20m

480m

Then I used the formula;

Area of a rectangle = length x width

Area = 480m x 20m

Area of a rectangle = 9600 meters.

I added 10 each time to the width and worked out the length and the area and I put all my results on this chart.

I then plotted my results on a graph.

This graph represents the results that I found when I investigated the largest area of a square and a rectangle with a perimeter 1000m. The graph shows that the largest area of a square or rectangle is 62500m²,which is a square.

Therefore I have proved that the four-sided shape with the maximum area is a square, which is the only regular quadrilateral.

I have proved that the square has a greater area than any rectangle with the perimeter of 1000 meters. From these results, I believe that regular shapes have greater areas than irregular shapes.

I will now prove this theory by trying some other irregular quadrilaterals. Another quadrilateral is a Parallelogram.

Area = base x perpendicular height

450m

50m

I know that the hypotenuse is always bigger than the adjacent and the opposite sides and therefore I know that since 50 meters is the hypotenuse, the height has to be less than 50 meters. This is because the hypotenuse is always bigger than the other two sides in a right-angled triangle and therefore the height will be less than 50 meters. Since the area of a parallelogram = height x base, the height will be less than in a rectangle and therefore have a smaller area.

The same theory applies to a Trapezium, as they are made up of diagonals and the heights will be less than the sides that are made up of hypotenuses.

This is because the hypotenuse is the longest side in a triangle and this is why the height is always going to be smaller.

To work out the length of a trapezium the following formula is needed:

½(a + b) x h

a

B

## Area of a trapezium = ½ (a + b) h

Kites and Rhombuses

I do not need to investigate these shapes, both because if I take an equilateral kite, it becomes a Rhombus and a Rhombus is a square. The square has the biggest area of all the quadrilaterals. This shows that you do not have to investigate Kites and Rhombuses as they all fall under the categories of the square.

Now I will investigate the areas of Triangles.

First I will try an equilateral triangle. It has all equal sides and the same angles.

There are three sides that all have to be the same length. The three lengths have to equal 1000 metres so you therefore have to divide 1000 by 3.

1000÷3 = 333.3333333 (to 7 d.p)

333.3333333m 333.3333333m

333.3333333m

To find the area of a triangle you have to use the formula;

Area of a triangle = ½ x base x height

The base is 333.3333333metres.

To find the height, we will divide the triangle into 2.

333.3333333 (hypotenuse) (c)

(a)

166.6666667

(b)

The base is half of 333.3333333 metres which;

Base = 333.3333333 ÷2

Base = 166.6666667 metres.

Now use Pythagoras’ Theorem,

It is a squared + b squared = c squared

C is always the hypotenuse, and the hypotenuse is always the angle opposite the angle

a ² + b ² = c ²

Find a

Therefore, rearrange the formula;

a ² = c ² – b ²

a ² = 333.3333333 ²- 166.6666667 ²

a ² = 111111.1111 – 27777.77779

a ² = 83333.33331

(Square route)

a = 288.6751346 metres.

Therefore,

The height is 288.6751346 metres.

The base is 333.3333333 metres.

Area of a triangle = ½ base x height.

Area = ½ x 333.3333333 x 288.6751346

Area = 48112.52243m²

The area of an equilateral of 1000 metres is 48112.52243m²

Now I am going to investigate an Isosceles Triangle:

Base = 50m

Side = 450m

Side = 450m

## Height² (h)= a²+b²

H² = 495²- 10²

H² = 245025-100

##

## H² = 244925

H = 489.8979

A= ½ base x height

A= ½ 20 x 498.8979

A= 4898.979m²

## Area for an Isosceles triangle = 4898.979m²

I am not going to do right angled-triangles because they are scalene triangles and because they are irregular triangles.

Polygons

I am now going to investigate Pentagon: As there are 5sides in a pentagon, I am going to divide 1000 by 5, which equals 200m per side.

Pentagons can be divided into 5 equal Isosceles triangles.

Exterior angle = 360°/n or 360°/5 = 72°

Interior angle =180° – 72° =108°

Area = ½ base x height x number of triangles

200m

For this we will use trigonometry:

Tan = opposite/ adjacent

Tan 36= 100/ adjacent

Adjacent = 100/ Tan36

Adjacent = 137.64

Height = 137.638

Area = ½ 200 x 137.638

Area = 13763.8192m²

13763.8192m x 5 =

The Area of the Pentagon is 68819.096m²

I am now going to investigate the area of a hexagon:

Exterior Angle = 360°/6 =60°

Interior angle = 180° -60°= 120°

## Tan30° = 83.33/H

H = 144.337m

Area = ½ base x height

Area = ½ 166 2/3 x 144.337

Area = 12028.083

12028.083 x 6 = 72168.65m²

Area of hexagon = 72168.65m²

From these two shapes I can conclude, that as the number of sides increase, so does the area.

