a² = 175,000
a = 418.3
So, the height of the triangle,
= 418.3cm²
150 75
Therefore,
½ BH
½ x base x height
½ x 150 x 418.3
½ x 62,745
Area of Triangle = 31,372.5m²
a² + b² = c²
a² = c² ─ b²
a² = 400² ─ 100²
400 400 400 a² = 160,000 ─ 10,000
== == a² = 150,000
a = 387.3
So, the height of the triangle,
= 387.3cm²
200 100
Therefore,
½ BH
½ x base x height
½ x 200 x 387.3
½ x 77,460
Area of Triangle = 38,730m²
a² + b² = c²
a² = c² ─ b²
375 375 375 a² = 375² ─ 125²
== == a² = 140,625 ─ 15,625
a a² = 125,000
a = 353.6
So, the height of the triangle,
= 353.6cm²
250 125
250 125 Therefore,
½ BH
½ x base x height
½ x 250 x 353.6
½ x 88,400
Area of triangle = 44,200m²
a² + b² =c²
a² = c² ─ b²
350 350 350 a² = 350² ─ 150²
a² = 122,500 ─ 22,500
== a == a² = 100,000
a = 316.2
So, the height of the triangle,
= 316.2cm²
Therefore,
½ BH
300 150 ½ x base x height
½ x 300 x 316.2
½ x 94,860
Area of triangle = 47,430m²
a² + b² = c²
a² = c² ─ b²
325 a² = 325² ─ 175²
325 325 a² = 105,625 ─ 30,625
== a == a² = 75,000
a = 273.9
So, the height of the triangle,
=273.9cm²
Therefore,
½ BH
350 175 ½ x base x height
½ x 350 x 273.9
½ x 95,865
Table: Area of triangle = 47,932.5m²
After careful calculation I have discovered that as the base increases and the height decreases, the area of an isosceles triangle increases. The graph above shows this. To do the calculation I used the formula ½ x base x height (Area of a triangle). To carry out this formula I was required to find the height, using “Pythagoras theorem” and the remainder two sides, how I did this is shown in the calculations. The answers presented are rounded up to 1 decimal point but the actual figures were used in the calculation. Also, the results show that to have the greatest area the base must be a length between 300 and 350. And in this graph it shows that the base with the length of 350 has the largest area.
THREE REGULAR SHAPES (POLYGONS)
In this section, I will be looking at three regular shapes (polygons). The shapes I will be looking at are; the regular pentagon, hexagon, and octagon. I will be finding out how they might be the apposite shape to solve the problem. I will ensue this by finding their areas, in other to this; I would have to break down the shapes into “Logical Order” (systematic reference) by breaking down into triangles, finding the area of the triangles and then carrying on from there.
Pentagon
32
200 200
360 ÷ 50 = 72
== == H
A
200
200
200 100
200
In this pentagon, I have split it into five triangles in other to find the area. I will then do the same thing I did for the triangles above; find their height in other to find their area.
Therefore, I’ll be finding the height of the right-angled triangle; unfortunately, I have got just one side of the right-angled triangle that is known. So I cannot find height using “Pythagoras theorem”, so I would be using trigonometry (SOH-CAH-TOA).
I am going to use TOA for the triangle because it is the appropriate formula to use now.
TOA = Tangent = Opposite
Adjacent
Tan 36 = 100
A
a = 100
Tan 36
a = 137.64
So, the height of the triangle = 137.6cm²
Therefore, to find out the Area of the triangle,
½ BH
½ x Base x Height
½ x 200 x 137.6
½ x 27,520
Area of triangle = 13,760m²
I will now find out the area of the pentagon. Since I divided the pentagon into five triangles; I’ll now multiply the area of the triangle by five in other to get the area of the pentagon.
Therefore;
Area of pentagon = 13,760 x 5
= 68,800m²
Hexagon _
166.7 360 ÷ 6 = 60
166.7
166.7
166.7 166.7
166.7
166.7
A H
83.35
In this hexagon, I have split it into six triangles in other to find the area. I will then do the same thing I did for the triangles above; find their height in other to find their area.
Therefore, I’ll be finding the height of the right-angled triangle; unfortunately, I have got just one side of the right-angled triangle that is known. So I cannot find the height using “Pythagoras theorem”, so I would be using trigonometry (SOH-CAH-TOA).
I am going to use TOA for the triangle because it is the appropriate formula to use now.
TOA = Tangent = Opposite
Adjacent
Tan 30 = 83.35
A
A = 83.35
Tan 30
A = 173.19
So, the height of the triangle = 173.2cm²
Therefore, to find out the Area of the triangle,
½ BH
½ x Base x Height
½ x 166.7 x 173.2
½ x 22,872.4
Area of triangle = 22,872.4m²
I will now find out the area of the hexagon. Since I divided the hexagon into six triangles;
I’ll now multiply the area of the triangle by six in other to get the area of the hexagon.
