succeeds that will continue my investigation.
1)
475 475
50
h 475
25
2)
h 450
50
3)
h 425
75
4)
h 400
100
5)
h 375
125
6)
h 350
150
7)
h 325
175
8)
h 300
200
^ The graph has a smooth curve,; however, unlike the scalene triangles, the isosceles triangles results aren’t “symmetrical”; they ascend to a point, and then decline with greater magnitude. The explanation for this is because the sum of the two sloping sides must be greater than that of the base. As the base values are increasing, the sloping lengths must increase in order to comply with this standard.
Using the same formulas, but only changing the base values in scale of 10, I have produced this table of data.
Evidently, we see that the data has revealed a higher value that the previous table. As there is still a gap around the largest triangle, I will produce another table of data, this time increasing the base values by an even smaller number; 2. Again, a graph depicting the areas will be displayed after the diagrams.
1)
h 350
150
2)
h 345
155
3)
h 340
160
4)
h 335
165
5)
h 330
170
6)
h 325
175
7)
h 320
180
8)
h 315
185
^ The graph resembles an “n” to a much greater extent than before; though it still, obviously, remains “unsymmetrical”.
The highest value has changed yet again; and, to make no room for error next, I will create another table of data with the base values being increased by 0.5. A final graph will be shown after the diagrams.
1)
h 340
160
2)
h 339
161
3)
h 338
162
4)
h 337
163
5)
h 336
164
6)
h 335
165
7)
h 334
166
8)
h 333
167
9)
h 332
168
10)
h 331
169
11)
h 330
170
12)
h 329
171
13)
h 328
172
14)
h 327
173
^ We can see that the shape of the graph is very similar to an “n” shape now. We know that it cannot be symmetrical – you can notice that there is no exact middle point in the graph. I have showed this on the graph by putting a line through the highest point (highest value).
Once again, the highest area has changed.
1)
h 335
165
2)
h 334.75
165.25
3)
h 334.5
165.5
4)
h 334.25
165.75
5)
h 334
166
6)
h 333.75
166.25
7)
h 333.5
166.5
8)
h 333.25
166.75
9)
h 333
167
10)
h 332.75
167.25
11)
h 332.5
167.5
12)
h 332.25
167.75
13)
h 332
168
14)
h 331.75
168.25
^ Once more, we have discovered a higher area between the gaps. Judging by the measurements, I can predict that the largest area would be that of an equilateral triangle.
This proves that my prediction is correct.
Triangles - Equilateral
Bearing in mind that the perimeter = 1000m, I have divided it by 3 to give us the one equilateral triangle.
Area = ½(b x h)
Because 1000 ÷ 3 = 333.33 (2d.p.)
333.33m 333.33m
333.33m
333.33m
166.67m
Since there is only one instance of the equilateral triangle, I surmise that there is no need for a graph to plot one point on. We can clearly see that the only area that an equilateral triangle will produce is 48112.63m²; this is the maximum area any triangle can hold while observing the 1000m perimeter.
Nevertheless, I can prove logically that neither triangle will hold the maximum potential area while abiding by the 1000m perimeter. This is because the shape of the triangle (regardless of whether it is isosceles, equilateral or even scalene) hinders it, as shown in various examples below. I indicate the “wasted space” which, if filled, not only gives us a bigger potential area, but tells us that quadrilaterals (specifically rectangles) have the ability to hold a higher maximum potential area (with the perimeter of 1000m).
Isosceles
Equilateral
“Wasted Space”
Scalene
Quadrilaterals – Parallelograms
Continuing on with my investigation, I will now examine the extent of which parallelograms will go to hold its maximum area, again, whilst abiding by the decided perimeter of 1000m. The following table I created in Excel show the areas of parallelograms, with each interior angle increased by 10.
Area = b x perpendicular height
Side B is set at 400, and the base (which has to comply in order to fulfil the 1000m perimeter.
a
b
a
Here is the table of data in formula view.
