Triangles
Scalene
350m 200m
a b
450m
c
As we already know everything there is to know to find the area of this triangle, the procedure is therefore very simple. The scalene triangle heights don’t need to be solved to work out their areas; all that is needed is one simple formula.
S = half the perimeter (in this case 500)
√ (S (S-a)(S-b)(S-c))
√500 (500-350)(500-200)(500-450)
√500 x 150 x 300 x 50
√1125000000
33541.02m2
Review
In the previous section:
I have noticed that the more regular the triangle is, the larger its area will be.
I discovered that an Equilateral triangle therefore has the largest area compared to the rest of the triangles.
End of Section 1
In the next section:
I will start investigating different Quadrilaterals.
These will be: Squares
Rectangles
Trapeziums
Parallelograms
There is only one possibility for a Square.
Quadrilaterals
Squares
250m
250m 250m
250m
Area = length x breadth
As we already know both the length and the width of the square, it is the easiest shape to solve the area of.
Area = length x breadth
Area = 250m x 250m
Area = 62,500m2
Quadrilaterals
Rectangles
400m
100m 100m
400m
Area = length x breadth
As we already have both the length and the breadth of the rectangle shown above, we can solve the area with no trouble whatsoever.
Area = length x breadth
Area = 400m x 100m
Area = 40,000m2
Quadrilaterals
Trapeziums
200m
200m 200m
400m
?
Area = ½ (base + top)
200m
200m 200m
100m 200m 100m
If both sides of the trapezium are 200m long, the top is 200 and the base is 400, then if I split the trapezium into a rectangle and two triangles, then the base of each of those triangles will be 100m long and so therefore I can use Pythagoras to solve the height, therefore I can work out the area.
Pythagoras’ Theorem must now be used to solve the perpendicular height of this trapezium if the area is to be found.
2002 = 1002 + x2
2002 – 1002 = x2
30000 = x2
√30000 = x
173.205m = x
Now that the perpendicular height of the trapezium has been solved, working out the area is an easy procedure.
Area = ½ (base + top) height
Area = ½ (400 + 200) 173.205
Area = ½ (600) 173.205
Area = 300 x 173.205
Area = 51,961.52m2
I see no reason to bother with the trapeziums though because the length of the perpendicular height will always be shorter than the length of one of the sides. Therefore the area is always less than a similar rectangle.
As I explained above, I can split the shape into a rectangle and two right-angled triangles and those two triangles have their hypotenuse as the two sides of the trapezium, and in a right-angled triangle, the hypotenuse is always the longest side.
Quadrilaterals
Parallelograms
If I take a regular rectangle and label the width “y” and the length “x”, then the area will be “xy”.
x x
y y
If I then take a similar parallelogram and also label the width “y” and the length “x” then the height of the parallelogram will always be smaller than the length of the side because if I split the shape into two right-angled triangles and a rectangle, the height will always be less than the side labeled “y” because y is the hypotenuse of the triangle (the longest side). Therefore if the height is smaller than y, the overall area will be smaller than xy.
Due to this discovery, I have realized that solving the area of a parallelogram is not necessary because it will always be less than that of a square or a rectangle (more regular shapes).
The same happened with the triangles, the Equilateral (most regular of the 3 triangles) had the largest area due to its regularity.
Review
In the previous section:
Once again, I have noticed that the more regular a shape is, the larger it’s are will be.
In this case it was the square that had the largest area out of all the Quadrilaterals.
So far in my investigation, I have found that regular shapes produce larger areas than those of irregular shapes, so therefore from now on, I will only use regular shapes in my investigations.
End of Section 2
In the next section:
I will be investigating the regular Pentagon, Hexagon, Decagon, Undecagon and a 20-sided polygon.
Regular Pentagon
200m 200m
200m 200m
200m
To work out the are of this polygon, I have to split it up into five equal triangles like so:
Working out the area of this pentagon led me onto discovering a very useful formula for solving the areas of regular shapes.
Regular Pentagon
n = number of sides
To work out the length of one side: 1000/n
To work out the length of half a side: 1000/2n
To solve the length of the perpendicular height, this angle must be found:
This is by doing: 360/2n
Now that we have the angle and half the length of a side we can solve the height.
Perpendicular Height: 1000/2n
tan36 = --------------
height (h)
tan36 x h = 100
h = 100/tan36
h = 137.638192m
Regular Pentagon
Multiplying this by ½ the base gives me the area of one triangle.
