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.
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 = 1/2 nar ⇔ a = 2A/nr ⇔ na = 2A/r ⇔ Perimeter = 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 = a/√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.
Related Links:
Area of a Trapezium
Area of a Trapezium - WorksheetArea of a Trapezium Area of a Polygon
Worksheet on Trapezium Worksheet on Area of a Polygon