• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month   # Find the biggest value of the ratio area / perimeter for some triangles.

Extracts from this document...

Introduction 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. 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.

B c                        a

A                    b                   C

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

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

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: 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:  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: 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

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Fencing Problem essays

1. ## Graphs of Sin x, Cos x; and Tan x

Question 1 Find the size of angle R. The Answer Did you get 25.4�? Well done! Teacher's Note If not, remember that Multiplying both sides by 4, we get = 0.4293... Therefore, R = 25.4� Question 2 Find the length of YZ. The Answer Did you get 4.40cm? Well done!

2. ## Investigating different shapes to see which gives the biggest perimeter

Results so far: I have now put my results in to a spreadsheet and produced a graph out of it. This helps me analyse my results so far. I have printed out my data on paper. Conclusion on rectangles: I have so far found out that a square has the biggest area out of all rectangles.

1. ## Geography Investigation: Residential Areas

However, the flaw with this is that it's my individual view of the area and someone else might argue I was unfair with the number I gave the street. Using the data collection method I did for the 'external questionnaire' made my results slightly biased because it was my own

2. ## To investigate the ratio of Area:Perimeter for triangles (2) ...

= 0.65 c = (a sin 60)/(sin 80) = 0.88 Perimeter = 1+.65+0.88 =2.53 The area = 1/2 bc sin A = 1/2*65*.88 * sin 80 = .282 m2 The ratio of area:perimeter is 0.282/2.53 = 0.111 If the size of the triangle alters will the ratio alter.

1. ## Regeneration has had a positive impact on the Sutton Harbour area - its environment, ...

To ensure a wide range of accurate data for this survey, we tried to cover the largest area possible within the study area. Beforehand, we also had to work out where each of the 10 surveys were to be carried out.

2. ## Difference in Japans Two Biggest Regions: Kanto and Kansai

Nara is the capital city of Nara prefecture in the Kansai region of Japan, near Kyoto. Nara was the capital of Japan for some time in the 8th Century. Most of Japanese society during this period was agricultural in nature, centered on villages.

1. ## Investigate the affects of the surface area: volume ratio on the cooling of an ...

Generally animals that live in cold regions tend to be bigger than related species that live in hot areas because they need to retain heat.

2. ## Fencing problem.

a right-angled triangle I shall consider taking the same procedures that have been taken into mind as before. I shall not need to use the Pythagoras theorem, as I shall know the height of the triangle. But to prove that the above triangle is a right-angled triangle I shall use the Pythagoras's theorem. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 