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Area of a hexagon is the space occupied within the boundaries (or sides) of the hexagon. Using the side and angle measurements, we can find the area of a hexagon. Hexagons can be commonly observed in different forms in our beautiful nature around.

1)

(Note: Diagonal is a line segment formed by joining any two non-adjacent vertices as shown in the figure on the right). Area of a regular hexagon can be easily calculated by considering the 6 equilateral triangles formed inside.

2)

A regular hexagon has all the 6 sides and 6 angles equal in measure. When diagonals passing through the center of the hexagon are drawn, 6 equilateral triangles of equal size are formed (as shown in the figure below). If the area of one equilateral triangle is calculated then we can easily calculate the area of the given regular hexagon.

Therefore, sum of all the interior angles of a convex regular hexagon S = (6 – 2) * 180°

Therefore, each interior angle of a convex regular hexagon = (6 – 2)/6 * 180°

Given a regular convex hexagon as shown in the figure above, where point ‘C’ is the center of the hexagon.

Triangle ABC is an equilateral triangle, as all the angles inside triangle ABC are equal to 60° (half of the interior angle 120°). Hence all its sides are also equal.

Therefore, let the side lengths of AB = BC = CA = s

CM is the perpendicular drawn to the side AB. Let CM = h

As ‘M’ becomes the midpoint of side AB, hence MB = s/2 (half of the side length of AB).

Now in triangle CMB, we can apply the Pythagorean Theorem to get the relationship between the height ‘h’ of the triangle, and the side length ‘s’.

Hence, h

So, h

Now, Area of triangle ABC = 1/2 * base * height.

This implies, Area A = 1/2 * s * h ==> A = 1/2 * s * (s * √3/2) ==> A = s

Therefore, Area of triangle ABC = s

Now, a regular hexagon consists of 6 such congruent equilateral triangles.

Hence, Area of a Regular Hexagon = 6 * s

Given that the side length, s = 8cm

Area of a regular hexagon, A = 3/2 * s

Hence, Area = 3/2 * 8

Given Area of a regular hexagon, A = √12 square feet and this can be further simplified as 2√3 ft

Area of a regular hexagon, A = 3/2 * s

Therefore, 3/2 * s

This gives: s

Hence, s

Therefore, the side length of the regular hexagon is 1.15ft (approximately)

Let us look at an example below where the information is given in the figure.

In the given figure, we observe that the side lengths are given and the lengths of the diagonals are also given.

We can see that the irregular hexagon is split into 4 triangles A, B, C and D.

Since the side lengths of each triangle are given, we can use Heron’s formula.

Triangle A:

s = (10+ 6 + 10)/2 ==> s = 13

Now Area of triangle A = √[s(s-a)(s-b)(s-c)] = √[(13* (13 -10) * (13 – 6) * (13 -10)]

Area of Triangle A = √(13 * 3 * 7 * 3)

Triangle B:

s = (10 + 11 + 7)/2 = 14

Area of triangle B = √[(14 * (14 – 10) * (14 – 7) * (14 – 11)] = √(14 * 4 * 7 * 3)

Similarly using Heron’s Formula as shown above, we get the areas of triangles C and D as well.

Area of Triangle C

Now, Area of the Irregular Hexagon = Area of Triangle A + Area of Triangle B + Area of Triangle C + Area of Triangle D

Area of the Hexagon = 28.6 + 34.3 + 16.9 + 19.9

Perimeter of a Regular Hexagon is equal to sum of all its side lengths.

Therefore, Perimeter of a Regular Hexagon of side length ‘s’ (as shown in the figure on the right) = s + s + s + s + s + s

This gives us Perimeter of Regular Hexagon = 6s

Perimeter of an Irregular Hexagon is also sum of all its side lengths.

If ‘a’, ‘b’, ‘c’, ‘d’, ‘e’, and ‘f’ are the side lengths of an irregular hexagon (as shown in the figure on the right),then Perimeter of an Irregular Hexagon = a + b + c + d + e + f