Though the formula shown earlier is the conventional formula for finding the area of a parallelogram, in this case we can alter it to our convenience. Due to the fact that to the find the height the operation involved is; Sin θ x 400 (because the sloping value is fixed to 400); and to find the area is to multiply the height by 100 each time (because the base value is fixed to 100), I deduce that the area formula can also be seen as;
Area = 400 x 100 x Sin θ
= 40000 x Sin θ
Below is the table as on the previous page, but in normal view.
The highest area in this table consists of a parallelogram with a base of 100, sloping height of 400, and interior angle of 90°. With this information I can deduce that the parallelogram with the largest area in the table is in the form of a rectangle. I have also realised that the values after the highest area are identical to that of the values before the highest area (but in descending order). This leads me to predict that the graph (which will be shown after the following diagrams) will have a perfect “n” shape, a smooth curve, and be comprised of “symmetrical” data. Following all of this, I will construct another table with the angle going up in a scale of 2; this is to see if a higher value rests within the gap between the highest area and the two areas around it.
1)
400
100
2)
400
100
3)
400
100
4)
400
100
5)
400
100
6)
400
100
7)
400
100
8)
400
100
9)
400
100
10)
400
100
=
^ We can see that the “symmetrical” data is very clearly being portrayed on the graph, and I have indicated the symmetry with a line through the highest point on the graph.
The highest area has not varied; yet to see that there is no room for error, I will include a final part in this stage of my investigation with another table of data. The angle will be increased in a scale of 0.5 this time, and from there I will finalise and evaluate my result. Firstly, here are the diagrams depicting what is shown in the table above. Naturally, a graph delineating the trend of my findings above will be included after the diagrams.
1)
400
100
2)
400
100
3)
400
100
4)
400
100
5)
400
100
6)
400
100
7)
400
100
8)
400
100
9)
400
100
10)
400
100
11)
400
100
^ The graph once again portrays the picture of the “symmetrical” data, and once again I have signified this with a line of symmetry through the highest point (highest value) on the graph.
Again, I have noticed the highest area has not varied. After the following diagrams, I will again display a graph corresponding to the areas in the table above; and I will conclude this stage of my investigation.
1)
400
100
2)
400
100
3)
400
100
4)
400
100
5)
400
100
6)
400
100
7)
400
100
8)
400
100
9)
400
100
^ I can now safely conclude that 40000m² is the highest area possible for a parallelogram to hold whilst keeping to the set perimeter of 1000m. I know that when I come across the largest area in each batch of parallelograms I examine, it always comes in the form of a rectangle.
Leading from this new discovery, I can now reflect back on my preceding stages; I have determined that the highest area for each of the shapes I have looked at so far all resemble a regular polygon. Consequently, this prompts me to explore regular polygons, as they evidently hold the highest areas.
Below are some diagrams which elucidate why parallelograms are not ideal for achieving the maximum potential area.
“Wasted Space”
As I have shown before in the triangles stage, the parallelogram (and also trapezium, as shown) does not reach its maximum potential area; as indicated, it can be “expanded” to form rectangles and attain its maximum capacity. The awkward shape of the parallelogram is the reason why its maximum area is less than that of the triangle’s.
As a result, I will now analyse rectangles in the next stage of my investigation.
Quadrilaterals - Rectangles
I will proceed to show the maximum possible area for a rectangle with perimeter = 1000m; again, presenting my results in a graph.
With the information I have gathered in the previous stages, I predict that the rectangle will hold the greatest area out of all the shapes examined so far.
I am varying the length/width of the rectangles in units of 50 accordingly.
Below is the table of data presenting my findings.
Already we can see that
Area = l x w
450
1) 50 x 450
= 22500 50
400
2) 100 x 400
= 40000 100
350
3) 150 x 350
= 52500 150
300
4) 200 x 300
= 60000 200
250
5) 250 x 250
= 62500
250
After the square, the 6th shape’s area would be equivalent to that of the 4th shape. This is because as the width is increased by 50, the length must decrease by 50 to comply with the perimeter of 1000; we can see that the area = 300 x 200 = identical to the 4th shape, but the length and width have interchanged. On the next page, you will see a graph depicting my findings.
