Length A = c2 – b2 = a2
= 3002 – 2002
= 500002
= √50000
= 223.60m
Therefore, the height of the triangle is 223.607m.
Area = base x height / 2
= 400 x 223.607 / 2
= 89442.8 / 2
= 44720m2
OR by using Heron’s Formula:
Area =
= 500 (500 – 300)(500 – 400)(500 – 300)
= 2000000000
= √2000000000
= 44720m2
The area of this isosceles triangle is 44720m2.
I will now work out the area of the next isosceles triangle:
Base of one triangle = 380 / 2 = 190m
Length A = c2 – b2 = a2
= 3102 – 1902
= 600002
= √60000
= 244.95m
Therefore, the height of the triangle is 244.95m.
Area = base x height / 2
= 380 x 244.95 / 2
= 93081 / 2
= 46540.5m2
OR Heron’s Formula:
Area =
= 500 (500 – 310)(500 – 380)(500 – 310)
= 2166000000
= √2166000000
= 46540.5m2
The area of this isosceles triangle is 46540.5m2.
The next isosceles triangle has two sides of length 320m and a base of 360m.
Base of one triangle = 360 /2 = 180m
Length A = c2 – b2 = a2
= 3202 – 1802
= 700002
= √70000
= 264.57m
Area = base x height / 2
= 360 x 264.57 / 2
= 95245.2 / 2
= 47622.6m2
OR Heron’s Formula
Area =
= 500 (500 – 320)(500 – 360)(500 – 320)
= 2268000000
= √2268000000
= 47622.6m2
The area of this isosceles triangle is 47622.6m2.
The next triangle has two sides of length 330m and a base of 340m:
Base of one triangle = 340 / 2 = 170m
Length A = c2 – b2 = a2
= 3302 – 1702
= 800002
= √80000
= 282.84m
Therefore, the height of the triangle is 282.843m.
Area = base x height / 2
= 340 x 282.843 / 2
= 96166.62 / 2
= 48082.8m2
OR Heron’s Formula:
Area =
= 500 (500 – 330)(500 - 340)(500 – 330)
= 2312000000
= √2312000000
= 48082.8m2
The area of this isosceles triangle is 48082.8m2.
The next triangle has two sides of length 340m and base of 320m:
Base of one triangle = 320 / 2 = 160m
Length A = c2 – b2 = a2
= 3402 – 1602
= 900002
= √90000
= 300m
Therefore, the height of the triangle is 300m.
Area = base x height / 2
= 320 x 300 / 2
= 96000 / 2
= 48000m2
OR Heron’s Formula:
Area =
= 500 (500 – 340)(500 - 320)(500 – 340)
= 2304000000
= √2304000000
= 48000m2
The area of this isosceles triangle is 48,000m2.
The next isosceles triangle has two sides of length 350m and a base of 300m:
Base of one triangle = 300 / 2 = 150m
Length A = c2 – b2 = a2
= 3502 – 1502
= 1000002
= √100000
= 316.22m
So, the height of the triangle is 316.22m.
Area = base x height / 2
= 300 x 316.22 / 2
= 94866 / 2
= 47433m2
OR Heron’s Formula:
Area =
= 500 (500 – 350)(500 – 300)(500 – 350)
= 2250000000
= √2250000000
= 47433m2
The area of this isosceles triangle is 47433m2.
Below, is a table of the areas of isosceles triangles which have bases of lengths ranging from 400m to 200m:
I will now create a line graph to represent these results (see next page).
From the results in the table and the graph on the next page I have worked out that the equilateral triangle which has all three sides of length 333.33m is the triangle which gives the maximum area, 48111.186m2, out of all the triangles investigated in this section.
Rectangles
The next set of shapes that I am going to be investigating is rectangles. They are shapes with four sides and two sets of parallel lines. In this section, I will be calculating the area of two types of rectangles:
A square is a type of rectangle where all four of its sides are equal and are the same length. An oblong is a rectangle which has two pairs of equal sides.
The only formula I will need to work out the areas of the rectangles is:
Area = length x width
Square
The square of 1000m that I am going to be investigating has four sides, each measuring up to 250m. This is the only square possible which matches the principle of having the perimeter of 1000m:
Perimeter = 250 + 250 + 250 + 250 = 1000m
The area of this square:
Area = length x width
= 250 x 250
= 62500m2
I can also calculate the area of a square by splitting it up into four equilateral triangles and finding the area of each triangle and then multiplying the answer to that by four:
I will have to split this triangle into half to find the height of the whole triangle:
Base of right - angled triangle = 250m /2
= 125m
Area of whole triangle = base x height / 2
= 250 x 125 / 2
= 31250 / 2
= 15625m2
Area of square = 15625 x 4
= 62500m2
So, the area of the square is 62500m2.
Oblongs
The first oblong that I will be investigating has a length of 50m and a width of 450m:
The perimeter of this rectangle:
Perimeter = 450 + 50 + 450 + 50
= 1000m
Area = length x width
= 50 x 450
= 22500m2
The next oblong has a length of 75m and a width of 475m:
Perimeter = 425 + 75 + 425 + 75
= 1000m
Area = length x width
= 75 x 425
= 31875m2
The next rectangle has a length of 100m and a width of 400m:
Perimeter = 400 + 100 + 400 + 100
= 1000m
Area = length x width
= 100 x 400
= 40000m2
The next rectangle has a length of 125m and a width of 375m:
Perimeter = 375 + 125 + 375 + 125
= 1000m
Area = length x width
= 125 x 375
= 46875m2
The next rectangle has a length of 150m and a width of 350m:
Perimeter = 350 + 150 + 350 + 150
= 1000m
Area = length x width
= 150 x 350
= 52500m2
Below, I have created a table showing the results of all the possible rectangles which could be investigated:
Using the results from the table above, I have created a graph representing the areas of the rectangles investigated (see next page).
