• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

To investigate the ratio of Area:Perimeter for triangles (2) To investigate the suggestion made by students that a 40,60,80 triangle would maximise the above mentioned ratio.

Extracts from this document...

Introduction

Natalie Pinfield Williams

Aims

(1)        To investigate the ratio of Area:Perimeter for triangles

(2)        To investigate the suggestion made by students that a 40,60,80 triangle would maximise the above mentioned ratio.

(3)        Should the students hypothesis be un-sustainable then find out what triangle would maximise the ratio

Important facts about triangles

Simple formula for the area        Area = ½ base *height

        General Formula                        Area = ½ ab sin C                

        Interior Angles                        A+B+C = 180        

        Sin rule                                 a          =         b   _        =         c  

                                                Sin A                Sin B                Sin c

        Cosine Rule                                a2 = b2 + c2 – 2bc Cos A

Student Suggestion

We will Investigate the claim made by the students that a 40,60,80 triangle would maximise the ratio of area:perimeter

Angles

The students suggested that a 40 60 80 would maximise the area : perimeter ratio.

We will make an initial assumption that the 40 6080 refers to the size of the angle in degrees. We accept that they could have been referring to lengths.

Is our assumption possible?

Yes as 40 + 60 + 80 =180. The interior angles do sum to 180 which they must for any triangle so the students could have been referring to the angles.

Calculation

...read more.

Middle

c         =        (a sin 60)/(sin 80)        =        0.88

Perimeter = 1+.65+0.88 =2.53

The area        =         ½ 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.

The students did not give us a length. We assumed a = 1m. Will the ratio alter if we increase the length? If we consider that the length of side a = x m we can produce a general formula for the are of a triangle with angels 40,60,80

If a=x, then we can see from the previous section that b=.65 x and c =.88 x

The perimeter will then be  = 2.53x

The area         = ½ bc sin A

=        .65 x * .88 x sin 80

        =         .573 x2 sin 80

        =        .564 x2

The ratio is therefore

=         0.564 x2 / 2.54 x        

=         0.044 x

The important thing to see is that ratio therefore varies as or is proportional to x. So the larger we make our triangle the higher the ratio.

There is therefore no maximum ratio. The ratio simply increases as the side length increases.

For a comparison of the ratio, where we change the angles, we need to define the perimeter.

...read more.

Conclusion

2 must be positive x2-a2 must be less than x2 hence the largest area enclosed within a quadrilateral is when a=0,  this means all four sidses are the same length. In other words this is a square.

It appears form the above that regular shapes enclose more area than non regular. From the above it would therefore be sensible to assume that an equilateral triangle, being the regular form of a triangle would enclose the most area with a set perimeter and hence produce the best ratio of area:perimeter.

Equilateral Triangle

If the perimeter is set at 1m then each side =1/3m

An equilateral triangle has angles of 60 degrees so

The area is therefore  =        1/2 *1/3 *1/3 * sin 60 = .047

Which produces a ratio =.047

Conclusion

Which ever way the students were thinking they were wrong. An equilateral triangle gives a better ratio than either of the ways their statement could be taken.

Perhaps the students were misheard and were not talking about triangle or even mathematics as this discussion seems remarkable unlikely. They were perhaps talking about the distance to a party or the ratio of the number CD’s that they own.

Perhaps the students should pay more attention to their mathematics teacher.

...read more.

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

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Medicine and mathematics

    Therefore, we have to proceed using logs. 10/300=0.6x 0.03.=0.6x Log0.03./Log0.6 = x 6.86(3 s.f.) = x 6.86 Hours. 60 hours 0.86 = 51.6 minutes 60 minutes x 0.6 = 36 seconds Meaning that after 6 hours, 51.6 minutes, and 36 seconds after administration there will be less than 10mg amount of active penicillin in the bloodstream.

  2. When the area of the base is the same as the area of the ...

    There is an odd thing also in the 18 by 18 line graph because the line for the volume and the line for the area of the base both meet up and drop form a high volume to a low volume this shows me that there are trends in the

  1. Investigating different shapes to see which gives the biggest perimeter

    So I can now apply the tan rule to work out height: Adjacent = Opposite/Tan 30 Height = 83.3/0.58 = 144.34m So height equals 144.34m. Therefore area of isosceles is: Area = 1/2 base x height = (1/2 x 166.6)

  2. The aim of this coursework is to investigate which shape gives the largest enclosed ...

    To work out the area for this triangle we have to again divide it into 2 to make it into a right angled triangle. At this point we do not no the height but by using trigonometry we can work it out.

  1. Geography Investigation: Residential Areas

    Relating this evidence back to the concentric theory it mildly proves that Basingstoke does relate to this theory. I have proved my hypothesis correct because I have found that Sarum Hill does in fact have the lowest area rating and that The Beaches has the highest.

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

    This created about 500 jobs. These things, amongst the building of modern, luxury accommodation, the free space that was created around the area became more enviable at just 10 minutes from the Central business District and the creation of Barbican Glass, meant that the City Council now had a larger income that could be spent on regenerating the area further.

  1. How can I make My local Area Sustainable?

    I then decided it would be a good idea to add some Recycling Areas because it is one of the governments and local authority's targets for the environment. It is also their target to get more people aware of recycling.

  2. The Fencing Problem. My aim is to determine which shape will give me ...

    far all my graphs prove that my statement was true and I have also realised that the increase between the shapes have decreased. Before the increase between the equilateral triangle and the square was 14,376 compared to the difference of the hexagon and heptagon, which is 1993.526.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work