In conclusion to isosceles triangles, my investigation shows that the largest isosceles triangle has a base of 333⅓m. Coincidentally, this triangle happens to be an equilateral triangle. This shows that an equilateral triangle has the largest area.
In conclusion, my investigation has shown that out of all the three types of triangle, equilateral has the largest surface area.
Now I shall use these results and plot them on a line graph to show that as the base goes wider, the area increases but when the base goes past the 333⅓m line, the area starts to decrease again:
This graph does not show that the maximum area is 333⅓m². The scale isn’t zoomed in enough to see whether the area is 333⅓m² or not. To show that it is, I have to plot another graph that is zoomed in so the base is going up by 0.1m each time. Then it will show if the area is 333⅓ m² or not.
As you can see the curved line is at its highest point when the base is at 333⅓m. This means that the area (which represents the line) is at its highest point when the base is 333⅓m long. This means that the triangle is an equilateral.
Quadrilaterals:
I am going to start off with the most general cases of quadrilaterals. I am going to use the same method as I did when I compared scalene triangles with isosceles to show that a rectangle would have the largest area. There are so many quadrilaterals with infinitely many different areas, that I will only work on the most basic ones; the trapezium and the parallelogram.
Quadrilaterals: Comparing a rectangle with a Trapezium and Parallelogram
Just by looking at this diagram, I think it is reasonable to assume that no matter how tall or wide the trapezium is, it will never be as big as a rectangle in terms of area. The same happens with a parallelogram but I still should show it anyway:
My two diagrams have shown that the rectangle is the shape that will have the largest area no matter what the height or width of the other quadrilateral is.
Quadrilaterals: The Rectangle
This is a basic example of a rectangle with a perimeter of 1000m. This is the formula to calculate the area of a rectangle: height x width
height x width = area
200 x 300 = 60000m²
Therefore, the area of the triangle above is 60000m². Now I must do this for lots of rectangles using the same method as I did for triangles; fill out some tables zooming in on each table every time, and then showing the results on a chart. I’m starting by going up by 50m each time:
Already the area has started to decrease somewhere around the 200m-300m spot so I will zoom in to find a closer answer:
The area is biggest around the 240m-260m spot so I shall zoom in again:
I have to zoom in one last time to finally figure out which rectangle has the largest area. This time I am going up in 0.1 of a metre:
In conclusion, I have investigated that the rectangle with the largest area (that has a perimeter of 1000m), has a base length of 250m and a height length of 250m. On the next page is a line graph showing the relationship between the base and the area of the rectangles. It also shows that the rectangle with the base of 250m will have the largest area:
Conclusion:
Just like the equilateral, coincidentally, this shape happens to be a square; which is a regular polygon. Having examined triangles and quadrilaterals, it seems to be that only regular polygons have the largest area. Therefore, I shall continue my investigation investigating only regular polygons.
The Regular Pentagon:
I believe that it is the regular pentagon that will have the larger area than all the other pentagons. Below is a diagram of a regular pentagon. It is split up into triangles because it will help me to explain how to find the area of it:
The lines aren’t properly aligned in the diagram but you will have to assume that they are so each triangle is equal in terms of height, width and area. I have worked out that the length of each side must be 200m because since there are five sides altogether, I have to divide 1000 by 5. To work out the area of the pentagon I must split the polygon up into smaller triangles like I have done so above. Before I start calculating the area, I must work out the interior angle of the pentagon because I will need it for the trigonometry that is to follow. To find out the interior angle of the pentagon I divide 360º (360º = a circle) by the number of sides, which in this case, is 72º. I then have to divide the triangle into two so I get a right-angle triangle (as you can see in the diagram above, the triangle which is split up has an interior angle of 36º; which is half of 72º). Now I need to find the height of the triangle and this is where trigonometry comes in. Comparing the length of one of the sides of the pentagon (200m) with the height of the triangle, I am estimating that the height will be about 140m. I have to remember now that because I have halved the triangle, the base is 100m, not 200m:
SOH|CAH|TOA
tan 36 = opp
adj
tan 36 = 100
x
x = 100
tan 36
height = 137.638m (3s.f.)
