FINDING THE AREA OF A REGULAR POLYGON USING SPECIAL RIGHT TRIANGLES

Use what you know about special right triangles to find the area of each regular polygon. Leave your answer in simplest form.

Example 1 :

Solution :

Since the given triangle is a regular polygon, every sides will have the same measure. So, it must be a equilateral triangle.

Number of small triangles created by joining 

In triangle ODC,

∠ODC = 90, ∠DCO = 30, ∠COD = 60

Smallest side = OD

In 30-60-90 special right triangle, the side which is opposite to 60 is √3(Smaller side).

hypotenuse (OC) = 2(Smallest side)

OC = 2OD

Then,

18 = 2OD and OD = 9

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

Perimeter = 3(18√3) ==> 54√3 cm

Area = (1/2) x 54√3 x 9

= 27√3 x 9

= 243√3 cm²

Problem 2 :

Solution :

Number of sides of a polygon = 6

by drawing lines from center to each vertex, we may draw six triangles of equal measure.

Angle measure of each triangle = 360/6 ==> 60

In triangle OAB,

OB = Hypotenuse, smaller side = AB

OB = 2(AB)

In special right triangle, the side which is opposite to 60 is √3(Smaller side).

OA = 4√3 then AB = 4 and OB = 2(4) ==> 8

Side length = 2(4) ==> 8

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

Perimeter = 6(8) ==> 48

Area = (1/2) x 48 x 4√3

= 24 x 4√3

= 96√3 cm²

Problem 3 :

Solution :

OA is perpendicular bisector, AB = 5 cm

∠AOB = 60, ∠ABO = 30

OA = Smaller side

AB = √3 (Smaller side)

5 = √3 (OA)

OA = 5/√3 (Apothem)

Side length = 10 cm

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

Perimeter = 3(10) ==> 30

height = 5/√3

Area = (1/2) x 30 x (5/√3)

= 15 x 5√3

= 75/√3

Rationalizing the denominator, we get

= (75/√3) x (√3/√3)

= 75√3/3

= 25√3 cm²

Problem 4 :

Solution :

Central angle of one triangle = 360/60

= 60 degree

OB = 8, AB = Smaller side

2AB = OB, then AB = 4

OA = 4√3 (Apothem)

Side length = 8

Perimeter = 6(8) ==> 48

Area = (1/2) x 48 x 4√3

= 24 x 4√3

= 96√3 square units.

Problem 5 :

A honeycomb is made up of regular hexagonal cells. The length of a side of a cell is 3 mm. What is the area of a cell?

Solution :

Inside the hexagon, we have six equal triangles and one angle measure is 60 degree.

In triangle ABD, <BAD = 30 degree, BC = 1.5 mm

Opposite side = BD = 1.5 mm, Hypotenuse = AB and Adjacent side = AD = a

tan θ = Opposite side / Adjacent side

tan 30 = BD / AD

1/√3 = 1.5/a

a = 1.5√3

Length of apothem = 1.5√3

Perimeter = 6 x side length

= 6 x 3

= 18 mm

Area of polygon = (1/2) x 18 x 1.5√3

= 9 x 1.5√3

= 13.5√3

= 13.5 x 1.732

= 23.38 mm²

So, the required area of the polygon is 23 mm²

Problem 6 :

You are painting a mural of colored equilateral triangles. The radius of each triangle is 12.7 in. What is the area of each triangle to the nearest square inch?

Solution :

Hypotenuse = 12.7 in

Opposite side = s/2

Adjacent side = ?

sin θ = Opposite side / Hypotenuse

Let x be the opposite side.

sin 30 = x/12.7

1/2 = x/12.7

12.7/2 = x

Opposite side = x = 6.35 inches

tan θ = Opposite side / Adjacent side

tan 30 = 6.35 / (s/2)

1/√3 = 2(6.35)/s

s = 2(6.35)(√3)

s = 2 x 6.35 x 1.732

s = 21.994

Side length = s = 22 inches

Perimeter = 3(22)

= 66 inches

Area of polygon = (1/2) x 6.35 x 66

= 33 x 6.35

= 209.55 square inches

So, the area of polygon is 210 square inches approximately.

Problem 7 :

Find the shaded area of the regular pentagon PENTA. The apothem measures about 2.0 cm.

Solution :

Since it is pentagon and it has 5 equal side, five equal triangles.

One angle is 72 degree and half of the angle is 36 degree.

Apothem = OM = 2 cm

Side length = EN

EM = x and MN = x/2

tan θ = Opposite side / Adjacent side

tan 36 = MN/OM

0.7265 = (x/2) / 2

0.7265 = x/4

x = 0.7265(4)

x = 2.906

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

= (1/2) x 2.9 x 2

= 2.9 cm²

Area of 3 equal triangles = 3 x 2.9

= 8.7 cm²