Polygon Formula

Each shape’s perimeter = 1000m

Therefore one side = 1000/n

Regular polygons are split up into triangles.

The formula for the small triangle inside the regular polygon

=½ base x height

Then you just multiply by the number of sides (n) :

n (½ base x height)

The final formula = n (½ x 1000/n x height)

We then had to find the height of the triangle, using tan. I divided the triangle into 2. 360/n, then became 180/n, as we divided the triangle into 2.

Tan= opposite/ Adjacent

Tan (180/n) = opposite/ adjacent

Height = ½ x 1000/n/ Tan (180/n)

A = n(( ½ x 1000/n) x ½ x1000/n/Tan (180/n))

A= n(500/n x (500/n /Tan( 180/n)))

A= n(250000/n)/(n x tan (180/n))

A= n(250000/n²xTan (180/n)

Formula for an n sided shape with a 1000 perimeter is:

A= 250000/(n x Tan (180/n)

If we put the formula into practice the rest of the polygons:

### Heptagon

Area = 250000/ (n x tan (180/n)

Area = 250000/ (7 x tan (180/7)

Area = 74161.478m²

Octagon

Area = 250000/ (n x tan (180/n)

Area = 250000/ (8 x tan (180/8)

Area = 75444.174m²

Nonagon

Area = 250000/ (n x tan (180/n)

Area = 250000/ (9 x tan (180/9)

### Area = 76318.717m²

#### Decagon

Area = 250000/ (n x tan (180/n)

Area = 250000/ (10 x tan (180/10)

### Area 76942.088m²

11- sided shape

Area = 250000/ (n x tan (180/n)

Area = 250000/ (11 x tan (180/11)

##### Area = 77401.983m²

Dodecahedron

Area = 250000/ (n x tan (180/n)

Area = 250000/ (12 x tan (180/12)

##### Area = 77751.058m²

20- sided shape

Area = 250000/ (n x tan (180/n)

Area = 250000/ (20 x tan (180/20)

##### Area = 78921.894m²

50 –sided shape

Area = 250000/ (n x tan (180/n)

Area = 250000/ (50 x tan (180/50)

##### Area = 79472.724m²

100- sided shape

Area = 250000/ (n x tan (180/n)

Area = 250000/ (100 x tan (180/100)

##### Area = 79551.290m²

150- sided shape

Area = 250000/ (n x tan (180/n)

Area = 250000/ (150 x tan (180/150)

##### Area = 79565.836m²

1000- sided shape

Area = 250000/ (n x tan (180/n)

Area = 250000/ (1000 x tan (180/1000)

##### Area = 79577.210m²

10000- sided shape

Area = 250000/ (n x tan (180/n)

Area = 250000/ (10000 x tan (180/10000)

Area = 79577.469m²

As the working out above shows, that the more sides there the larger the area is. But the increase is very small.

My final investigation is the Circle:

Perimeter = 2Πr

2Πr = 1000m

Πr = 1000 ÷ 2

Πr = 500m

r = 500 ÷ Π

r = 159.155

AREA = Πr²

= Π (159.155) ²

= Π (25330.296)

Area of a Circle = 79577.47155m ²

I have proven my hypothesis as the circle has the largest area with a perimeter of 1000m.

Radians

I am going to develop on the area of a circle as it is very similar to the area of a regular polygon which,

Area = 250000 ÷ (n x tan (180/n)

Radius = 1000 ÷ 2Π

From this I can see that I can get the formula of a regular polygon. This is how:

Place the above equation into a formula for a circle and you get

Area = Π x 500² / Π²

250000 / Π

Eventually the Polygon will turn into a circle, as a circle has an infinitve number of sides.

To develop on the idea of

250000/ n x tan (180/n) I am changing the formula into radians

Π / 180 is the method I found out for radians.

250000/nx tan (180/n) x (Π /180)

From this I can see that 180 and 180 cancel and as tan doesn’t make a difference in the equation so I am left with

25000/ n(Π/n)

## The n’s cancel out…

250000/ Π

Which shows again that the formula for a circle and polygon is very similar

I can also see that I can make an equation to use any perimeter.

If I take my original formula:

A = n(( ½ x 1000/n) x ½ x1000/n/Tan (180/n))

A = n(( ½ x p/n) x ½ x p/n/Tan (180/n))