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# Area of a Polygon

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# Area of a Polygon:

We split a quadrilateral into triangles and find its area. Similar methods can be used to find the area of a polygon.

Polygon refers to a closed 2D shape which is made up of a finite number of line segments.

## Steps to find the area of a polygon:

1. Divide the polygons into parts (triangles, trapezium, rectangle, square, and parallelogram) to find out its area.

2. Adding the area of all triangles and trapeziums.

### Some of the ways to find the area are shown below:

1)

By constructing two diagonals AC and AD the pentagon ABCDE is divided into three parts.
So, Area ABCDE = Area of ∆ABC + Area of ∆ACD + Area of ∆AED.

2)

By constructing one diagonal AD and two perpendiculars BF and CG on it, pentagon ABCDE is divided into four parts.
So, Area of ABCDE = Area of right-angled ∆AFB + Area of trapezium BFGC + Area of right-angled ∆CGD + Area of ∆AED.

3)
Area of hexagon MNOPQR = Area of Triangle MNO + Area of Triangle RPQ + Area of Rectangle MOPR.

4)

Area of this polygon = area of 2 trapeziums.

#### Example

Polygon ABCDE is divided in different parts as shown in figure. If AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and BF = 2 cm, CH = 3cm, EG = 2.5 cm. Then find the area of the polygon.

Area of polygon ABCDE = Area of right ΔAFB + Area of trapezium BFHC + Area of right ΔCHD + Area of right ΔEGD + Area of right ΔAGE

= 1/2 xx "AF" xx "BF" + 1/2 xx ["BF + CH"] xx "FH" + 1/2 xx "HD" xx "CH" + 1/2 xx "GD" xx "GE" + 1/2 xx "AG" xx "GE"

= 1/2 xx 3 xx 2 + 1/2 xx [2 + 3] xx [6 – 3] + 1/2 xx [8 – 6] xx 3 + 1/2 xx [8 – 4] xx 2.5 + 1/2 xx 4 xx 2.5

= 3 + 7.5 + 3 + 5 + 5

= 23.5 cm2

#### Example

The area of a trapezium-shaped field is 480 m2, the distance between two parallel sides is 15 m and one of the parallel sides is 20 m. Find the other parallel side.
One of the parallel sides of the trapezium is a = 20 m, let another parallel side be b, height h = 15 m.
The given area of trapezium = 480 m2.
Area of a trapezium = 1/2 h(a + b)
So, 480 = 1/2 × 15 × (20 + b)
or (480 xx 2)/15 = 20 + b
or 64 = 20 + b
or b = 44 m.
Hence the other parallel side of the trapezium is 44 m.

#### Example

There is a hexagon MNOPQR of side 5 cm. Aman and Ridhima divided it into two different ways. Find the area of this hexagon using both ways.
Aman’s method:
Since it is a hexagon so NQ divides the hexagon into two congruent trapeziums.
Now area of trapezium MNQR = 4 xx ((11 + 5))/2 = 2 xx 16 = 32 cm2
So the area of hexagon MNOPQR = 2 × 32 = 64 cm2.

Ridhima’s method:
∆MNO and ∆RPQ are congruent triangles with an altitude of 3 cm.
Area of ∆MNO = 1/2 xx 8 xx 3 = 12 cm2 = Area of ∆ RPQ
Area of rectangle MOPR = 8 × 5 = 40 cm2.
Now, area of hexagon MNOPQR = 40 + 12 + 12 = 64 cm2.
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Problems on Finding Area of a Polygon - Part 1 [00:23:30]
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##### Series: Area of a Polygon
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