These formulas can all be combined and simplified to give one main formula in order to find the area:
Area = ½ ( variable base * height)
= ½ (variable base * √ (c²-A²))
= ½ (2(√a²) - √ (c²- A²))
= A * (√(C² - A²)
The formula above can be used to find the area of other isosceles triangles with different bases. To make calculations faster and more efficient I have decided to do my calculations in ‘Microsoft Excel’, this way I can get more results and can also construct a graph to show my results and also observe/find out what lengths on the triangle are needed to get the maximum area for a three-sided shape.
I have now found the dimensions of the triangle which give the maximum possible area. I shall now move on and investigate ‘quadrilaterals’ and thus find the dimensions of the quadrilaterals which will give me the maximum area for a quadrilateral! If I was to have a fixed base (like in the case for triangles) then again the area would depend on the height of the quadrilateral. To find the area of a quadrilateral you need to calculate; (base * height). I am NOT going to investigate parallelograms, but only rectangles. In the case of a parallelogram if you have a fixed base then the height is always going to be smaller than that of a rectangle, hence it will give smaller area.
Now that parallelograms have been eliminated, I can just concentrate on finding the maximum area for a rectangular shape using different base sizes. The formula is easier to work out for quadrilaterals than that of triangles: AREA = (base * height).
AREA = (variable base * ((1000- (2*variable base))/2)
From the last two investigations on triangles and quadrilaterals, the maximum area has been found by using the regular shape, i.e. an equilateral, a square; where all the sides are equal. Therefore for the remainder of my investigation I realise that it would be sensible to just concentrate on the regular shape. Pentagon is the name given to a five-sided shape and the area of this is found by splitting the shape into 5 regular triangles. The area of one of these triangles is worked out by splitting it further in half, once this is worked out it can be multiplied by five and the area of the total shape can be calculated.
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I have found the maximum area for a five-sided shape, so now I am going to try another shape. I have decided to investigate on an eight-sided shape, just in case there is a pattern that follows on from pentagons. This way I can get another area of a shape with an even number of sides. The name for an eight-side shape is octagon and to find the area of a regular octagon it has to be split into equal triangles; (same principle as a regular pentagon) in this case it has to be split into eight equal triangles, and then the area is calculated by splitting it further in half, once this has been found it can be multiplied by 8 and the total area of an octagon can be found.
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Form doing two different sided-shapes I have come to the conclusion that the area continues to get higher and higher as the number of sides of the shape increase. Therefore, I have to find a formula that tells me how to find the area of any-sided shape. I have also realised that the procedure in finding the area for any regular polygon is exactly the same; where the shape is divided into equal triangles and the area of one triangle is calculated; then multiplied by the number of sides to find total area. This method can be used to find a formula to find the maximum area for a shape with any number of sides. I have decided to use ‘n’ as the number of sides the shape has.
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I have successfully figured out an algebraic formula to find out the maximum area for any sided shape! I now have to test my formula with the results I have already calculated for triangles, squares, pentagons and octagons.
I can now go further and investigate what shape would have the maximum area using 1000 metres of fencing.
My analysis shows that as the number of sides increase so does the maximum area. However, with increasing number of sides, the rate of rise in area gets smaller and smaller. As the number of sides increase the shape becomes increasingly circular. Therefore, I am going to see what the area of a circle values at.
Circumference of circle = 2πr
1000 = 2πr
500 = πr
Radius = 500/ π
Radius = 159.1549431 metres
Area of Circle = πr²
= π(159.1549431²)
Area of Circle = 79577.47155m²
It seems to be that there is an ASYMPTOTE in this problem. Looking at my graph there seems to be a number which the graph is trying to reach but will never actually get there. This asymptote is the area of a circle; this can be proven mathematically using explanation involving radians. Firstly the area of a circle can be linked to the formula I worked out to find the maximum area of any sided shape; this is because as the number of sides in a shape increase the actual structure becomes more and more like a circle.
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AREA OF CIRCLE ≈ AREA OF ANY SIDED SHAPE!
Π * 500/π² ≈ 500²/ (n (tan (180/n))
500²/π ≈ 500²/ (n (tan (180/n))
CONTINUING:
500²/π ≈ 500²/ (n (tan (π/n))
THERFORE:
500²/π ≈ 500²/ (n (π/n)
500²/π ≈ 500²/π