I have calculated that the rectangle which gives the maximum area is the square (length = 250m, width = 250m). I have noticed that the area amplifies when the value of the length is increased and reduces when the value of the length is decreased. The area is the same even when the lengths and widths are swapped around.
Regular Polygons
In this section, I will investigate a number of regular polygons such as the pentagon, hexagon and heptagon. The reason for why I am going to be investigating regular polygons and not irregular polygons is because so far it has been the shapes with all its sides equal that have given me the maximum area. If I investigate irregular polygons then they may not give me the maximum areas that I want and it may give me an awkward value. Therefore, I will be working out the areas of regular polygons:
Pentagon
A pentagon is a polygon which had five sides. A regular pentagon has all five sides equal of length. This is the first polygon that I am going to be investigating in this piece of coursework.
The pentagon above has a perimeter of 1000m. I have to split the pentagon into five triangles so that I can calculate the area of the pentagon.
The length of one side of the pentagon, i.e. base of one triangle, would be:
Base of one triangle = 1000 / 5
= 200m
And the interior angle of one triangle would be:
Interior angle = 360 / 5
= 72˚
As this is an isosceles triangle, the bottom two angles would be the same. Therefore, to calculate the angles of the two “Xs”, I would do:
Angle x = 180 – 72
= 108
= 108 / 2
= 54˚
In order to work out the area of the triangle, I will need to find the height of the triangle first. So, I will split the triangle into half:
Using the information above, I will now decide whether to use the formula of sin, cos or tan. SOH – CAH – TOA states that:
- sin˚ = opposite / hypotenuse
- cos˚ = adjacent / hypotenuse
- tan˚ = opposite / adjacent
I have decided to use the formula of “tan”:
tan˚ = opposite / adjacent
tan 54˚ = opposite / 100
opposite = tan 54˚ x 100
= 137.638192m
Therefore, the height of the triangle is 137.638192m. Now, the area of the whole triangle can be calculated:
Area = base x height / 2
= 200 x 137.638192 / 2
= 27527.63841 / 2
= 13763.8192m2
So, the area of the whole pentagon:
Area = 13763.8192 x 5
= 68819m2
Formula For Regular Polygons
1 x (1000 x Tan ((180 – 360) / 2) x 500) x n
2 ( n (( n ) ) n )
This formula can be simplified:
1 x (1000 x Tan (90 – 180) x 500) x n
2 ( n ( n ) n )
1 x (1000 x Tan (90 – 180) x 500)
2 ( ( n ) n )
500 x Tan (90 – 180) x 500
( n ) n
n = number of sides
This would be the shorter version of the previous calculation. Now that I have worked out the area of the pentagon, I can use this formula to check the answer by converting the “Ns” into the number of sides in the shape and in this case, its 5.
Tan ((90 – 180) x 5002)
(( 5 ) 5 )
Area of pentagon = 68819m2
This formula has just been proved to be right as the answer it gave for the area of a regular pentagon is the exactly the same as the answer calculated by using the longer method. Now, I will use this formula to find the areas of the other regular polygons.
Hexagon
A hexagon is a polygon which has six sides. A regular hexagon has all six sides equal and the same length.
I will calculate the area of the hexagon using the formula for regular polygons:
Tan (90 – 180) x 5002
( 6 ) 6
Area of hexagon = 72176m2
Heptagon
A heptagon is a polygon with seven sides. A regular heptagon has seven equal sides.
The area of the heptagon using the formula:
Tan (90 – 180) x 5002
( 7 ) 7
Area of heptagon = 74179m2
Octagon
An octagon is a polygon with eight sides. A regular octagon has eight equal sides.
The area of the octagon using the formula:
Tan (90 – 180) x 5002
( 8 ) 8
Area of octagon = 75444m2
Nonagon
A nonagon is a polygon with nine sides. A regular nonagon has nine equal sides.
The area of the nonagon using the formula:
Tan (90 – 180) x 5002
( 9 ) 9
Area of nonagon = 76331m2
Decagon
A decagon is a polygon with ten sides. A regular decagon is a shape with ten equal sides.
The area of a decagon using the formula:
Tan (90 – 180) x 5002
( 10 ) 10
Area of Decagon = 76942m2
As the number of sides the shape has increases, the area increases as well. Now, I will produce a table showing the results of the polygons’ areas I have already calculated and also some other polygons:
So far, the shape which has given me the maximum area is the shape with 1000000 sides. However, there is still one more shape we need to find the area of and that is the circle.
Circle
The last shape that I will calculate the area of is the circle.
The circle above must have a circumference of 1000m. To find the area, I will need to use the formula:
Area = πr2
But before I can work out the area of the circle, I need to calculate the radius of the circle which is until now, unknown.
Diameter of circle = circumference / pi
= 1000 / 3.141592654
= 318.3098862
Radius of circle = diameter / 2
= 318.3098862 / 2
= 159.1549431m
Now I can calculate the area of the circle:
Area = πr2
= π x (159.15494312)
= π x 25330.29591
= 79577.471m2
A graph of all the shapes that I have investigated in this coursework:
Evaluation
There is a trend in the areas of all the shapes that I have investigated. As the number of sides of the shapes increases, the area increases. However, there is a point of the graph where the area cannot get any bigger. This assumption concerns the last three shapes that I investigated; the 100000 sided shape, 1000000 sided shape and circle. This tells me that the area that these three shapes obtained is the maximum area that can be achieved using a shape which has a perimeter of 1000m.
Therefore, I can come to a conclusion that the farmer can fence off a plot of level land which has the shape of a circle and a circumference of 1000m.
