Area of a Polygon


In area of a polygon we will learn about polygon, regular polygon, central point of the polygon, radius of the inscribed circle of the polygon, radius of the circumscribed circle of a polygon and solved problems on area of a polygon.


Polygon: A figure bounded by four or more straight lines is called a polygon. 


Regular Polygon: A polygon is said to be regular when all its sides are equal and all its angles are equal.

A polygon is named according to the number of sides it contains. 

Given below are the names of some polygons and the number of sides contained by them. 

  • Quadrilateral - 4 
  • Pentagon - 5 
  • Hexagon - 6 
  • Heptagon - 7 
  • Octagon - 8 
  • Nonagon - 9 
  • Decagon - 10 
  • Undecagon - 11
  • Dodecagon - 12 
  • Quindecagon -15 

  • Central Point of a Polygon: 

    The inscribed and the circumscribed circles of a polygon have the same centre, called the central point of the polygon. 


    Radius of the Inscribed Circle of a Polygon: 

    The length of perpendicular from the central point of a polygon upon any one of its sides, is the radius of the inscribed circle of the polygon. 
    The radius of the inscribed circle of a polygon is denoted by r


    Radius of the Circumscribed Circle of a Polygon: 

    The line segment joining the central point of a polygon to any vertex is the radius of the circumscribed circle of the polygon. The radius of the circumscribed circle of a polygon is denoted by R

    In the figure given below, ABCDEF is a polygon having central point O and one of its sides a unit. OL ⊥ AB. 
    Then, OL = r and OB = R 



    Area of a polygon of n sides 

            = n × (area ∆OAB) = n × ¹/₂ × AB × OL 

            = (ⁿ/₂ × a × r) 

    Now, A = \(\frac{1}{2}\) nar ⇔ a = \(\frac{2A}{nr}\) ⇔ na = \(\frac{2A}{r}\)

     ⇔ Perimeter = \(\frac{2A}{r}\)

    From right ∆OLB, we have:

    OL² = OB² - LB² ⇔ r² = {R² - (ᵃ/₂)²}

                          ⇔ r = √(R² - (a²/4)

    Therefore, area of the polygon = {n/2 × a × √(R² - a²/4) square units.





    In area of a polygon some of the particular cases such as;



    (i) Hexagon: 

        OL² = (OB² - LB²)

               = {a² - (a/2)²} = (a² - a²/4) = 3a²/4

    ⇒ OL = {(√3)/2 × a}

    ⇒ Area ∆OAB = 1/2 × AB × OL

                           = {1/2 × a × (√3)/2 × a}

                           = (√3)a²/4

    ⇔ area of hexagon ABCDEF  = {6 × (√3)a²/4} square units

                                                   = {3(√3)a²/2} square units.

    Therefore, area of a hexagon = {3(√3)a²/2} square units.


    (ii) Octagon:

        BM is the side of a square whose diagonal is BC = a.

    Therefore, BM =  \(\frac{a}{\sqrt{2}}\)

    Now, OL = ON + LN

                  = ON + BM = (a/2 + a/√2)

    ⇔ Area of given octagon

        = 8 × area of ∆OAB = 8 × 1/2 × AB × OL

        = 4 × a × (a/2 + a/√2) = 2a² (1 + √2) square units.

    Therefore, area of an octagon = 2a² (1 + √2) square units.


    We will solve the examples on different names of the area of a polygon.

    Area of a Polygon

    1. Find the area of a regular hexagon each of whose sides measures 6 cm.

    Solution:

    Side of the given hexagon = 6 cm.

    Area of the hexagon = {3√(3)a²/2} cm²

                                     = (3 × 1.732 × 6 × 6)/2 cm²

                                     = 93.528 cm².



    2. Find the area of a regular octagon each of whose sides measures 5 cm.

    Solution:

    Side of the given octagon = 5 cm.

    Area of the octagon   = [2a² (1 + √2) square units

                                    = [2 × 5 × 5 × (1 + 1.414)] cm²

                                    = (50 × 2.414) cm²

                                    = 120.7 cm².



    3. Find the area of a regular pentagon each of whose sides measures 5 cm and the radius of the inscribed circle is 3.5 cm.

    Solution:

    Here a = 5 cm, r = 3.5 cm and n = 5.

    Area of the pentagon    = (n/2 × a × r) square units

                                       = (5/2 × 5 × 7/2) cm²        

                                       = 43.75 cm².



    4. Each side of a regular pentagon measures 8 cm and the radius of its circumscribed circle is 7 cm. Find the area of the pentagon.

    Solution:

    Area of the pentagon    = {n/2 × a × √(R² - a²/4) square units

                                       = {5/2 × 8 × √(7² - 64/4)} cm²

                                       = {20 × √(49 - 16)} cm²         

                                       = (20 × √33) cm² 

                                       = (20 × 5.74) cm²        

                                       = (114.8) cm².


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