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 = \(\frac{1}{2}\) 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 - Worksheet**

**Worksheet on Area of a Polygon**

**8th Grade Math Practice**** ****From Area of a Polygon to HOME PAGE**

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