The diagram shows that for the same perimeter, and the same base, the isosceles has the longest height. Also the isosceles has two equal sides of the same length therefore once I know the base of the triangle; I can work out the length of the sides.
To work out the area of an isosceles triangle, I will use the following formula:
Area = ½ Base x Height
So before I can work out the area, I need to work out the height of the isosceles triangle. To find out the height I must half the triangle to get a right-angled triangle from the isosceles and then I will use Pythagoras’s theorem to work out the height.
Normal Isosceles Triangle:
When I cut the triangle in half, I get a right-angled triangle like this:
I can now use Pythagoras’s theorem to work out the height. Once I know the height, I can then work out the area of the isosceles triangle.
Pythagoras’s Theorem:
Pythagoras’s theorem suggests that a² (height) + b² (base) = c² (side). If we rearrange the equation, we can work out the height if we know the base and the side. The equation will now look like a² = c² – b² (in an algebra formula). This could also be written as height2 = side² – base². So to work out the height, we simple need to square root the equation. The final equation looks like this,
Height = √Side² – (Base/2)². Notice we have to divide the base in two because when I halved the isosceles, the base is halved. The side and height remain the same and therefore do not require to be halved.
Triangle 1:
Above is my first isosceles triangle which I am going to investigate. The length of the side is 425m and base is 150m. But to work out the area of the triangle, I need to know the height. I am going to do this by cutting the triangle in two halves to get a right-angled triangle.
Note, the base is 75m because the isosceles has been halved so the base of the isosceles is 150m while the base of the right-angled triangle is 75m. I can now use Pythagoras’s theorem to work out the height of the triangle (a).
a² = c² – b²
= 4252 – 752
= 180625 – 5625
= 175000
a = √175000
= 418.3m
So the height of the isosceles triangle is 418.3m because the height of the isosceles is the same as the height of the right-angled triangle. Now using the formula, Area = ½ base x height, I can now work out the area of the isosceles triangle.
Area = ½ base x height
= (½ x 150) x 418.3
= 75 x 418.3
= 31374.8m²
Triangle 2:
Cut it in half to get a right-angled triangle so I can apply the Pythagoras’s theorem:
a² = 375² - 125²
= 140625 – 15625
= 125000
a = √125000
= 353.6m
So the height of the isosceles is 353.6m. I can now work out its area:
Area = ½ base x height
= (½ x 250) x 353.6
= 125 x 353.6
= 44194.2m²
Triangle 3:
Cut it in half to get a right-angled triangle so I can apply the Pythagoras’s theorem:
a² = 325² - 175²
= 105625 – 30625
= 75000
a = √75000
= 273.9m
So the height of the isosceles is 273.9m. I can now work out its area:
Area = ½ base x height
= (½ x 350) x 273.9
= 175 x 273.9
= 47925.7m²
Spreadsheet:
I am now going to obtain more results using a spreadsheet-software and draw a graph out of it.
The first spreadsheet printout shows that the base with the biggest area is between 330m and 340m. I have therefore done a second spreadsheet which shows that the base with the longest area is between 333m and 334m. I believe a regular (or equilateral) triangle, with base 333.3m (recurring), will give the biggest area out of all triangles because a regular rectangle gives the biggest area out of all rectangles. To find out if this prediction is correct, I am now going to draw an equilateral triangle and work out its area below:
Equilateral triangle:
Cut it in half to get a right-angled triangle so I can apply the Pythagoras’s theorem:
a² = 333.3² - 166.6²
= 111088.89 – 27755.56
= 83333.3
a = √83333.3
= 288.7m
So the height of the isosceles is 288.7m. I can now work out its area:
Area = ½ base x height
= (½ x 333.3) x 288.7
= 166.6 x 288.7
= 48112.5m²
So a regular triangle has a bigger area than isosceles triangles with bases 333m and 334m long. To find out if it has a bigger area than isosceles triangles with bases 333.32m and 333.34m, I have extracted some rows from a spreadsheet software:
Conclusion on triangles:
The table proves and verifies that a regular triangle has the biggest area out of all isosceles triangles and therefore proves my prediction which I made because regular rectangles have the biggest area out of all rectangles so regular triangles should have the biggest area as well. This is why I have done a third and final printout of a spreadsheet which proves that a regular triangle has the biggest area out of all triangles with an area of 48112.52m².
The graph shows that the area of an isosceles triangle increases as the length of the base increases; however only up to the base length of a regular triangle with equal sides. At this point, this is the biggest triangle with a perimeter of 1000m and area of 48112.52m². After the base of a regular triangle, is reached the area of the triangle decreases as the length of the base increases.
Summary so far:
I have so far found out that the area of a shape increases as the number of sides of the shape increases. This is because the rectangle with the biggest area had a bigger area than the triangle with the biggest area. So for this reason I am now going to investigate polygons that have more sides. I am going to keep the sides of the next shapes, I am going to investigate, the same because I have found out from my investigation on regular rectangles and regular triangles that shapes with equal sides have the biggest area. So now all of my next shapes will have equal sides.
Polygons:
I am now investigating shapes with more than 4, and equal, sides.
