# 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.

• 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.

Area of a Trapezium

Area of a Trapezium

Area of a Polygon

Area of a Trapezium - Worksheet

Worksheet on Trapezium

Worksheet on Area of a Polygon

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