^This, clearly is the only batch of rectangles I can study; I have depicted the “symmetrical” data on the graph using a line of symmetry. To make sure that 62500 is really the highest value, I will test the values closest to each length and evaluate the results.
i.e. 249.999 x 250.001 = 62499.99999…
This confirms that 62500 is the highest value.
Regular Polygons – Pentagon
I will now investigate regular polygons (as I have mentioned earlier). I will begin this stage with the pentagon. As shown, I have divided the polygon into triangles, and found the area of one of the triangles; using trigonometry (tan) to find the height, and then substituting the values into the conventional triangle formula (A = ½bh). I then multiply the area of the triangle by n (the number of sides the polygon has) and the resulting value is the area.
Area = {½[(1000 ÷ 5) x h]} x 5
1000 ÷ 5 = 200
200 200 360 ÷ 5 = 72°
200 200
200
h
200 100
Regular Polygons – Hexagon
As shown, I have divided the polygon into triangles, and found the area of one of the triangles; using trigonometry (tan) to find the height, and then substituting the values into the conventional triangle formula (A = ½bh). I then multiply the area of the triangle by n (the number of sides the polygon has) and the resulting value is the area.
Area = {½[(1000 ÷ 6) x h]} x 6
166.67 1000 ÷ 6 = 166.67
360 ÷ 6 = 60°
166.67 166.67
166.67
166.67
166.67
h
166.6666667 83.33333333
Regular Polygons – Octagon
As shown, I have divided the polygon into triangles, and found the area of one of the triangles; using trigonometry (tan) to find the height, and then substituting the values into the conventional triangle formula (A = ½bh). I then multiply the area of the triangle by n (the number of sides the polygon has) and the resulting value is the area.
Area = {½[(1000 ÷ 8) x h]} x 8
125
125 125
125 125
125 125
125
h
125 62.5
Regular Polygons – Decagon
As shown, I have divided the polygon into triangles, and found the area of one of the triangles; using trigonometry (tan) to find the height, and then substituting the values into the conventional triangle formula (A = ½bh). I then multiply the area of the triangle by n (the number of sides the polygon has) and the resulting value is the area.
Area = {½[(1000 ÷ 10) x h]} x 10
100
100 100
100 100
100 100
100 100
100
h
100 50
Regular Polygons – Pentadecagon (15-sided-shape)
As shown, I have divided the polygon into triangles, and found the area of one of the triangles; using trigonometry (tan) to find the height, and then substituting the values into the conventional triangle formula (A = ½bh). I then multiply the area of the triangle by n (the number of sides the polygon has) and the resulting value is the area.
Area = {½[(1000 ÷ 15) x h]} x 15
66.67 66.67
66.67
66.67
66.67
66.67
66.67
66.67
66.67
66.67
66.67
66.67
66.67
66.67 66.67
h
66.66666667 33.33333334
Regular Polygons – Icosagon (20-sided-shape)
As shown, I have divided the polygon into triangles, and found the area of one of the triangles; using trigonometry (tan) to find the height, and then substituting the values into the conventional triangle formula (A = ½bh). I then multiply the area of the triangle by n (the number of sides the polygon has) and the resulting value is the area.
Area = {½[(1000 ÷ 100) x h]} x 100
50 50
50 50
50 50
50 50
50 50
50 50
50 50
50 50
50 50
50 50
h
50 25
Regular Polygons – Hectogon (100-sided-shape)
As you can see, I have not incorporated a diagram of a Hectogon because of its close resemblance to a circle (at this scale factor). However, the calculations of the base and height of the triangles are shown, and of course the calculations.
Area = {½[(1000 ÷ 100) x h]} x 100
Regular Polygons – Chiliagon (1000-sided-shape)
As you can see, I have not incorporated a diagram of a Hectogon because of its close resemblance to a circle (at this scale factor). However, the calculations of the base and height of the triangles are shown, and of course the calculations.
Area = {½[(1000 ÷ 1000) x h]} x 1000
Regular Polygons – Myriagon (10,000-sided-shape)
As you can see, I have not incorporated a diagram of a Hectogon because of its close resemblance to a circle (at this scale factor). However, the calculations of the base and height of the triangles are shown, and of course the calculations.
