I am now going to look at triangles and see if they will produce an even larger area. Firstly I am going to look at the isosceles triangle; this is a triangle in which 2 of the sides and 2 of the angles are the same. As I am using triangles I can use the equation:
Area (A) = ½ of the base (b) x height (h)
Which abbreviates to:
A=½bh
I do not know the measurement of the height so I will split my triangle into 2 by bisecting the angle as shown below to make this easier (all shapes drawn to scale 1cm:50m):
To find the height I will need to use Pythagoras’ Theorem:
h2 = √ hypotenuse2 - b2
h2 = √ 3002 - 2002
h2 = √ 90000 – 40000
h2 = √ 50000
h = 223.6068m to 4 decimal places (4 d.p) to make sure that my work is accurate.
Now I need to find out what ½ of the base is so I can put that into the equation, I find this out by dividing the base by 2:
b/2 = 400/2 = 200m
The equation to find the area of my triangle is like so:
A = ½bh
A = 200 x 223.61
A = 44721m2
I would have looked at a right-angled triangle, but the lengths that add up to 1000, do not fit easily into Pythagoras’ Theorem. I will leave that and go onto look at an equilateral triangle, the two main reasons why I am going straight to this and not looking at other triangles such as scalene triangles (ones in which all the sides and angles are different) are because:
- To find the area of a scalene triangle, I will have to draw it out to scale which would not give me an accurate enough answer because human measurement i.e. rulers will not give me up to 4 decimals places like I am looking for.
- From looking at quadrilaterals, both regular and irregular, I found out that the one with the biggest area was the regular one. Therefore I will only look at the regular shapes from now on.
I am going to start with the regular triangle, also known as an equilateral triangle; this means that all the angles are the same size and all the sides are the same length.
To divide the fencing equally for each side I will divide the fencing by the number of sides (n) which is:
1000/n = 1000/3 = 333 ⅓m
I now need to find out what the angles in my triangle add up to so I can work out the area. I already know from previous work that all the angles in a triangle add up to 180° so to find out what each angle is I need to divide the 180° between the 3 angles:
180/3 = 60°
If I split my triangle in half I will find it easier to use the equation A=½bh. I do not know what the height is though so I will have to use trigonometry to come across this. To find out whether I will need to use sine (sin), cosine (cos) or tangent (tan) I will need to draw out my triangle to see what angles/sides I have/need:
From looking at my diagram I can see that I need to find the opposite (opp) and I have the adjacent (adj), this means that I use tan because tan = opp/adj. I already know that my angle is 60° so all I need is to find half of my base which means I divide the existing base by 2:
Base (b)/2 = 133 ⅓ / 2 = 166 ⅔m
Now I can work out the height of my triangle:
Tan60 = h / 166 ⅔
h = 166 ⅔ x Tan60
h = 288.6751m (4 d.p)
Now I have all of the information to work out the area of my triangle:
A = ½bh
A = 166 ⅔ x 288.67
A = 48112.5224m2 (4 d.p)
To compare my work on triangles I will draw a table:
I was correct in saying that the regular shapes have the biggest area; therefore I will only look at regular shapes from now on. Now I will have a look at regular pentagons so I can see how their area’s compare to those which I have already looked at.
I will start with the regular 5 sided shape known as a pentagon. From reading my Letts GCSE Mathematics Revision Guide, I know that the easiest way to find the area of a regular polygon is to split the shape into isosceles triangles starting from the centre and connecting this point to each of the vertices.
To find the length of each side, I use the same equation as before:
1000/n = 1000/5 = 200m
As I will be using the formula A = ½bh, I need to divide the length into 2 as it will be the base of my triangle:
b/2 = 200/2 = 100m
Now I need to find the size of the angles in each of these triangles, starting with the exterior. You do this by dividing 360° by 5, the number of isosceles triangles:
Exterior Angle = 360°/n = 360°/5 = 72°
I then need to find the interior angle, which is the exterior angle subtracted from 180° and then divided by two because the degrees in a straight line add up to 180° and I have two angles which between them, add up to 180°:
Interior Angle = 180°- exterior angle = 180° - 72° = 108°/2 = 54°
At the centre of the pentagon I have 5 angles of 72°. If I dissect each of these angles into 2, I will end up with 10 right-angled triangles, I will then be able to find the height (h) by using tan:
Tan54 (from the 54 degree angle) = opp (what I am trying to find)/adj (half of the base)
This then equals:
Tan54 = h/100
This can be rearranged to:
h = tan54 x 100
h = 137.6381m (4 d.p)
I can then put all of this information into the formula to find the area of a triangle:
A = ½bh
A = 100 x137.6381
A = 13763.8192m (4 d.p)
Next I multiply this answer by the number of isosceles triangles, which is also the number of sides:
A = 13763.8192 x 5 = 68819.0960m2 (4 d.p)
To compare my work on quadrilaterals, triangles and pentagons, I will draw a graph to see if I can see any patterns that may help me in my work:
From looking at my graph I can now predict that as the amount of sides increase, so does the area. This has happened so far, but to support my prediction, I will look at the regular hexagon, a 6-sided shape.
