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 = ^{1}/_{2} nar ⇔ a = ^{2A}/_{nr} ⇔ na = ^{2A}/_{r} ⇔ Perimeter = ^{2A}/_{r}

From right ∆OLB, we have:

OL^{2} = OB^{2} - LB^{2} ⇔ r^{2} = {R^{2} - (^{a}/_{2})^{2}}

⇔ r = √(R^{2} - (a^{2}/4)

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

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

(i) **Hexagon: **

OL^{2} = (OB^{2} - LB^{2})

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

⇒ OL = {(√3)/2 × a}

⇒ Area ∆OAB = 1/2 × AB × OL

= {1/2 × a × (√3)/2 × a} = (√3)a^{2}/4

⇔ area of hexagon ABCDEF = {6 × (√3)a^{2}/4} square units

= {3(√3)a^{2}/2} square units.

Therefore, area of a hexagon = {3(√3)a^{2}/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) = 2a^{2} (1 + √2) square units.

Therefore, area of an octagon = 2a^{2} (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)a^{2}/2} cm^{2}

= (3 × 1.732 × 6 × 6)/2 cm^{2}

= 93.528 cm^{2}.

**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 = [2a^{2} (1 + √2) square units

= [2 × 5 × 5 × (1 + 1.414)] cm^{2}

= (50 × 2.414) cm^{2}

= 120.7 cm^{2}.

**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) cm^{2} = 43.75 cm^{2}.

**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 × √(R^{2} - a^{2}/4) square units

= {5/2 × 8 × √(7^{2} - 64/4)} cm^{2}

= {20 × √(49 - 16)} cm^{2} = (20 × √33) cm^{2}

= (20 × 5.74) cm^{2} = (114.8) cm^{2}.

**Related Links:**

**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**

8th Grade Math Practice

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