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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.

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.

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).

 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

 Rectangle Area Perimeter 2x1 2 6 2x2 4 8 2x3 6 10 2x4 8 12 2x5 10 14

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=

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

Middle

L =

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 =  For values of L2

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 triangle

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.

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

But sin 60 =

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

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

Area =

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.

This would make the volume of the cube = A³

This would make the Surface area of the cube = 6A²

If then:

Volume = surface area

A³ = 6A²       (A²)

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.

Volume of sphere =

Surface Area of sphere = 4

R³ = 4R²         (4R²)

R = 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 = R²H

The surface area of a closed cylinder = 2R² + 2RH

R²H = 2R² + 2RH      (R)

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

RH – 2R = 2H                          (Factorise)

R(H-2) = 2H

R =

H2

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

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