Pentagons:
Pentagons have 5 sides. I am now going to investigate a pentagon with 5 equal sides. Each side should have a base of 200m. The pentagon will have a perimeter of 1000m.
To find out the area of this pentagon, I will need to break the pentagon up into 5 triangles of the same area. This is how the triangles with the pentagon will look like:
All the triangles within the pentagon are isosceles triangles which means I can work out the area of each triangle using the formula area = ½ base x height as I did before. After I have worked out the area of one triangle, I can then work out the area of the pentagon by multiplying the area of one triangle by 5. This is because all triangles have the same area within a regular pentagon.
However, to work out the area of a triangle, I will need to work out its height. I was able to use Pythagoras’s theorem to work out the height in isosceles triangles however I cannot use Pythagoras’s theorem in this case. This is because I don’t know the hypotenuse of the triangle. For this matter, the solution is to use trigonometry. If I am able to work out one side and one angle within a right-angled triangle, I will be able to use trigonometry to find out the height. I know that the base of each side of the pentagon is 200m long. Therefore, the base of each regular sided triangle is 200m long and the base of each right-angled triangle is 100m long. So I need to know the size of angle adjacent to the height of the triangle. I can work this out because the dot at the centre of the pentagon equals 360°. So the size of each angle adjacent to the height of the triangle equals 360 ÷ 10 = 36°. The following diagram helps explain this
So to work out the area of each isosceles triangle, I need to work out the height of it first. I have the angle, 36°, the opposite value, 100m, and I need to work out the height on the adjacent to the angle. So I can now apply the tan rule which is adjacent = opposite/tan 36°. If I substitute the values from the pentagon into the formula I get, h = 100/tan 36°. So height is equal to 137.64m.
I can now work out the area of the isosceles using the formula ½ base x height:
Area = ½ base x height
= (½ x 200) x 137.64m
= 100 x 137.64
= 13763.8m²
So the area of one isosceles triangle within the pentagon is 13763.8m². The area of the pentagon is 5 times the value of one isosceles triangle. Therefore the value of the pentagon is 13763.8 x 5 = 68819.1m². This gives a bigger area than the biggest area possible with a rectangle or triangle and so proves my prediction of the area increasing as the number of sides increase.
Hexagons:
I am now going to investigate hexagons, 6 sided shapes. Again I am going to keep all sides equal to achieve the biggest area possible with a perimeter of 1000m. A 6 sided equilateral hexagon shape looks like this:
To work out the area of this hexagon, I will need to carry out the same steps I used in pentagons. This is because I can create isosceles triangles within the hexagon just like in the pentagon. The only difference this time is that I will need to multiply the area of one isosceles triangle by 6 instead of 5 because there are 6 sides in this shape and so I can create up to 6 triangles within the shape. So the hexagon will now look like this:
Note the size of each side is 166.6m because 1000/6 = 166.6. Therefore the base of each right-angled triangle is 166.6/2 = 83.3m. The angle of each right-angled triangle is 360/6 = 60/2 = 30. So I can now apply the tan rule to work out height:
Adjacent = Opposite/Tan 30
Height = 83.3/0.58
= 144.34m
So height equals 144.34m. Therefore area of isosceles is:
Area = ½ base x height
= (1/2 x 166.6) x 144.34
= 83.3 x 144.34
= 12028.13m²
So the area of the hexagon is:
12028.13 x 6 = 72168.8m²
So again the area has increased as the number of sides increased.
Octagon:
I am now going to investigate 8 sided shapes, Octagons. Again the sides will be equal and therefore I should receive the biggest area possible in octagons. The method to work out the area is the same as the method to work out pentagons and hexagons, use trigonometry to work out height then do ½ base x height.
Each side equals 125m (1000/8) and each base of right-angled triangle equals 62.5m (½ x 125) and each angle within the right angled triangle is equal to 22.5° (360/16). So I can now work out the height of the right-angled triangle:
Adjacent = Opposite/Tan 22.5
Height = 62.5/0.41
= 150.89m
So area equals:
Area = ½ base x height
= (1/2 x 125) x 150.89
= 62.5 x 150.89
= 9430.52.m²
Area of Octagon = 9430.52 x 8 = 75444.174m²
Decagon:
The final polygon I am going to investigate is the Decagon, a 10 sided shape. The sides’ lengths are going to be equal yet again and I believe this final polygon should give me the biggest area out of all the shapes I have done so far. The perimeter will also be the same, 1000m.
The above regular decagon has 10 sides. Each side is 100m long. The base of each right-angled triangle found when the isosceles is cut is 50m long. The angle within the right-angled triangle is equal to 18°. Using trigonometry, I can now work out the height of the triangles.
Adjacent = Opposite/Tan 18
Height = 50/0.32
= 153.88
So area equals:
Area = ½ base x height
= (1/2 x 100) x 153.88
= 50 x 153.88
= 7694.21m²
Area of Decagon = 7694.21 x 8 = 76942.1m²
The area of the Decagon is yet again the largest of all shapes investigated so far. I have now proven that the more the number of sides in a regular shaper, the bigger its area. I am now going to find out how big the area can be in any regular shape. To make it easier and quicker, I am going to put this in spreadsheet software and produce a chart to present the data better.
