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  • Level: GCSE
  • Subject: Maths
  • Word count: 2221

Find the biggest value of the ratio area / perimeter for some triangles.

Extracts from this document...

Introduction

image00.png

The scenario for my coursework is as follows:

Two students are discussing a problem that was set by their teacher to find the biggest value of the ratio area / perimeter for some triangles.

One student indicates that this was done with measurements of 40, 60 and 80 which where given by their teacher, but forgets to say which units were used and whether they were angles or sides.

I am going to try and solve the above problem and find the biggest value of the ratio area / perimeter. I am going to assume that all lengths are measured in centimetres and angles are measured in degrees.


image01.png

I am going to start by thinking of all the possible triangles that can be drawn.

Below is a triangle to show how I will address the angles and sides of the triangles I am going to draw.

                                             Bimage02.png

                                     c                        a

                             A                    b                   C

Below is a table to show the possible triangles using the numbers 40, 60 and 80:

image06.png

I will now try to draw the highlighted 6 of the above triangles. I will bear in mind that some will be the same as each other and some will be impossible to draw due to lack of information or simply impossibility to draw:



The triangles that are impossible to draw are; l, m, o, p, v, w, x, y, z, ab and ac. I have only managed to find one triangle that is that same as another.

...read more.

Middle

        = 733.76

Ratio        = a/p

        = 733.76 ÷ 178.57

        = 4.11

Triangle i:

Perimeter         = 40 ÷ Sin60

        = 46.19

        = Cos60 x 46.19

        = 23.1

        = 23.1+46.19+40

        = 109.29

Area   =  s        = p ÷ 2

        = 109.29 ÷ 2

        = 54.65

        = square root of s(s-a)(s-b)(s-c)

= square root of 54.65x(54.65-40)x(54.65-23.1)x(54.65-46.19)

        = square root of 54.65x14.65x31.55x8.46

        = square root of 213696.55

        = 462.27

Ratio        = a/p

        = 462.67 ÷ 109.29

        = 4.23

Triangle j:

Perimeter         = 40 ÷ Cos60

        = 80

        = 80+80+40

        = 200

Area   =  s        = p ÷ 2

        = 200 ÷ 2

        = 100

        = square root of s(s-a)(s-b)(s-c)

        = square root of 100x(100-80)x(100-40)x(100-80)

        = square root of 100x20x60x20

        = square root of 2400000

        = 1549.19

Ratio        = a/p

        = 1549.19 ÷ 200

        = 7.75

Triangle k:

Perimeter         = Sin60x40

        = 34.64

        = 34.64+80+40

        = 154.64

Area   =  s        = p ÷ 2

        = 154.64 ÷ 2

        = 77.32

        = square root of s(s-a)(s-b)(s-c)

        = square root of 77.32x(77.32-34.64)x(77.32-80)x(77.32-40)

        = square root of 77.32x42.68x-2.68x37.32

        = square root of –330059.84

        = impossible to square root a negative

Ratio        = a/p

Triangle l:        

                !Impossible!

Triangle m:

                !Impossible!

Triangle n:

Perimeter         = Sin80x60

        = 59.09

        = 59.09+40+60

        = 159.09

Area   =  s        = p ÷ 2

        = 159.09 ÷ 2

        = 79.55

        = square root of s(s-a)(s-b)(s-c)

        = square root of 79.55x(79.55-59.09)x(79.55-40)x(79.55-60)

        = square root of 79.55x20.46x39.55x19.55

        = square root of 1258458.98

        = 1121.81

Ratio        = a/p

        = 1121.81 ÷ 159.09

        = 7.05

Triangle o:

                !Impossible!

Triangle p:

                !Impossible!

Triangle q:

Perimeter         = Sin60x80

        = 69.28

        = 69.28+40+80

        = 189.28

Area   =  s        = p ÷ 2

        = 189.28 ÷ 2

        = 94.64

        = square root of s(s-a)(s-b)(s-c)

        = square root of 94.64x(94.64-69.28)x(94.64-40)x(94.64-80)

        = square root of 94.64x25.36x54.64x14.64

        = square root of 1919887.36

        = 1385.6

Ratio        = a/p

        = 1385.6 ÷ 189.28

        = 7.32

Triangle r:

Perimeter         = 40 ÷ Sin80

        = 40.62

        = Cos60 x 40

        = 20

        = 40+20+40.62

        = 100.62

Area   =  s        = p ÷ 2

        = 100.62 ÷ 2

        = 50.31

        = square root of s(s-a)(s-b)(s-c)

        = square root of 50.31x(50.31-40)x(50.31-20)x(50.31-40.62)

        = square root of 50.31x10.31x30.31x9.69

        = square root of 152343.07

        = 390.31

Ratio        = a/p

        = 390.31 ÷ 100.62

        = 3.88

Triangle s:

Perimeter         = 40 ÷ Cos80

        = 230.35

        = 60+40+230.35

        = 330.35

Area   =  s        = p ÷ 2

        = 330.35 ÷ 2

        = 165.18

...read more.

Conclusion

(Scale = 1:2)

The measurements are:

a = 43.6cm

b = 50

c = 31cm

Perimeter        = 43.6 + 50 + 31

                = 124.6cm

Area   =  s        = p ÷ 2

        = 124.6 ÷ 2

        = 62.3

        = square root of s(s-a)(s-b)(s-c)

        = square root of 62.3x(62.3-43.6)x(62.3-50)x(62.3-31)

        = square root of 62.3x18.7x12.3x31.3

        = square root of 448517.2

        = 669.71

Ratio        = a/p

        = 669.71 ÷ 124.6

        = 5.37


Below is a table to show the results of the ratios of the triangles that I found. The table will show any triangles that are the same or anomalous:

image07.png

I will now draw a graph to show the above results in a more graphical way. This will allow me to see if any triangles are the same by looking at the ratios:


image03.png


image04.png

I have found that 13 triangles are either impossible to draw or are impossible to calculate. These are k, l, m, o, p, s, v, w, x, y, z, ab and ac. I have found that  two pairs of triangles are the same. These are b & ae and ag & ai.

The largest ratio I have found, using the numbers 40, 60 and 80 are the triangles b and ae. This is one pair of triangles that are the same. Below is a diagram of triangle b with the numbers 40, 60 and 80:

image05.png

                                40                                         80

                                                     60

To extend this piece of work further, I would like to try using numbers that follow the same pattern as the numbers 40, 60 and 80. I would choose numbers like 10, 30 & 50 and 20, 40 & 60 etc. Then I would increase the gap between the numbers and use numbers like 10, 40 & 70, and 30, 60 & 90. Also, instead of changing numbers, I could change the shapes using such shapes as square, pentagon, hexagon, etc.

Simon Rodgers        Page         Center: 48107

...read more.

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