Now that I have worked out the height of the triangle, I need to calculate the whole area of the triangle:
(h x b) ÷ 2
(137.638192 x 200) ÷ 2
27527.6384 ÷ 2
area = 13,763.8192m²
Now that I know the area of one triangle I can easily work out the area of the whole pentagon by multiplying the area of the triangle by 5 because there are 5 triangles:
13,763.8192 x 5
area of regular pentagon = 68819.096m²
I can compare the area of a regular pentagon with the area of a square and the pentagon turns out to have the largest area. So far, the regular pentagon is on top of the area list in my investigation.
The Regular Hexagon:
The regular hexagon is very similar to the pentagon in terms of finding the area:
The reason I am repeating this is because a hexagon is made up of six equilateral triangles. Again because the shape is drawn out on computer, the lines couldn’t be exactly joined up but we will have to assume that they are. I have worked out the interior angle, the length of the sides, and all the other calculations that are needed to perform the trigonometry:
SOH|CAH|TOA
tan 30 = opp
adj
tan 30 = 83⅓
x
x = 83⅓
tan 30
height = 144.338m (3s.f.)
Now I can work out the area of the triangle:
(h x b) ÷ 2
(144.3375673 x 166.6666667) ÷ 2
24056.26122 ÷ 2
area = 12028.13061m²
Now I multiply my answer by six because there are six triangles:
12028.13061 x 6 = 72168.784m²
So the area of a hexagon is 72168.784m² (3s.f.)
The Regular Octagon:
This is the last time I will be drawing out a regular polygon because it uses the same method as the pentagon and hexagon. So for the last time, I will show the method on how to find the area of the octagon:
SOH|CAH|TOA
tan 22.5 = opp
adj
tan 22.5 = 62.5
x
x = 62.5
tan 22.5
height = 150.888m (3s.f.)
Now I can work out the area of the triangle:
(h x b) ÷ 2
(150.8883476 x 125) ÷ 2
18861.04346 ÷ 2
area = 9430.521728m²
Now I multiply my answer by eight because there are eight triangles:
9430.521728 x 8
area of octagon = 75444.17382m²
Investigating regular polygons with a large number of sides:
The method for finding the area of a polygon with even 1000 sides is still the same as the method I used for the pentagon, hexagon and the octagon. It seems as if as the numbers of sides are increasing, the area is also increasing. Now I will work out the areas of polygons with a large number of sides but it will take me forever if I carry on splitting up the shape, working out the angles, working out the lengths, and especially working out the height using trigonometry. Therefore, I need a formula that will help me find the area of a regular polygon with n sides so I can use that formula to find the area of any sided shape.
The Formula:
Now I will work out the formula systematically using algebra, and regular pentagon as an example:
The example diagram shows how I got all the lengths and angles. So in this case, the ‘n’ would be replaced by ‘5’ because ‘n’ equals the number of sides. I have drawn arrows with numbers pointing to each fraction because below I will describe what each fraction means:
- This is probably the easiest one. It is 1000 ÷ n because in order to find the length of one side, you need to divide the perimeter by the number of sides. In the pentagon’s case, it would be 1000 ÷ 5.
- If the length of one side is 1000 ÷ 5, then the length of half of one side would be 500 ÷ n.
- There are 360 degrees in a circle. So if I want to find the interior angle of just one triangle, then it would be 360 ÷ n. In the pentagon’s case, it would be 360 ÷ 5.
- If the interior angle of one triangle is 360 ÷ n, then the interior angle of half of one triangle would be 180 ÷ n.
I have only showed how we can use the fractions to work out the angles and lengths of any regular sided polygon. Now I will explain how to find the height of the triangle and then find the area of it.
