Properties Of A Regular Polygon

REGULAR POLYGONS Introduction A polygon is a closed figure bounded by three or more lines called its sides. A polygon having the least number of sides is a triangle. Thus a triangle has three sides. Quadrilateral has four sides; a Pentagon has five sides and so on. But when a polygon has all the sides equal and all the angles of the same measure, then it is called a regular polygon.

Properties of a Regular Polygon 1. It is an equilateral that is, having all the sides equal in length, and equiangular polygon. 2. The incircle and the circumcircle of a regular polygon have the same centre. 3. The radius of the circumcircle and the radius of the regular polygon are equal. 4. All the apothems are equal in a regular polygon 5. All the interior as well as exterior angles of a regular polygon are equal.

6. A radius of a regular polygon bisects the angle to which it is drawn. 7. The length of all the apothems of a regular polygon are equal. 8. An apothem of a regular polygon, when drawn on a side of a regular polygon bisects the side to which it is drawn. Thus in figure 9-1, AE=ED Relation between angles and sides of a regular polygon

Let us measure the exterior and interior angles of regular polygons. On actual measurement of ∠CAB of the above equilateral triangle, it was of 60o . Measure of exterior ∠CAD was found to be 120o . Since all the interior angles of an equilateral triangle are equal to 60o , therefore, sum of its interior angle is 60o × 3 = 180o . Also, an equilateral triangle has three exterior angles. The measure of each exterior angle is 120o . Since, each exterior angle of a regular polygon is equal, therefore, sum of three exterior angles is 120o × 3 = 360o Now, let us measure the interior and exterior angles of a regular pentagon. By actual measurement of ∠EAB using a protractor, it was found to be 108o and its exterior angle, ∠CBF , was found to be of 72o . Since, a regular pentagon has five interior angles, therefore, their sum is 108o × 5 = 540o . Since, each exterior angle is 72o , therefore, sum of all the five exterior angle is 72o × 5 = 360o .

Now, in both the polygons, we find that the sum of the interior angles differs but the sum of the exterior angles is the same. Hence, the exterior angle of a regular polygon is more fundamental in finding a relationship between the angles and sides of regular polygons. We have seen that each exterior angle of an equilateral triangle 120o and the sum of all the three exterior angles = 360o Thus, 120o + 120o + 120o = 360o or 3(120o ) = 360o or 3 =

360o 120o

… (1)

3=3 which is true where ‘3’ is the number of exterior angles. Since each side of the triangle has one exterior angles.

Therefore ’3’ can be taken as the number of sides. Hence from (1), we find that No of sides =

360o exterior angles

… 92)

Now, let us verify pentagon Number of sides of pentagon = 5 Exterior angle of a pentagon = 72o Substituting these values in (2), we find 5=

360o 72o

or 5=5 which is true.

Thus, it is true for pentagon also. Hence we conclude that number of sides of a regular polygon =

360o exterior angles

If ‘n’ is the number of sides of a regular polygon 360o Then n = exterior angles or exterior angle of a regular polygon =

360o n

If we denote the exterior angle by the letter ‘E’ Then E =

360o n

… (3)

Now, let us find the relation between the interior angle and the number of sides of regular polygon From relation (3), we have E=

360o n

But E = 180o -internal angle If we denote internal angle by the letter ‘I’ we have E = 180o − I Replacing ‘E’ by 180o − I in the relation E =

180o − I =

360o n

360o , we have n

or n(180o − I ) = 360o or 180o n − In = 360o or In = 180o n − 360o or In = 180o (n − 2) 180o (n − 2) or I = n

… (4)

180o (n − 2) Hence, internal angle of a regular polygon = I = , where n is the n number of sides of the regular polygon. SOLVED EXAMPLES 1. Find the internal and external angles of a regular octagon Sol: We know that an octagon is a regular polygon having eight sides Therefore

n=8

Substituting the value of n in the relation I =

I=

180o (8 − 2) 8

I=

180o × 6 8

180o (n − 2) n

I = 135o

Now substituting the n = 8 in the relation E =

360o , we have n

360o E= 8 E = 45o Therefore Internal angle of a regular octagon is 135o and external angle is 45o 2. Find the internal and external angles of a regular polygon having 18 sides Sol: Using the relation I =

I=

180o (18 − 2) 18

I=

180o ×16 = 160o 18

E=

180o (n − 2) , we have n ( Since n = 18 )

360o n

360o E= 18 E = 20o

3. A regular polygon was found to have its internal angle as 156o . Find its number of sides. Sol: Here I = 156o Substituting the value of I in the relation I =

156o =

180o (n − 2) n

180o (n − 2) , we have n

or 156o n = 180o (n − 2) or 156o n = 180o n − 360o or 180o n − 156o n = 360o or 24o n = 360o or n =

360o = 15 24o

Therefore The regular polygon has 15 sides Area of a regular polygon In order to find the area of a regular polygon we divide the regular polygon in triangles as shown below in figure 9-3

Let the radius of the hexagon be ' r ' and its side be ' 2a ' . Thus, in figure 9-3 AB = 2a and OA = r . As the apothem OG bisects the opposite side AB, so that 1 AG = × 2a = a 2

Now, in right Triangle OAG, using Pythagoras Theorem, we have

AG 2 + OG 2 = OA2 or OG 2 = OA2 − AG 2 or OG 2 = r 2 − a 2 or OG = r 2 − a 2 We know that the area of a triangle =

=

1 × base × apothem 2

1 × 2a × r 2 − a 2 2

Since there are 6 triangles in a hexagon

Therefore Area of hexagon =

1 × 2a × r 2 − a 2 × 6 = 6a r 2 − a 2 2

Since a is half the side Therefore 6a is half the perimeter Hence area of a regular polygon =

1 × perimeter × apothem 2

In this chapter, we will use this formula to find the area of a regular hexagon SOLVED EXAMPLES 1. Find the area of a regular pentagon whose radius is 4 cm and the length of the side is 7 cm Sol: Area of a regular polygon =

1 × perimeter × apothem 2

Now perimeter of the pentagon = 5 × side = 5 × 7 = 5 7 cm

r 2 − a 2 and here r = 4cm

Apothem =

⎛ 7⎞ Therefore r − a = 4 − ⎜⎜ ⎟⎟ ⎝ 2 ⎠ 2

2

r 2 − a 2 = 16 −

2

2

7 4

r 2 − a2 =

64 − 7 4

r 2 − a2 =

57 4

r 2 − a2 =

57 2

Therefore Area of the pentagon =

1 57 5 399 2 ×5 7 × = cm 2 2 4

2. Find the area of a regular hexagon whose each side is 6 cm and the radius is also 6 cm Sol: Here r = 6 cm and a =

6 = 3 cm 2

Now hexagon has six sides

Therefore Area of the given hexagon =

=

1 × 36 × 36 − 9 2

= 18 27

1 × (6 × 6) × 62 − 32 2

= 54 3cm 2

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