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  • Level: GCSE
  • Subject: Maths
  • Word count: 2554

I have been asked to find out the isoperimetric quotients of plane shapes using the formula: I.Q = 4ð x area of shape

Extracts from this document...

Introduction

Maths Coursework: Isoperimetric Quotients

I have been asked to find out the isoperimetric quotients of plane shapes using the formula:

I.Q = 4π x area of shape

(Perimeter of shape)₂

I will use the answers to find a pattern between the I.Qs of plane shapes. I will start with a circle as it only has one side, and then try other shapes increasing the number of sides each time.

To help me find a pattern I will make a table of results containing both the answers with π in it, and the answer worked out to a number to 2 decimal places.

 I.Q = 4π x area of shape

        (Perimeter of shape) ²

        = 4π x πr²

            (π d) ²

        = 4π x 12.566370614359172953850573533118

             157.91367041742973790135185599802

        = 157.91367041742973790135185599792

            157.91367041742973790135185599792

I.Q = 1

 I.Q = 4π x πr²

           (Πd) ²

     = 4π x 113.09733552923255658465516179806

             1421.2230337568676411121667039822

 = 1421.2230337568676411121667039814                                                                                                                                1421.2230337568676411121667039814

I.Q = 1

After trying the I.Q formula on 2 different sized circles, the I.Qs were the same. Using algebra, I will try and prove that the I.Q is the same for all circles:

I.Q = 4π x area of shape

       (Perimeter of shape) ²

      = 4π x πχ²

          (π2χ) ²

      = 4π x πχ²

      π² x 2² x χ²

     = 4π x πχ²

       π² x 4 χ²

    = π²χ²

      π²χ²

I.Q = 1

Therefore, the I.Q of a circle is always 1.

After finding the I.Q of a 1-sided shape, I will now see if the formula has the same effect on 3-sided shapes:

image00.png

Height of triangle: sin60 = 0

                                              10

               Height = sin60 x 10

Area of triangle = ½ b x h

                          = 5 x sin60 x 10

                          = Sin60 x 5

I.Q = 4π x area of shape

        (Perimeter of shape) ²

       = 4π x sin60 x 50

                  900

       = 2π x sin60

                 9

       = 0.604599788

I.Q = 0.60 to 2.d.p

Height of triangle = sin60 x 200

Area = ½ b x h

         = 100 x sin60 x 200

...read more.

Middle

4π x Tan60

                 24

      = π x Tan60

                6

      = 0.906899682

I.Q = 0.91 to 2d.p

Using Algebra:

Now I will use algebra to see if the I.Q of a hexagon is always the same:

Height of one triangle = Tan60 x ½χ

Area of one triangle = Tan60 x 0.25χ

Area of hexagon = Tan60 x 1.5χ²

I.Q = 4π x area of shape

      (Perimeter of shape) ²

      = 4π x Tan60 x 1.5χ²

                  36χ²

= 4π x Tan60

                  24

        = π x tan60

                 6

I.Q = 0.906899682

= 0.90 to 2d.p

Therefore the I.Q of a regular hexagon is always 0.90 to 2d.p

I will now try the I.Q formula on a 7-sided shape because

                                                             Height of one triangle = Tan64.28…  x 1

                                                                 Area of one triangle = Tan64.28… x 1

                                                                     Area of Heptagon =Tan64.28… x 7

I.Q = 4π x Tan64.28… x 7

                     196

      = 4π x Tan64.28…

                   28

      = π x Tan64.28…

                   7

      = 0.931940623

I.Q = 0.93 to 2d.p

  Height of one tri = Tan64.28… x 175

                                                                       Area of one tri = Tan64.28… x 25

                                                                          Area of Heptagon = Tan64.28… x 175

                                                        I.Q = 4π x Tan64.28… x 175

                                                                            4900

                                                               = 4π x Tan64.28…

                                                                           28

                                                               = π x Tan64.28…

                                                                            7

                                                               = 0.931940623

I.Q = 0.93 to 2d.p

Using Algebra:

I will now use algebra to find out if the I.Q of a heptagon is always the same:

Height of one triangle = Tan64.28… x ½ χ

Area of one triangle = Tan64.28… x 0.25χ

Area of heptagon = Tan64.28… x 1.75χ²

I.Q = 4π x area of shape

     (Perimeter of shape) ²

      = 4π x Tan64.28… x 1.75χ²

                        49χ²

    = 4π x Tan64.28…

                 28

    = π x Tan64.28

                7

    = 0.931940623

I.Q. = 0.93 to 2d.p

Therefore, the I.Q of a regular heptagon is always 0.93to 2d.p.

I will now carry on increasing the number of sides of plane shapes and find the I.Q of octagons:

                                                      Height of one triangle = Tan67.5 x 1

...read more.

Conclusion

I can also see that the more sides a shape has, the nearer it’s I.Q is to the I.Q of a circle, which has the highest I.Q of 1.

This could be because the more sides there are to a shape; the more it will resemble a circle. So if we carried on adding sides to a shape, eventually we would get to a circle, which has an infinite number of sides.

So you can see that when an equilateral triangle is fitted into a circle, there is a lot of space where the shapes don’t meet.image01.png

But when we put a decagon inside a circle, there is not much space. Therefore, if we kept putting shapes with more sides inside circles, the space would gradually decrease until it became non-existent, and it would’ve become a circle, with an I.Q of one.image02.png

Further investigation – irregular shapes

As I have used the formula for finding the IQ of a shape only on regular shapes, I will now try it on rectangles, irregular shapes.

                                               I.Q = 4π x area of shape

                                                     (Perimeter of shape) ²

= 4π x 8

                                                           144

                                                     =

                                                         18

                                                     =0.69813170079773183076947630739545

=0.70 to 2d.p

I.Q = 4π x 12

             256

      =

     21.3333

=0.589048622548086232211745634365

=0.59 to 2d.p

                                                               I.Q = 4π x 16

                                                                            400

                                                                      =

                                                                         25

I.Q = 0.50265482457436691815402294132472

      = 0.50 to 2d.p

I have noticed that as a rectangle becomes more stretched, its IQ decreases. I have shown this by finding the IQ of rectangles length = 2 times the width, length = 3 times the width, and length = 4 times the width.

Jenna Farmer

...read more.

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