# Area of Pentagon

The area of a pentagon is the total amount of space enclosed by all its five sides.

Since a pentagon is a two-dimensional shape, its area also lies in a two-dimensional plane. Its area is expressed in square units such as sq. m, sq. cm, sq. in or sq. ft.

## Formulas

### With Apothem

The basic formula to find the area of a regular pentagon is given below:

Let us solve an example

Find the area of a pentagon whose side is 7 cm and apothemÂ  is 4.817 cm.

Solution:

As we know,
Area (A) = 5/2(s Ã— a), here s = 7 cm, a = 4.817 cm
= 5/2 (7 Ã— 4.817)
â‰ˆ 84.3 sq. cm

Finding the area of a pentagon when PERIMETER and APOTHEM are known

Calculate the area of a regular pentagon whose perimeter is 70 cm and apothem is 9.634 cm.

Solution:

Here we will use an alternative formula,
Area (A) = 1/2 Ã— P Ã— a, here P = 70 cm, a = 9.634 Â cm
= 1/2 Ã— 70 Ã— 9.634
= 337.19 sq. cm

### With Side Length

The formula to find the area of a regular pentagon with side length is given below:

Let us solve an example

What is the approximate area of a regular pentagon of side 8 cm?

Solution:

As we know,
Area (A) = 1/4 âˆš5(5 + 2âˆš5)s2, here s = 8 cm
= 1/4 âˆš5(5 + 2âˆš5)82
â‰ˆ 110.11 sq. cm

Finding the area of regular pentagon inscribed in a circle when the RADIUS is known

Work out the area of regular pentagon inscribed in a circle whose radius is 9 cm.

Solution:

As we know,
Area (A) = 5/2 Ã— r2sin(72Â°), here r = 9 cm
= 5/2 Ã— 92 Ã— sin(72Â°)
= 192.59 sq. cm

## Area of an Irregular Pentagon

Till now, we have learned how to find the area of a regular pentagon. However, we do not get to find the area of regular polygons in the practical world. Let us see how to find the area of an irregular pentagon with an example.

Finding the area of an irregular pentagon when the SIDES and DIAGONALS are known

Find the area of an irregular pentagon-shaped land given below. Given are its 5 sides and 2 diagonals.

Solution:

The above irregular pentagon is divided into 3 separate triangles – Î”1, Î”2 , and Î”3.
Area of âŒ‚ABCDE = Area of Î”1 + Area of Î”2 + Area of Î”3
Here, we will use Heronâ€™s formula to find the area of each triangle.
Area (A) = âˆšs(s – a)( s- b)(s – c), here s = Â½ (a + b + c); a, b, and c are the 3 sides of any triangle
In Î”1, s = Â½(a + b + c), here a = 30 ft, b = 20 ft, c = 40 ft
= Â½(30 + 20 + 40)
= 45 ft
Area (A) of Î”1 = âˆšs(s – a)(s- b)(s – c), here s = 45 ft
= âˆš45(45 – 30)(45 – 20)(45 – 40)
= 43.3 sq. ft
In Î”2, s = Â½(a + b + c), herea = 40 ft, b = 25 ft, c = 33 ft
= Â½(40 + 25 + 33)
= 49 ft
Area (A) of Î”2 = âˆšs(s – a)(s- b)(s – c), here s = 49 ft
= âˆš49(49 – 40)(49- 25)(49 – 33)
= 58.79 sq. ft
In Î”3, s = Â½(a + b + c), herea = 33 ft, b = 18 ft, c = 32 ft
= Â½(33 + 18 + 32)
= 41.5 ft
Area (A) of Î”3 = âˆšs(s – a)(s- b)(s – c), here s =41.5 ft
= âˆš41.5 (41.5 – 33)(41.5 – 18)(41.5Â  – 32)
= 43.56 sq. ft
Now, Area of âŒ‚ABCDE = Area of Î”1 + Area of Î”2 + Area of Î”3
= 43.3 + 58.79 + 43.56
= 145.65 sq. ft