As we know, tan = opp ÷ adj. Every time we need to find the height of any sided regular polygon, we will always have to use tan. So if tan = opp ÷ adj, in algebraic terms, it is:
Height = 500 ÷ tan 180
n n
In the pentagon’s case, the formula would be (500 ÷ 5) ÷ (tan (180 ÷ 5))
Now that I have worked out a formula to work out the height of one of the triangles in the polygon, I need to work out how to find the area of it:
Area of triangle = ½b x h
The ‘½b’ means that it’s half the length of the base. As we know, in algebraic terms, half the base is ‘500 ÷ n’. Now I have everything that I need to put the area of a triangle into a formula in terms of ‘n’:
Area of triangle = (500 ÷ tan 180) x 500
n n n
The final bit of the formula is to multiply the area of the triangle by the number of sides to get the area of the whole polygon. So the formula would be:
Area of any regular sided polygon = (500 ÷ tan 180) x 500 x n
n n n
I can simplify this formula a bit by cancelling out the two double underlined ‘n’s. The final formula is:
A = (500 ÷ tan 180) x 500
n n
Now that I have the formula, I can find the area of any sided polygon. Below is a table that shows the number of sides and the areas of polygons with a very large number of sides. I worked the area out with the formula that I just worked out:
My table shows that as the number of sides increase, the area also increases. Below are three graphs that show the area against the number of sides. The first graph has the number of sides increasing by 1. The second has it increasing in 50s and the third in 1000s.
So the area is definitely increasing as the sides are increasing but let’s take a look at polygons with larger number of sides:
Finally let’s have a look at a graph in which the base increases by 1000 sides each time.
Although it looks like there are still huge gaps in between each number of sides, we have to take in account that we are going up by 1000 sides each time. Also, if you look at the values on the y axis, you can see that the gaps are actually really small. This graph is zoomed in a lot to make it look as if the gaps are big. These graphs easily show that as n increases, so does the area. There will be a time where n is towards reaching its limit. It will never reach it, but it will always get closer and closer to it. The limit in this case is when the shape gets to an infinite number of sides; which is a circle.
The Circle:
A circle isn’t in fact a polygon and I cannot find the area of it by using my formula because you cannot put infinite into a formula. If you think about it, as the shape is increasing in the number of sides, it is forming more and more into a circle. A square; which has four sides, looks nothing like a circle but a ten thousand sided shape looks a lot more like a circle. My graph was getting more and more closer to the limit. Now I will work out the area of the circle with a perimeter of 1000m:
Circumference of a circle = 2πr
1000 = 2πr
r = 1000
2π
r = 159.1549431m
Now that I have the radius I can work out the area of the circle:
area of a circle = πr²
area = π x 159.1549431²
area of circle = 79577.47155m²
Below I will use diagrams to help me explain how and why the circle has the largest area:
Assuming that all of these shapes have a perimeter of 1000m, the octagon or pentagon can never have the larger area. I have already justified that as n increases the shape starts to look more like a circle. A square will stick out of the circle more than a pentagon would because the side lengths are decreasing and slowly becoming into a circle. Because they are decreasing, they won’t stick out of the circle as much. This is why I have made the pentagon stick out of the circle more.
The pentagon has five sides, and is leaving the circle with a larger area than the eight sided octagon is. This means that as the number of sides increase, the ‘shaded’ part of the circle decreases.
Comparing the formula for the area of a circle with the area of n
There is an interesting relationship between the two formulas. The formula for the area of n is:
A= 250,000
n tan (180)
n
The formula for the area of a circle that has a 1000m perimeter is:
A= πr² = π x 100,000
4π²
= 250,000
π
There is a relationship between the denominators. Let’s take 5 as an example for n. the denominator for the first equation would then be:
5 tan (180)
5
The answer to this is:
3.149591887
The value of π goes on forever in terms of decimal places but on a calculator, it is:
3.141592654
As you can see the values are very close between the two denominators. Now I shall try substituting n with a lower number than 5. I will try using 3:
3 tan (180)
3
The answer is:
3.144466757
As you can see the answer is getting closer to π but it isn’t going lower. Below is a table showing that as the numbers are decreasing, the value is getting closer to π but never reaching it. It is acceptable to even use numbers like 0.1 to represent n because we know that a circle doesn’t have 0.1 sides, it has an infinite number of sides:
Even when n is 0.1, the value doesn’t reach the same as π. We can then say that π is a limit which cannot be reached but can get closer and closer to being reached.
The final part of this investigation will be a graph showing the ‘shaded’ part of the circle against the number of sides. To find the area of the shaded part, I have to subtract the area of the regular polygon from the area of the circle:
In conclusion, my investigation has shown that a farmer should have a regular circular fence as the shape of the plot.
