Because the last two shapes have had the largest areas when they are regular, I am going to use regular shapes from now on.
The next shape that I am going to investigate is the pentagon.
Because there area 5 sides, I can divide it up into 5 segments. Each segment is an isosceles triangle with the top angle being 72º. This is because it is a fifth of 360º. This means that I can work out both the other angles by subtracting 72 from 180 and dividing the answer by 2. This gives 54º each. Because every isosceles triangle can be split into 2 right-angled triangles, I can work out the area of the triangle, using trigonometry. I also know that each side is 200m long, so the base of the triangle will be 100m. GnJr27sHS from GnJr27sHS student GnJr27sHS central GnJr27sHS co GnJr27sHS uk
Using SOH CAH TOA I can work out that I need to use Tangent.
H = 100 tan54 = 137.638
O = 100
T = tan 36
This gives me the length of H so I can work out the area.
Area = ½ × b × H = ½ × 100 × 137.638 = 6881.910
I now have the area of half of one of the segments, so I simply multiply that number by 10 and get the area of the shape.
Area = 6881.910 × 10 = 68819.096m²
All of the results that I have got so far have shown that as the number of sides increase, so to does the area. Using a spreadsheet and formula I have created a table that shows my prediction is right. This is show on the next page.
The formulae for the spreadsheet are:
To work out the base of a polygon you divide the perimeter of the polygon by the number of side (n)
To put this equation in to a spreadsheet, you must type the following:
=(1000/A3)
To work out the height of the triangle on a polygon, the equation is:
To put this equation in to a spreadsheet, you must type the following:
=(500/A3)/TAN(3.14/A3)
The equation to work out the area of the triangle is:
wwcd cdw stcdcdud ecd cdnt ccd encdtral cdcocd uk.
To put this equation in to a spreadsheet, you must type the following:
=(B3*C3)/2
To work out the area of the polygon, the equation is:
To put this equation in to a spreadsheet, you must type the following:
=(A3*D3)
From the method that I used to find the area for the pentagon I can work out a formula using N as the number of sides. To find the length of the base segment I would divide 1000 by the number of sides. Also on the next page is a graph showing the number of sides against area.
As you can see from the graph, the line straightens out as the number of side’s increases. Because I am increasing the sides by large amounts and they are not changing I am going to see what the result is for a circle. Circles have an infinite number of sides, so I cannot find the area using the equation for the other shapes. I can find out the area by using π. To work out the circumference of the cir le the equation is πd. I can rearrange this so that diameter equals circumference/π. From that I can work out the area using the πr² equation.
DIAMETER = 1000 / π = 318.310
RADIUS = 318.310 / 2 = 159.155
AREA = π × 159.155² = 79577.472m²
From this I have concluded that a circle has the largest area when using a similar circumference. This means that the farmer should use a circle for her plot of land so that she can gain the maximum area.
Aim:
A farmer has brought 1000 metres of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the biggest size. I will investigate different shapes the fencing can make to achieve the largest area.
I am going to start investigating different shape rectangles because they are the easiest shapes to work put the perimeter all of these shapes will have a perimeter of 1000 metres. Below are 2 rectangles (to scale) showing how different shapes with the same perimeter can have different areas. I will use a scale of 1cm:100m.
1) 2)
400 metres (4 centimetres)
Height 300m(3cm)
200m (2cm)
100 metres (1 centimetre)
Width
I will work out the area of both rectangles by using the formula below. Both rectangles have a perimeter of 1000m
1)
Area of rectangle = height * width
Area of rectangle = 400m * 100m
Area of rectangle = 40000 m²
2)
Area of rectangle = height * width
Area of rectangle = 300m * 200m
Area of rectangle = 60000m²
As you can notice the areas of the rectangles 1 and 2 are different though the perimeters of both are 1000m.
Now I will put the areas widths and lengths of rectangles. I will change the value of the widths and go up in increments of 10m.
I will not use negative numbers for they are realistically impossible. Mathematically negative lengths are possible but I know that investigating this wont give me the answers I want.
I will put my results in a table now.
The highlighted row gives the biggest area; after I go past this row I start to repeat my self.
A square of perimeter 1000m gives me the largest area. I will further investigate this because I was going up in increments of 10. I will go into the decimal widths and lengths.
Below is a table of results.
