shape with the largest area is a circle. The table below illustrates this fact.
Shape
Number of Sides
Area (m²)
Equilateral Triangle
3
48,112.5
Square
4
62,500
Regular Pentagon
5
68,819.096
Regular Hexagon
6
72,166.662
Regular Heptagon
7
74,161.478
Regular Octagon
8
75,444.173
Regular Nonagon
9
76,318.817
Regular Decagon
0
76,942.088
Regular 15 sided polygon
5
78,410.501
Circle
n
79,577.471
The graph below shows that the largest area is found when the 1000m perimeter forms a circle. It also shows that the increase in area lessens each time the regular polygon has an increased number of sides. For example from 3 sides to 4 sides the area increases by 14,387.5 whereas between 9 and 10 sides the area increases by 623.271.
Craig Banfield - 1 of 33
