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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 = 4&amp;eth; x area of shape

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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:

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

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

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.

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.

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

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