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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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Shaalaa.com | Problems on Finding Area of a Polygon - Part 1

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