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  • Level: GCSE
  • Subject: Maths
  • Word count: 1463

Equable shapes

Extracts from this document...

Introduction

Section one

Introduction

During this project I will be investigating equable shapes. An equable shape is a shape which has an equal perimeter and area. An example of this is a rectangle of sides 6x3.

image39.pngimage00.png

Section two

Squares

First I shall investigate squares, trying to find equable squares. The way I will be doing this is by setting out a table.  The table below shows the perimeter and area of each square with a width and length of one to six.

Square

Area

Perimeter

1x1

1

4

2x2

4

8

3x3

9

12

4x4

16

16

5x5

25

20

6x6

36

24

Now I will use my table of results to produce a graph. image01.png

image40.png

Section Three

Rectangles

I started investigating rectangles by filling in different tables. The first table I did kept a fixed width of 1 (shown below is the table showing my results).

image12.png

Rectangle

Area

Perimeter

1x1

1

4

1x2

2

6

1x3

3

8

1x4

4

10

1x5

5

12

1x6

6

14

1x7

7

16

1x8

8

18

1x9

9

20

image22.png

image33.png

Rectangle

Area

Perimeter

2x1

2

6

2x2

4

8

2x3

6

10

2x4

8

12

2x5

10

14

image49.pngimage35.pngimage36.pngimage34.png

By looking at this graph we can see that the linear graphs for area and perimeter are parallel. This suggests that we can’t have an equable rectangle with a width of 2. Now I will try to prove this algebraically.

Area = 2L

Perimeter = 2L+4

      2L=2L+4

2L-2L=4

      0L=4

        L=image54.png

Clearly, you can’t have a rectangle that is infinitely long.

...read more.

Middle

L = image78.png

L = 2.1

Area = 42 x 2.1 = 88.2

Perimeter = 2(42 + 2.1) = 88.2

This shows that this rectangle is equable as the area and the perimeter are the same. This can work for any rectangle with a width greater than 2, as long as you know the width (W). Now I will make W (width) the subject.

          LW = 2L+2W

LW – 2W = 2L

   W (L-2) = 2L

W = image79.png For values of Limage80.png2

Using this formula, we can take the length (L) of any rectangle (assuming that the length is greater than two) and calculate the width needed to make the rectangle equable.

Section 4

Triangles

All Regular polygons have some traits that are the same. All the sides have the same width and all interior angles are the same. A regular triangle is otherwise known as an equatorial triangle. We also have to remember to get equality the area and perimeter has to be equal.

First I will look at a triangleimage09.png

image11.pngimage10.png

image13.png

image15.pngimage14.png

In algebraic form, the perimeter of this equilateral triangle is 3X

We do not know the area of the equilateral triangle so we will split it in half and find the formula for a right angle triangle.

image41.png

Area of one right-angle triangle = image42.png                                 (equation 1)

But sin 60 = image43.png

Rearrange this to get H = X sin 60                                         (equation 2)

Now I will substitute formula two into formula one to get:

Area = image44.png

...read more.

Conclusion

Section six

Three dimensional shapes

For a three dimensional shape to be equable we need to change the conditions for equability. The volume of the shape now needs to be the same as the surface area, instead of the area being equal to the perimeter. I started off with a cube.

I am going to let the width of a cube be A. image25.png

This would make the volume of the cube = A³

This would make the Surface area of the cube = 6A² image26.png

If then:image27.png

Volume = surface area

A³ = 6A²       (image70.pngA²)image28.png

A = 6

This shows that the cubes length, width and weight to be 6 for the cube to be equable.

Now I will find equability with a sphere.

(R = radius)image71.png

Volume of sphere = image73.pngimage74.png

Surface Area of sphere = 4image74.png

image73.pngimage74.pngR³ = 4image74.pngR²         (image70.png4image74.pngR²)

image75.pngR = 1                 (x3)

          R = 3

This shows that for a sphere to be equable the radius has to be 3.

Now I will be doing a closed cylinder. A closed cylinder is a cylinder with a base AND a lid.

The volume of a closed cylinder = image74.pngR²H

The surface area of a closed cylinder = 2image74.pngR² + 2image74.pngRHimage29.pngimage31.pngimage30.pngimage32.png

image74.pngR²H = 2image74.pngR² + 2image74.pngRH      (image70.pngimage74.pngR)

         RH = 2R + 2H                 (Make R the subject)

RH – 2R = 2H                          (Factorise)

   R(H-2) = 2H          

           R = image76.png

Himage77.png2

This shows that for a closed cylinder to be equable the height has to be greater than two.

...read more.

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