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

Authors Avatar

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. This is triangle q, which is the same as triangle k.

To work out the areas, perimeters and ratios of the triangles, I will use the below formulas:

Perimeter = a + b + c

Area = square root of s(s-a)(s-b)(s-c), s=perimeter/2

        = ½ abSinC (for triangles “u” and “aa” only)

Ratio= area/perimeter

I will now try to work out the areas, perimeters and ratio of area / perimeter:

Triangle a:

I have found that triangle “a” is different from the others and because of this I have found that it is incredibly difficult to find out the perimeter and area of the triangle. I have decided to tackle this problem at the end of my calculations when I have more experience of similar problems.

Triangle b:

Perimeter         = 40 + 60 + 80

        = 180

Area   =  s        = p ÷ 2

        = 90

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

        = square root of 90x(90-40)x(90-60)x(90-80)

        = square root of 90x50x30x10

        = square root of 1350000

        = 1161.90

Ratio        = a/p

        = 1161.90 ÷ 90

        = 12.91

Triangle c:

Perimeter         = 60÷Sin40

                = 93.34

                         = 93.34+80+60

                         = 233.34

Area        =  s  = p÷2

                        = 233.34÷2

                        = 116.67

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

        = square root of 116.67x(116.67-60)x(116.67-80)x(116.67-93.34)

        = square root of 116.67x56.67x36.67x23.33

        = square root of 5656373.24

        = 2378.31

Ratio        = a/p

        = 2378.31 ÷ 233.34

Join now!

        = 10.19

Triangle d:

Perimeter         = Sin40x80

        = 51.42

        = 51.42+80+60

        = 191.42

Area   =  s        = p ÷ 2

        = 191.42 ÷ 2

        = 95.71

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

        = square root of 95.71x(95.71-51.42)x(95.71-60)x(95.71-80)

        = square root of 95.71x44.29x35.71x15.71

        = square root of 2378094.08

        = 1542.11

Ratio        = a/p

        = 1542.11 ÷ 191.42

        = 8.06

Triangle e:

Perimeter         = 80 ÷ Sin 40

        = 124.46

        = Sin60x80

        = 69.28

        = 69.28+124.46+80

        =273.74

Area   =  s        = p ÷ 2

        = 273.74 ÷ 2

        = 136.87

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

= square root of 136.87x(136.87-80)x(138.67-69.28)x(138.67-    124.46)

...

This is a preview of the whole essay