Regular polygons are the polygons which sides and interior angles are equal. Equilateral triangle, Square, regular pentagon etc are Regular polygons. It has importance in some algorithms and specific applications. Here I am introducing formulas for finding longest and shortest diagonal of regular polygon. So many formulas are available for finding diagonals but the advantage of my formula is only one side of the polygon and the number of sides is needed to find the diagonals
INTERIOR ANGLE
If ‘n’ is the number of sides of regular polygon,We know that the sum of the interior angles of regular polygon = (n-2)180.So one interior angle = (n-2)180/n
SHORTEST DIAGONAL
Let as consider the bellow regular polygon for derive the equation of shortest diagonal. if the polygon have ‘n’ sides each interior angle will be equal to (n-2)180/n.Here the shortest diagonal is ‘AC’ and ‘AE’ as shown in the figure
In all regular polygons the shortest diagonal will make an isosceles triangle like triangle ABC as shown in the figure. Let as consider the triangle for deriving the shortest diagonal
Consider the triangle AOB and draw a perpendicular bisector OC as shown in the figure
bellow
Here AB=2 x AC………………………………………………………….. (6)
Cos (angle OAB) =AC/OA
Substituting (3), (4), (5) & (6) in the above equation
We get
Cos ((n-2)90/n) =a/2r
r = (a x Sec ((n-2)90/n))/2………………………………………… (7)
Substituting (7) in (2) we get
Longest diagonal= a x Sec ((n-2)90/n)
Where a= one side of the regular polygon
n= number of side (even only)
Consider the regular polygons with number of side are odd as shown in the bellow figure.
Here AB & AC is the longest diagonal. In all this type of the regular polygons the longest
diagonals make such an isosceles triangle ABC as shown in the figure.
Here angle BAC = interior angle/ (n-2)…………………………………. (8)
In all regular polygons the shortest diagonal will make an isosceles triangle like triangle ABC as shown in the figure.
Let as consider the triangle for deriving the shortest diagonal
We know that the interior angle B is (n-2)180/n.let us draw a perpendicular bisector BP from B to P. It will bisect the shortest diagonal AC
There fore shortest diagonal=2 x AP …………………………. (1)
Let us find the value of AP from right angle triangle APB
Angle ABP = (n-2)180/2n (Bisecting interior angle)
= (n-2)90/n
Sin ((n-2)90/n) = AP/AB
AP= a x Sin ((n-2)90/n) (AB = a = one side of regular polygon)
Substituting in (1) we get
Shortest diagonal= (2a) x (Sin ((n-2)90/n))
Where a= one side of the regular polygon
n= number of side
LONGEST DIAGONAL
The longest diagonal of regular polygon can be find by two formula. One formula is for n=even and other is for n=odd
LONGEST DIAGONAL ‘n’ IS EVEN
Longest diagonal of Regular polygon with number of sides is even is equal to the diameter of the cicumcircle of the regular polygon. so let us derive circumcircle radius
Longest diagonal=diameter of circumcircle =2 x Radius (r)…………………………… (2)
Consider the bellow regular polygon. Here OA and OB represent the radius of
circumcircle. So the triangle ABO is isosceles triangle.
Here angle OAB is half of the interior angle= (n-2)90/n………………………… (3)
AB=a (one side of polygon)……..........…………………………………...………………. (4)
OA=r (radius)…………………....................…………………………………………………. (5)
bellow
Here AB=2 x AC………………………………………………………….. (6)
Cos (angle OAB) =AC/OA
Substituting (3), (4), (5) & (6) in the above equation
We get
Cos ((n-2)90/n) =a/2r
r = (a x Sec ((n-2)90/n))/2………………………………………… (7)
Substituting (7) in (2) we get
Longest diagonal= a x Sec ((n-2)90/n)
Where a= one side of the regular polygon
n= number of side (even only)
LONGEST DIAGONAL ‘n’ IS ODD
Here AB & AC is the longest diagonal. In all this type of the regular polygons the longest
diagonals make such an isosceles triangle ABC as shown in the figure.
Here angle BAC = interior angle/ (n-2)…………………………………. (8)
Let us consider triangle ABC for deriving the longest diagonal. We can draw a perpendicular bisector from B to P as shown in the figure
Consider the right triangle APC.
Here
AC is the longest diagonal=d…………………………………………………. (9)
PC is half of the side of the polygon = (a/2)………………………………….. (10)
Angle PAC=angle BAC/2…………………………………………………….. (11)
Sin (angle PAC) = PC/AC
Substituting (8), (9), (10) & (11) in the above equation
We get
Sin (90/n) = a /2d
d = (a/2) x ((cosec (90/n))
Longest diagonal = (a/2) x ((cosec (90/n))
Where a = one side of the regular polygon
n = number of side (odd only)
DERIVED FORMULAS
SHORTEST DIAGONAL = (2a) x (Sin ((n-2)90/n))
Where a = one side of the regular polygon
n = number of side
LONGEST DIAGONAL = a x Sec ((n-2)90/n)
Where a = one side of the regular polygon
n = number of side (even only) (n=4, 6, 8……….)
LONGEST DIAGONAL = (a/2) x ((cosec (90/n))
Where a = one side of the regular polygon
n = number of side (odd only) (n=3, 5, 7………….)
LONGEST DIAGONAL OF PENTAGON = a x 1.6180
LONGEST DIAGONAL OF HEXAGON = a x 2
LONGEST DIAGONAL OF HEXAGON = a x 2.2469
where a= one side of the regular polygon
Where a = one side of the regular polygon
n = number of side
LONGEST DIAGONAL = a x Sec ((n-2)90/n)
Where a = one side of the regular polygon
n = number of side (even only) (n=4, 6, 8……….)
LONGEST DIAGONAL = (a/2) x ((cosec (90/n))
Where a = one side of the regular polygon
n = number of side (odd only) (n=3, 5, 7………….)
SPECIAL FORMULAS
We can derive special formulas for triangle, square, pentagon etc by substituting value of
‘n’
SHORTEST DIAGONAL OF PENTAGON = a x 1.6180
SHORTEST DIAGONAL OF HEXAGON = a x 1.7320
SHORTEST DIAGONAL OF HEPTAGON = a x 1.801
where a= one side of the regular polygon
LONGEST DIAGONAL OF HEXAGON = a x 2
LONGEST DIAGONAL OF HEXAGON = a x 2.2469
where a= one side of the regular polygon
