How to Find Area of Regular Polygon With Side Length

Problem 1 :

What is the area of regular nanagon with 10 cm sides.

Solution :

Number of sides of Nanagon = 9

∠PCR = 360/9 ==> 40

Drawing perpendicular bisector CS, we get

∠PCS = 40/2 ==> 20 and PS = PR/2 ==> 5 cm

The area of the regular nonagon is about 618 cm².

Problem 2 :

Octagon with side length 6 cm

Solution :

Number of sides = 8

∠AOB = 360/8 ==> 45

∠COB = 45/2 ==> 22.5

Side length = 6 cm

Length of an apothem,

Perimeter = sides × length

= 8 × 6

P = 48 cm

Area = 1/2 × aP

= 1/2 × 7.246 × 48

A = 173.91 cm²

The area of the regular octagon is about 174 cm².

Problem 3 :

Decagon with side length 4 yd

Solution :

Number of sides = 10

Side length = 4 yd

Length of an apothem,

AC = 2, BC = 2, Apothem = OC

∠AOB = 360/10 ==> 36

∠COB = 36/2 ==> 18

OC = Apothem

tan θ = BC/OC

tan 18 = 2/a

a = 2 / tan 18

a = 2/0.325

a = 6.153

Perimeter = sides × length

= 10 × 4

P = 40 yd

Area = 1/2 × aP

= 1/2 × 6.153 × 40

A = 123.06 yd²

The area of the regular decagon is about 123 yd².

Problem 4 :

The Pentagon in Arlington, Virginia, is one of the world’s largest office buildings. It is a regular pentagon, and the length of each of its sides is 921 ft. what is the area of land that the Pentagon covers to the nearest thousand square feet?

Solution :

Number of sides = 5

Side length = 921 ft

∠AOB = 360/5 ==> 72

∠COB = 72/2 ==> 36

Length of an apothem,

tan θ = BC/OC

tan 36 = 921/a

a = 921/1.452

a = 634.3

Perimeter = sides × length

= 5 × 921

P = 4605 ft

Area = 1/2 × aP

= 1/2 × 634.3 × 4605

A = 1460475.75 ft²

The area of the regular pentagon is about 1460475 ft².

Problem 5 :

find the area of the shaded region.

Solution :

Area of the shaded region = area of circle - area of pentagon

Radius of the circle = 6

Area of circle = πr²

= 3.14(6²)

= 113.04 square units

Area of pentagon :

Perimeter = 5(6)

= 30

∠BOC = 72/2 ==> 36

AB = 6, BC = 3 units

tan 36 = BC/OC

0.726 = 3/OC

OC = 3/0.726

OC = 4.13

Area of pentagon = (1/2) x apothem x perimeter

= (1/2) x 4.13 x 30

= 15 x 4.13

= 61.95 square units

Area of shaded portion = 113.04 - 61.95

= 51.09 square units.

Problem 6 :

Solution :

Area of the shaded region = area of circle - area of hexagon

Area of hexagon :

∠BOC = 360/6

= 60

tan θ = opposite side / adjacent side

tan 60 = BC/OC

1.732 = 2/OC

OC = 2/1.732

OC = 1.15

sin θ = opposite side / Hypotenuse

sin 60 = BC/OB

1.732/2 = 2/OB

0.866 = 2/OB

OB = 2/0.866

OB = 2.3

Radius = 2.3 units

Perimeter of hexagon = 6(4)

= 24 units

Area of shaded portion = πr² -  (1/2) x apothem x perimeter

= 3.14(2.3)² -  (1/2) x 1.15 x 24

= 16.61 - 12 x 1.15

= 16.61 - 13.8

= 2.81 square units

Problem 7 :

Solution :

Area of the shaded region = area of sector - area of triangle

Area of sector = (θ/360) x πr²

= 45/360 x 3.14 x 10² 

= 39.25 square units

∠BOC = 45/2

= 22.5

sin θ = opposite side / Hypotenuse

sin 22.5 = BC/OB

0.386 = BC/10

BC = 10 (0.386)

= 3.86

AB = 2(BC)

= 2(3.86)

= 7.72

cos θ = Adjacent side / Hypotenuse

cos 22.5 = OC/OB

0.923 = OC/10

OC = 10 (0.923)

= 9.23

Area of triangle = (1/2) x base x height

= (1/2) x AB x OC

= (1/2) x 7.72 x 9.23

= 3.86 x 9.23

= 35.62 square units