Table of Contents
Last modified on August 3rd, 2023
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.
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 apothem is 4.817 cm.
As we know,
Area (A) = 5/2(s × a), here s = 7 cm, a = 4.817 cm
= 5/2 (7 × 4.817)
≈ 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.
Here we will use an alternative formula,
Area (A) = 1/2 × P × a, here P = 70 cm, a = 9.634  cm
= 1/2 × 70 × 9.634
= 337.19 sq. cm
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?
As we know,
Area (A) = 1/4 √5(5 + 2√5)s2, here s = 8 cm
= 1/4 √5(5 + 2√5)82
≈ 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.
As we know,
Area (A) = 5/2 × r2sin(72°), here r = 9 cm
= 5/2 × 92 × sin(72°)
= 192.59 sq. cm
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.
The above irregular pentagon is divided into 3 separate triangles – Δ1, Δ2 , and Δ3.
Area of ⌂ABCDE = Area of Δ1 + Area of Δ2 + Area of Δ3
Here, we will use Heron’s formula to find the area of each triangle.
Area (A) = √s(s – a)( s- b)(s – c), here s = ½ (a + b + c); a, b, and c are the 3 sides of any triangle
In Δ1, s = ½(a + b + c), here a = 30 ft, b = 20 ft, c = 40 ft
= ½(30 + 20 + 40)
= 45 ft
Area (A) of Δ1 = √s(s – a)(s- b)(s – c), here s = 45 ft
= √45(45 – 30)(45 – 20)(45 – 40)
= 43.3 sq. ft
In Δ2, s = ½(a + b + c), herea = 40 ft, b = 25 ft, c = 33 ft
= ½(40 + 25 + 33)
= 49 ft
Area (A) of Δ2 = √s(s – a)(s- b)(s – c), here s = 49 ft
= √49(49 – 40)(49- 25)(49 – 33)
= 58.79 sq. ft
In Δ3, s = ½(a + b + c), herea = 33 ft, b = 18 ft, c = 32 ft
= ½(33 + 18 + 32)
= 41.5 ft
Area (A) of Δ3 = √s(s – a)(s- b)(s – c), here s =41.5 ft
= √41.5 (41.5 – 33)(41.5 – 18)(41.5 – 32)
= 43.56 sq. ft
Now, Area of ⌂ABCDE = Area of Δ1 + Area of Δ2 + Area of Δ3
= 43.3 + 58.79 + 43.56
= 145.65 sq. ft
Last modified on August 3rd, 2023