Area = 137.638192 x 100
Area = 13763.8192m2
Multiplying this by 5 will then give me the area of the whole pentagon.
Area = 13763.8192 x 5
Area = 68819.09602m2
Putting these small steps together gave me this formula:
250,000 (see formula page to see steps taken
--------------- to reach this formula)
n tan (180/n)
This formula works with all regular shapes.
Regular Hexagon
166.6m
166.6m 166.6m
166.6m 166.6m
166.6m
I will work out the area by using my new formula:
250,000
-------------
6 tan (180/6)
250,000
--------------
3.46410162
72168.78365m2
Regular Decagon
Each side will be 100 metres long.
Once again, I will use my formula:
250,000
-------------
10 tan (180/10)
250,000
-------------
3.249196962
76942.08844m2
I have noticed that as the number of sides increase, so does the area of the shape. I will try an Undecagon, and a 20-sided polygon and if the areas increase then I am correct.
Regular Undecagon
Each side will be 90.90 metres long.
I will once again use my formula.
250,000
-------------
11 tan (180/11)
250,000
-------------
3.229891422
77401.98271m2
Regular 20-sided Polygon
Each side will be 50 metres long.
I will again use my superb formula:
250,000
-------------
20 tan (180/20)
250,000
-------------
3.167688806
78921.89395m2
Review
In the previous section:
I discovered a fantastic new formula that works for all regular polygons.
I discovered that the more sides a shape has, the larger its area will be.
In that section, the 20-sided polygon had the largest area, but a more numerous sided shape would have a larger area than that.
End of Section 3
In the next section:
I will look at the circle and then come to a final conclusion about the coursework. (I have chosen to look at the circle last because the more and more sides a shape has, the more it starts to look like a circle). I will also show the build up to my final formula.
Formula
For all regular shapes:
As you saw in the section about pentagons, I showed how to work out each small bit of one of the triangles in the pentagon.
I put all of these small parts together to form one simplified formula that works for all regular shapes and where ‘n’ is the number of sides.
To work out the height of the triangle is:
(see page 22 to see where these
1000÷2n expressions come from)
tan (360÷2n)
To work out the area of 1 triangle is:
1000÷2n x 1000
tan (360÷2n) 2n
To work out the area of the whole pentagon is therefore:
1000÷2n x 1000
n x (tan (360÷2n) 2n )
This whole thing can be simplified to:
500÷n x 500
n x (tan (180÷n) n )
Simplified again to:
500 x 500 x 1 .
n x ( n n tan(180÷n) )
Simplified again to:
500 1 .
500 x n x tan(180÷n)
Simplified again to:
250,000 . 1 .
n x tan(180÷n)
Finally simplified to:
250,000
n tan (180÷n)
Above is my final formula and the steps that I took to reach it.
Circle
For the circle, I cannot use my formula because a circle is an infinity sided shape.
1000m
To work out the diameter I have to use: 1000/π
To work out the radius I have to use: 1000/2π
Radius = 159.1549431m
Area = πr2
Area = 79577.47154m2
The relationship between the formula for regular shapes and the formula for a circle
I have noticed that the bottom line of my formula for regular shapes gets closer to π as the number of sides increases. I will explain why in this section.
Since 180o is the same as π in radians, my formula can also be written as:
250,000 (π÷n is the same as 180÷n in
n(tan(π÷n)) radians)
180 divided by n is the expression for the angle.
As ‘n’ increases, the angle gets smaller. When the angle is smaller, the tangent of a small angle is the same as the angle in radians, therefore there is no need for tan (in the case of very small angles, the tangent of the angle makes no difference to the angle itself, therefore as n increases and the angle gets small there is no need for tan. Therefore tan can be discounted and the formula will therefore be written as):
250,000
n(π÷n)
Now the n’s can cancel out and I’m left with:
250,000
π
which is the same as the area for a circle.
Review
In the previous section:
I discovered that the circle had the largest area out of all the other shapes that I investigated.
I couldn’t use my formula for the circle because a circle does not have any sides, it is a curved shape.
End of Section 4
Conclusion
The More regular shapes are, the larger their areas will be.
The more sides a shape has, the larger its area will be.
I have noticed that the more sides a shape has, the closer and closer the number on the bottom line of my formula gets to π (as explained in the previous page).
In conclusion, the circle is the largest shape of the lot and the farmer should use a circular fence for her farm.
End of Coursework