Area = {½[(1000 ÷ 10000) x h]} x 10000
I have explored various regular polygons so far, and we can see now that as the number of sides is increased, the area is larger each time. I can now assume that a circle will finally give us the largest area possible, while keeping to the 1000m perimeter (or circumference, in this case).
Circle
I will now aim to prove my prediction; I will use the conventional area formula (A = πr²). To find out the radius, I will rearrange the formula for finding out the circumference (since we have a known variable already, C = 1000).
Finally, we can see that my prediction is indeed correct, and the circle holds the most area whilst abiding by the set rule of the perimeter being 1000m. On the next page I will show a graph depicting the highest area of each of the shapes I have investigated. This will make it clearer for us to see that the circle is the highest, and also help us see that as the number of sides increase, the area becomes larger.
General Formula for N-Gon (Polygon with n number of sides)
In the previous stages, finding the areas for polygons with more than 4 sides was straightforward; the same formula was used, and n (number of sides) was changed accordingly for each one. Triangles, however, could either use the conventional formula, ½bh (however, this only works for any triangle but scalene), or use Hero’s Formula which is applicable to any triangle.
The key similarity in working out the areas of polygons with more than 4 sides is splitting the shape into equal triangles. Once I find the area of one of the triangles, I only have to multiply the area by the number of triangles (determined by the number of sides of the shape i.e. n) to obtain the area of the whole polygon.
To find a general formula for finding the area for a regular polygon with any number of sides, I will proceed to carry out a step-by-step explanation with reasons for my calculations and methods.
1)
To find out each side (n = no. of sides)
= 1000 ÷ n
also = the base of each triangle
A B
1000 ÷ n
2)
To find out length AC;
AC = AB ÷ 2 = 0.5 x (1000 ÷ n) = 500 ÷ n
3)
To find out the interior angle AÔC
AÔB = 360 ÷ n AÔC = 0.5 x (360 ÷ n) A B
= 180 ÷ n 1000 ÷ n
2
4)
To acquire the perpendicular height;
O
Tan AÔC = AC ÷ h Tan θ A
∴ h Tan AÔC = AC
∴ h = AC ÷ Tan AÔC
= 500 ÷ n
Tan (180 ÷ n)
= 500
n Tan (180 ÷ n)
5)
Step 4) must be multiplied by 1000 ÷ n (base of triangle), as the area formula for a triangle is ½bh.
[ Continued overleaf ]
Continued…
= ½ 1000 x 500
n n Tan (180 ÷ n)
= 500
n² Tan(180 ÷ n)
6)
Now that I have found the general formula for one of the triangles in the polygon, all I must do is multiply the area of that triangle by the total number of triangles/sides the polygon has.
= n x 500²
n² Tan (180 ÷ n)
Final simplification;
= 250,000
n Tan (180 ÷ n)
I will now proceed to test (authenticate) the formula by testing it on a hexagon (n = 6) and a 1000-sided-shape (n = 1000).
1) Hexagon:
= 250,000
6 x Tan (180 ÷ 6)
= 250,000 = 72168.78 (2d.p.)
6 x Tan 30
2) 1000-sided-shape:
= 250,000
1000 x Tan (180 ÷ 1000)
= 250,000 = 79577.21 (2d.p.)
1000 x Tan 0.18
Now that I have verified that the formula works I am able to use it to find the formula of any regular polygon by substituting n with the number of sides the polygon has.
In Conclusion;
To conclude, I will summarise what I discovered during this investigation.
I found that as n (number of sides) increases, so does the area; this, in turn, led me to investigate the circle which, as I predicted, held the largest area whilst keeping to the set rule of the perimeter/circumference being 1000m.
I have also found out and showed in stages a general formula in which the area of any regular polygon can be found, just by substituting n for the number of sides the polygon has.
I have also shown a graph which shows clearly the maximum areas for each polygon, and this also reinforces the idea that as the number of sides increase, so does the maximum potential area.