To find the area of my hexagon, I will use the same method as I did for my pentagon. First I will find out the length of each side:
1000/n = 1000/6 = 166⅔m
Then I will find out the exterior angle:
Exterior angle = 360°/n = 360°/6 = 60°
Then the interior angle:
180° - exterior angle = 180° - 60° = 120°/2 = 60°
Then I bisect my 6 isosceles triangles into 12 right-angled triangles to make my working out easier. As I will be using the triangle formula A = ½bh I will need to find half of the base:
b/2 = 166 ⅔ /2 = 83 ⅓m
Now I find the height by using tan again:
Tan = Opp/Adj
Tan 60 = h/83 ⅓
h = 83 ⅓ x tan60
h = 144.3376m (4 d.p)
Then I find out the area of my isosceles triangle:
A = ½bh
A = 83 ⅓ x 144.3376
A = 12028.1306m2 (4 d.p)
Then I multiply my answer by 6, the number of isosceles triangles in my hexagon:
A = 12028.1306 x 6 = 72168.7837m2 (4 d.p)
I will plot this information as well as the information from the other graph onto a new graph showing number of sides against area:
This graph shows that my prediction is still correct as the area is getting larger as the amount of sides increase. I now want to try a heptagon, 7-sides, to see if this makes a difference to my results. Before I do that though, I want to find a formula which will give me the area of a regular polygon. I want to do this because when I looked back at my work, I realised that I was doing the same working each time, only changing a few numbers according to the number of sides in the shape (n).
At the end I had to multiply my answer by the number of sides so I will also need to use n in my formula. Each time I was dividing the 1000m by the number of sides to get the length of each side so I can use 1000/n in my formula, I will need to put this in brackets though because I am multiplying n by the answer that 1000/n gives me:
n(1000/n)
I split my isosceles triangles into right-angled triangles so there were twice as many triangles after I done this, so I will need to square my answer:
n(1000/n)2
I divided my base by 2 so I could use this to work out the area of my triangles, but I also had to divide the opposite by the adjacent when I used tan to find the height which equals:
(b/2)/h
This cancels down to:
b/2h
I need to put the tan into this part of my workings to find out the measurement of the height in my triangle, and this needs to take priority over the dividing so I will put it in brackets:
h=L/(2tan)
Something needs to go after the tan for it to make sense. I divided 360° by the number of sides to get isosceles triangles so I could use 360°/n but since I then divided this answer by 2 to get right-angled triangles, so I can put 180/n to get a more simple equation:
L/(2tan(180/n))
The 180/n is also in brackets because I am multiplying that answer by 2tan, and then the whole of that answer is will be divided into L. The length of the base for each of the triangles is also L/2:
L/(2tan(180/n))(L/2)
This can be simplified to:
/(4tan(180/n))
Then I add my beginning part to it and get:
A = n(1000/n)2/(4tan(180/n))
I am going to check my formula to see if I get 68819.0960m2 for the area of a regular pentagon:
A = n(1000/n)2/(4tan(180/n))
A = 5(1000/5)2/(4tan(180/5))
A = 68819.0960m2
This shows that my formula has given me the same answer as what I got when I worked this out the long way. I am now going to check my formula and my prediction by working out the area of regular heptagons (7 sides), octagons (8sides), nonagons (9 sides), decagons (10 sides) and then 11 and 12 sided shapes:
It looks as though my formula works for the regular shapes with up to 12 sides. Later I want to see if it still works for shapes with thousands of sides. First I want to check my prediction by another graph to show number of sides against area:
My graph shows that the area is still increasing as the amount of sides increase, even though it is not increasing nearly as much as it did at the start. I now want to go back and check to see if my formula works with shapes that have a lot more sides than those that I have already looked at:
From looking at my chart you can see that as the sides go up the area is still going up, just. This just about proves that my prediction is correct but there is one thing that will definitely prove that as the number of sides increase, so does the area. That is to find the area of a circle with the circumference of 1000m. The reason why I say this is because circles have infinite (or endless) amount of sides.
Although I will not be able to use my formula to work out the area of the circle, I can find it out by rearranging the formula to find the circumference of a circle which is:
Circumference (C) = Pi (3.14) x Diameter (d)
C = 3.14 x d
This can be rearranged to:
d = C/3.14
This then works out as:
d = 1000/3.14
d = 318.3099m (4 d.p)
I then divide this by two since I need the radius (r), which is half of the diameter:
r = 318.3099/2 = 159.1550 (4 d.p)
I can then use the formula, A = 3.14 x r2 to work out the area of the circle:
A = 3.14 x r2
A = 3.14 x 159.15502
A = 79577.4716m2 (4.d.p)
This proves that my prediction is correct as the area does in fact increase and the number of sides increases as shown below:
I think that the reason why circles have the biggest area is because they do not have any obvious vertices (corners) like a square does, therefore there is less wastage with the area because it all spreads out evenly from the one centre point where as in a square it starts to spread out but then has to go off into the corners. This is why when more sides were added, the bigger the area became, basically because there was less wastage.