Therefore;
Area of hexagon = 22,872.4 x 6
= 137,234.4m²
OCTAGON
125 360 ÷ 8 = 45
125 125
125 125
125 125
125 125
H
A
62.5
In the octagon above, I have split it into eight triangles in other to find the area. I will then do the same thing I did for the triangles above; find their height in other to find their area.
Therefore, I’ll be finding the height of the right-angled triangle; unfortunately, I have got just one side of the right-angled triangle that is known. So I cannot find the height using “Pythagoras theorem”, so I would be using trigonometry (SOH-CAH-TOA).
I am going to use TOA for the triangle because it is the appropriate formula to use now.
TOA = Tangent = Opposite
Adjacent
Tan 22.5 = 62.5
A
A = 62.5
Tan 22.5
A = 150.89
So, the height of the triangle = 150.9cm²
Therefore, to find out the Area of the triangle,
½ BH
½ x Base x Height
½ x 125 x 150.9
½ x 18,862.5
Area of triangle = 18,862.5m²
I will now find out the area of the octagon. Since I divided the octagon into eight triangles; I’ll now multiply the area of the triangle by eight in other to get the area of the octagon.
Therefore;
Area of octagon = 18,862.5 x 8
= 150,900m²
I have completed triangles and I went on to research regular shapes. I selected the number of sides randomly and I have chosen 5 sides, 6 sides and 8 sides. To discover the area of the shape I was required to divide the shape into equal isosceles triangles. I would find the area of the triangles and then times the area by the number of triangles altogether. The number of triangles should be equal to the number of sides. I have used trigonometry to discover the height of a triangle and used the formula ½ x base x height to discover the area. The numbers shown are rounded up but the correct figures were used in the calculation. It can be seen that as the number of sides increases so does the area.
The graph above shows this; has the octagon (8-sided) has the largest area out of all the other polygons.
THE GENERATED FORMULA (N-SIDED)
This formula is just an easier way of working out the area of polygons without going through all the process I went through in the previous ones I’d done. I am putting in my investigation, because our teacher just thought us, and how it is used. So I’ve not used it for the first polygons, so that I can show you how it is derived, its advantage and the difference in how useful it is. The derivation of this formula is a bit stressing, very long, and confusing, but after it has been derived, you shall see its capability. I am going to use a 6-sided (hexagon) so that it doesn’t get too complicated. I’ll work it out in the normal other it is done and at the same time be comparing it to an n-sided hexagon.
Now that I’ve derived the formula, I am now going to use it to see how fast and easy it can be. Below is an example of a polygon that I’ll find its area using the formula; I’ll find the area of a 99-sided polygon (imaginary actually) to let you know that the formula is not limited (as long as you have a scientific calculator), and it still will not get complicated a bit.
To find the area of the shape you simply substitute the number of sides with n. when this formula is inserted into a scientific calculator then the area is instantly calculated.
Thus;
To find the area of the 99-shaped polygon = 500²
n x tan (180/n)
Area = 500²
99 x tan (180/99)
Area of Decagon (after inserted into the calculator)
= 79,550.8m²
As you can see, instead of splitting the polygon into triangles, and then finding the area, by first finding the height and.........., it has all being in about five lines by just one unproblematic formula.
CIRCLE
In this part of my coursework, I’ll be looking at how a circle might be able to solve the problem. I will also using the same old procedure; finding the area of the circle.
Thus;
1000m
First of all, in other to find the area of the circle, I’ll first find the circumference. In the previous different shapes I’ve done, I looked for the height, but in the circle, the height is the circumference; so there are already changes (only in the circle). I fill there is something different about this circle; I believe it will bring out a different answer, therefore leading to the remedy of this problem.
Therefore;
Formula for circumference of a circle = 2πr
= 2 x π x radius
Thus;
= 1000m =2πr
= 1000m = r
2π
r = 1000
2π
r = 1000 (π = 22/7 == 3.142)
2 x 3.142
r = 1000
6.284
Radius = 159.13
Therefore;
To find the area of the circle;
= πr²
The above is the formula for finding the area of the circle.
Area of circle = πr²
A = π159.13²
= π x 159.13²
= π x 25,322.4
= 3.142 x 25,322.4
Area of circle = 79,562.10m²
Surprisingly, according to the graph above and all the other shapes I’ve studied, it seems that the Octagon is the shape with the largest area, according to statistics, the circle should have the largest area, but according to my own investigation, the octagon has the largest area.
CONCLUSION
In conclusion, the results in my graph show that an 8- sided shape or an octagon has the largest area out of all the shapes studied. This shows that the farmer will fence off the plot in the shape of an octagon of 1000 metres to fence off the plot of the land which contains the maximum area. The illustrations and calculations above show the different sided shapes that could have been a possibility. These include quadrilaterals, triangles, regular shapes and the circle. But the octagon shows us that out of all the shapes I have mentioned the octagon still has the largest area.
The reason for my result could still be the reason that the octagon is almost close to a circle, so the shape with largest area could therefore be octagon.
