## Area of a Pentagon

A polygon with five sides is called a pentagon. It can be a simple polygon or a self-intersecting one. In a regular pentagon, the five sides are equal in length and the internal angles are equal – 108^{0}. The exterior angles are 72^{0}. In an irregular pentagon, this is not the case- the sides may not be equal and the angles can be different. There can be five diagonals in a regular pentagon. A convex pentagon is one in which the five vertices point outward. In a concave pentagon, the opposite happens. Let us learn the various methods to find out the area of a pentagon.

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## The apothem of a pentagon

The apothem is a line from the centre of the pentagon drawn to the midpoint of a side such that it is perpendicular to the side. (this is different from the radius. The radius extends from the centre to the vertex.)

Calculating the area of a regular pentagon.

- The Pentagon is divided into five triangles, by drawing five radiuses from the centre to radiate out to each of the vertices.
- Now, the apothem is drawn, which is the height of each triangle.
- Thus, the area of each triangle is calculated by A = ½ * base * height, where base = side length of the Pentagon and height = length of the apothem of the triangle.
- Area of Pentagon = perimeter × apothem / 2
- The small triangle is right-angled and so we can use sine, cosine and tangent to find how the
**side**,**radius**,**apothem**and**5**(number of sides) are related:

sin(π/n) = (Side/2) / Radius | → | Side = 2 × Radius × sin(π/5) |

cos(π/n) = Apothem / Radius | → | Apothem = Radius × cos(π/5) |

tan(π/n) = (Side/2) / Apothem | → | Side = 2 × Apothem × tan(π/5) |

If only the side length is known and apothem is not known, we proceed differently.

- Divide the pentagon into triangles.
- Each triangle is subdivided into two right-angled triangles.
- The angle of the centre of the pentagon is 36
^{0}(360^{0}/10). - Thus, tan 36 = base/height.
- Here, base = ½ of the length of the side of the pentagon.
- Thus, height, or length of apothem = base length/ tan 36 = (side length/2)/ tan 36
- Now if height is known, area of the triangle = ½ * base * height = ½ * side * apothem.
- Thus, area of octagon = 5*1/2*s*a = (5*s
^{2})/(4*tan 36^{0}) = (5*s*^{2}) / (4√(5-2√5)).

If only the perimeter is known

- P = 5s
- S= p/5 where p = perimeter and s = length of the side.
- Area of the pentagon = ½* p * a where a = apothem length.

If only the radius is known,

- Area = (5/2)
*r*^{2}sin(72º), where*r*is the radius.

If it is an irregular pentagon, the easiest way is to divide it into a number of geometric figures, right angled triangles, squares or otherwise, and then proceed using appropriate formulas.

**Examples**

**Question 1: **Find the area of a pentagon of side 5 cm and apothem length 3 cm.

**Solution: **

Given,

s = 5 cm

a = 3 cm

Area of a pentagon

= 5/2 s*a

= 5/2 * 5 * 3 cm^{2}

= 75/2 cm^{2}

= 37.5 cm^{2}

**Question 2: **Find the area of a pentagon of side 12 cm and apothem height 7 cm.

**Solution: **

Given,

s = 12 cm

r = 7 cm

Area of a pentagon

= 5 / 2 * s *a

= 5/2 * 12 * 7 cm^{2
}

= 5∗12∗7 /2 cm^{2
}

= 420 / 2 cm^{2
}

= 210 cm^{2}

^{ }

**Question:** Find the area of a pentagon of side 10 cm and apothem length 5 cm.

**Solution:
**Given,

s = 10 cm

a = 5 cm

Area of a pentagon

= 5 / 2 * s *a

= 5 / 2 × 10 × 5 cm^{2
}= 125 cm^{2}

** **Consider one triangle formed by joining two of the adjacent vertices with the center of a pentagon.

The interior angle O = 360*o*5 = 72º.

Since the triangle AOB is an isosceles triangle (AO = BO) ≤ A =≤ B = n.

So the measure of angle A = >

In triangle ∆ AOB = 72º + n + n = 180º

2n = 180º – 72º

2n = 108º.

n = 108*o*2 = 54º.

Now use the trigonometric ratio to get the value of h (height of the triangle which is same as the apothem of the pentagon.)

tan A = *h*(*half* *of* *AB*)

So h = 12 s × tan 54º

= 0.3369 s

So area of the triangle = 12 s h = 12 s ×(0.3369 s)

=0.16845 *s*2

So we can represent the area of a regular pentagon only in terms of its side length

= 5 × Area of each triangle

= 5 × (0.16845 *s*2)

= 0.84225 *s*2

**Question:** Find the area of the given regular pentagon whose each side measures 8 cm.

**Solution:**

Apply the formula for area with side using trigonometry

A = 0.84225 *s*2

= 0.84225 × (8)2

= 53.094 sq cm.

Check out our article on Geometry formulas here.