Steyning & District U3A
Discovering Mathematics
Session 36
Areas of Polygons
Pascal’s Triangle & Combinatorics
Area of a Polygon
Derive a general formula for the summation of the area of an n sided regular polygon.
Clue: the polygon consists of n isosceles triangles. The base = side length of polygon.
The formula will include the number of sides n, the side length s & the angle 2θ
subtended by each triangle at the centre.
The angle θ = 360o/2n = 180o/n
Thus for a pentagon; θ = 36o and for hexagon θ = 30o
θ
Tanθ = side opposite/side adjacent = s/2/h = s/2h
Transposing; 2h = s/tanθ or h = s/2 tanθ
h
s/2
Area of a triangle is; half base x height = s/2* s/2tanθ = s2/4tanθ
There are n isosceles triangles in a polygon with n sides.
And for any n sided regular polygon; Area = ns2/4tanθ
Maximised Use of Fence Panels
A farmer has 60 fence panels each 3m long to use to construct a compound in which to
graze goats.
He can construct a compound of any regular polygonal shape up to one having 12 sides,
but he must use all 60 fence panels.
Which shape should he choose to obtain the maximum grazing area & what is it?
If he chooses a triangular area, each side will utilize 20 panels, each of length 3m.
For a square area, each side will have 15 panels, for a pentagon 12 panels etc.
Using our derived formula; Area = ns2/4tanθ.
For a triangular compound: n=3, s = 60m and θ = 60o, Area = 3*602/4tan60o
tan 60o = √3. Therefore, Area = 3*3600/4*√3 = 900*√3 = 1559 m2
For a square compound: s = 45m and Area = 45*45 = 2025 m2
For a pentagonal area; n = 5, s = 36m and θ = 36o and Area = 5*362/4tan 36o
Tan 36o = 0.7265. Therefore Area = 5*9*36/0.7265 = 2230 m2
Area of Polygons (cont)
For a hexagonal area, n = 6, s = 30m & θ = 30o and Area = 6*302/4tan 30o
Tan 30o = 0.5773. Therefore Area = 6*900/4*0.5773 = 2338 m2
Polygons of 7, 8, 9 or 11 sides cannot use all of the fence panels, but 10 & 12 are feasible
For 10 sides, n = 10, s = 18 and θ = 18o and Area = 10*182/4tan 18o
Tan 18o = 0.3249. Therefore Area = 10*81/0.3249 = 2493 m2
For 12 sides, n = 12, s = 15m and θ = 15o and Area = 12*152/4tan 15o
Tan 15o = 0.2679. Therefore Area = 12*15*15/4*0.2679= 2519 m2
By inspection we see that the area increases in line with the increase of n, but that the
value of the increase is gradually diminishing.
The limit could be determined by differentiating the formula; A = ns2/4*Tanθ with
respect to n,
In our problem s = 60*3/n and θ =180/n, so A = n*(180/n)2/4*Tan(180/n)
where dA/dn = 0 for Max. or Min. values of area.
The formula can be simplified to A = 8100/nTan(180/n), but is still a rather complex
differentiation to attempt here.
Area of Polygons (Cont. 2)
3000
2500
2000
1500
Area m2
1000
500
0
0
10
20
30
40
50
60
70
n sides
A plot of Area against number of sides shows that the curve trends to an asymptopic
value of about 2580 m2.
It would therefore be rather pointless for the farmer to construct a compound with more
than, say. 10 or 12 sides.
Combinatorics & Pascal’s Triangle
Pascal's Triangle - TED ED
Permutations & Combinations
Binomial Expansion - Khan Academy
That’s it Folks
Suggestions for future topics?