From this table I can see that the perfect square of perimeter 1000m produces the largest area.
From this graph I can see that the highest point is on the 250m mark. The graph is a parabola and is symmetrical. At the centre point of the graph I reach the highest value for the area, after this I just repeat my self.
In a rectangle, any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m * 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side. To work out the area of a rectangle with a base length of 200m, I subtract 200 from 500, giving 300 and then times 200 by 300. I can put this into an equation form.
Area = x(500 – x)
Key: x = width of rectangle
All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas. wwbc bcw stbcbcud ebc bcnt cbc enbctral bccobc uk;
Now I will investigate triangles. I will first investigate isosceles triangles because triangles such as scalene have more than one different variable so there are millions of possible combinations. If I know the base length I can work out the length of the other two sides because they are the same.
For example if the base is 200m. I can take this away from 1000m and that answer I can divide by 2. Or more simply:
Side = (1000 – 200) ÷ 2 = 400
To work out the area I need to know the height of the triangle. To work out the height I can use Pythagoras’ theorem. Below is the formula and area when using a base of 200m.
H2 = h2 – a2
H2 = 4002 – 1002
H2 = 150000
H = 387.298
½ * 200 * 387.298 = 38729.833m.
On the next page there is a table of results for triangles.
From the table I can see that the regular shape of the family has given me the largest area.
I can see that the regular shapes of these families give the largest area.
Because the last 2 shapes have had the largest areas when they are regular, I am going to use regular shapes from now on. This would also be a lot easier as many of the other shapes have millions of different variables.
The next shape that I am going to investigate is the pentagon.
Because there are 5 sides, I can divide it up into 5 segments. Each segment is an isosceles triangle, with the top angle being 720. This is because it is a fifth of 360. This means I can work out both the other angles by subtracting 72 from 180 and dividing the answer by 2. This gives 540 each. Because every isosceles triangle can be split into 2 equal right-angled triangles, I can work out the area of the triangle, using trigonometry. I also know that each side is 200m long, so the base of the triangle is 100m.
Tanq = opposite/adjacent
Tan54 = o / a
O = tan54 * a
O = tan54 *100
O = 137.638192m
137.638192m * 200 =27527.63841 s97VJf4i from s97VJf4i student s97VJf4i central s97VJf4i co s97VJf4i uk
27527.63841/2=13763.81921
13763.81921*5=68819.096m²
All of the results that I have got so far have shown that as the number of side’s increases, the area increases. I am going to investigate this further with a regular hexagon (6 sides) and a regular heptagon (7 sides).
I am going to work out the area of the 2 shapes using the same method as before.
Hexagon:
1000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3.
360 ÷ 6 = 60 ÷ 2 = 30
Area = ½ * b * H = ½ * 83 1/3 * 144.338 = 6014.065
6014.065 * 12 = 72168.784m2
Heptagon:
1000 ÷ 7 = 142.857 ÷ 2 = 71.429m
360 ÷ 7 = 51.429 ÷ 2 = 25.714 degrees
Area = ½ * b * H = ½ * 71.429 * 148.323 = 5297.260
5297.260 * 14 = 74161.644m2
My predictions were correct and as the number of side’s increases, the area increases. Below is a table of the number of sides against area
To find the general formula for an n sided regular polygon I am going to construct a formula.
To find the length of the base of a segment I would divide 1000 by the number of sides, so I could use 1000/s (s=number of sides), but because I want to find half of one segment I use (1000/s)/2. To find the interior angle of the n sided polygon I can use, (360/s)/2. I am finding half of the interior angle of a segment. Then I have to find the length of the perpendicular in the segment. To do this I use the tangent formula. Adjacent = opposite/ Tanq
A = [(1000/n)/2]/ Tan (360/s)/2
Then to find the area of the regular polygon I have to multiply the value of the line A by half of the base (1000/s)/2 then I have to multiply that by the number of sided the regular polygon has. So basically it’s A * base *0.5 * number of sides.
As the table above shows as the number of sides go up the area goes up. Now I am going to investigate the area of a perfect circle. I predict that this will ultimately have the largest area because it has an infinite amount of sides.
pR²
3.14 * 159.235 (2 d.p) * 159.235 (2 d.p) = 79617.16561m²
I can see that this produces the largest area.
