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
As no side lengths have been give then we could assume any one side is any length we want. By defining the length of one side the others are obviously defined. We can calculate the other length by use of the sine rule.
We will make a decision: Let the length of side ‘a’ be 1m.
The perimeter is = a + b + c:
By the sine rule
a = b = c
Sin 40 sin 80 sin 60
Therefore b = (a sin40)/(sin80) = 0.65
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. For the remainder of this study we will use a perimeter of 1m.
Side Length
If we assume the numbers given are side lengths then for us to work out the area we need to obtain one of the angles.
For our triangle let a=40,b=60,c=80
To get one angle we need to use the cosine rule. This can be rearranged it to get to.
Cos A = b2+c2-a2
2bc
= 3600+6400-1600
9600
= 0.875
A = cos-1(.0875)
A = 28.9 degrees
Area of the triangle = ½ bc sin A
= 1.2 * (60 * 80) *sin 28.9
= 1160 to 3 sf
Ratio = 1160/180
= 6.44
Clever bit.
We have shown from the above section that the ratio is proportional to the perimeter. To compare these two ratios we need to have the same perimeter. The first was 2.53 the second was 180. As seen before the ration is proportional to the perimeter so to get a fair comparison we must divide by the perimeter.
The ratio with the same perimeter is (6.44/180) = 0.035
This ratio for sides of length 40,60,80 is lower than that for the triangle with angles 40,60,80 so we must assume that the student was talking about angles rather than sides length when she said that 40.60,80 triangle would maximise the ratio. Or of course the student’s hypothesis was simply wrong.
More thoughts
We have decided to confine our investigation to triangles with a perimeter of 1 m. We can see that if you have an isosceles triangle with a small base and tall sides its area will be very small. In fact if we make very, very narrow with the long sides approximately = .5 and the base close to 0 then the area will approach 0.
Similarly if the angle at the top becomes large the triangle becomes long and flat so the area once again becomes very small. (Remember that we have constrained the perimeter to 1)
In both the above cases the area becomes very small and therefore the ration of the area/perimeter also becomes very small. These are not the sort of triangles we need to maximise the ratio.
It is obvious that the best ratios are to be gain with triangles with sides of similar lengths.
a
Regular shapes
It is generally accepted, though I cannot prove it, that the most area that it is possible to enclose with a perimeter of a set length is enclosed within a circle. If we consider a quadrilateral it is clear that a rectangle encloses more area than a parallelograms or trapezium.
If we have a perimeter of ‘4x’ and the quadrilateral was a square then each side would be of length x and the area would be x2. If we consider a reactangle such that the long side was increased from x to x+a then the shorter side would become x-a if the perimeter was kept at 4x
The area would therefore be:
(x-a)(x+a) = x2-a2 Difference of squares.
As a2 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.