Properties of Hexagon: Types, Formula and Derivation

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Hexagons, is a type of polygon polygon. A polygon is a closed chain of a finite number of line segments. The area bounded by these line segments is called a polygon. Polygons are 2-dimensional figures. Whereas a hexagon is a 6-sided polygon with six corners. The sum of internal angles of a hexagon is 720°. A hexagon having all its sides of the same length and equal vertex angles are called a Regular Hexagon. 


Types of Hexagon 

There are 4 different types of hexagon, based on the length of its sides and the angles. These types are as follows: 

  • Regular Hexagon: The hexagon having equal sides and angles is called a Regular Hexagon. The sums of the interior and exterior angles of a regular hexagon are 720° and 360° respectively. 
  • Irregular Hexagon: An irregular hexagon has unequal sides and angles. The sum of the interior angles of an irregular hexagon is still 720° but each angle does not measure 120°.
  • Convex Hexagon: The hexagon having the interior angles measure less than 180° is a Convex Hexagon. It can either be regular or irregular, depending on the length of the sides and the angles. 
  • Concave Hexagon: The hexagon having at least one interior angle more than 180° is a Concave Hexagon. It is always an irregular hexagon. 

Properties of Hexagon 

Some of the important properties of the hexagon are as follows: 

  • A regular hexagon has 6 equal sides and 6 equal vertices. 
  • A regular hexagon has 9 diagonals. 
  • The sum of the interior angles of a regular hexagon is 720° and each interior angle is 120°.
  • The sum of the exterior angles of a regular hexagon is 360° and each exterior angle is 60°.
  • There are 6 axes of symmetry in a regular hexagon, with 3 of them passing through the diagonals and the other 3 passing through the midpoint of opposite sides. 
  • A regular hexagon can be divided into 6 equilateral triangles.
  • The circumcircle (a circle that passes through all the vertices of a polygon) of the hexagon has its centre overlap with that of the hexagon and its diameter is the same as the diameter of the hexagon. 
  • The incircle (a circle that passes through the mid-points of the sides of a polygon) of the hexagon also has its centre overlap with that of the hexagon. The incircle is tangential to all the sides of the hexagon. 

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Formulas for the Perimeter and Area of Hexagon 

For a regular hexagon, the formula for its perimeter is:

Perimeter = 6a

where  

a = Length of a side 

The formula for the area of a hexagon is: 

Area = (3\(\sqrt{3}\)/2) a^2 

where  

a = Length of a side

Derivation of the Formula for the Area of Hexagon 

Let’s have a hexagon with side “a” and divide it into 6 equal parts. This will give us 6 equilateral triangles.

Derivation of the Formula for the Area of Hexagon 

To find the area of the hexagon, we will find the area of one equilateral triangle and multiply it by 6.

the area of the hexagon, we will find the area of one equilateral triangle and multiply it by 6

Let’s take the equilateral triangle BXC. Let the ∠BXC be θ. As there is a complete angle (360°) at X and each triangle is congruent, 360° will be divided into 6 equal angles. 

Therefore, θ = 360° / 6 = 60°

Now, let’s drop a right-angle from X on the side BC. Let XY be h.

let’s drop a right-angle from X on the side BC

XY bisects the angle θ and the side BC. Therefore, ∠BXY = 30° and BY = a/2. 

Now, we know that: 

tanθ = Perpendicular / Base 

tan(∠BXY) = BY / XY 

tan(30°) = (a/2) / h

√3/3 = (a/2) / h

h = a/2 * 3/\(\sqrt{3}\)a

Now, the area of triangle BXC = ½ bh

Area = ½ x a x a/2 X 3/\(\sqrt{3}\)

Area = (3\(\sqrt{3}\)) (a^2)/4

As mentioned above, the area of hexagon = 6 x area of the triangle 

Area of Hexagon = 6 x (3\(\sqrt{3}\)) (a^2)/4

Area of Hexagon = (3\(\sqrt{3}\)/2) a^2


Hexagon is a 6-sided polygon with six corners. The sum of internal angles of a hexagon is 720°

  • A hexagon having all its sides of the same length and equal vertex angles are called a Regular Hexagon. 
  • An irregular hexagon has unequal sides and angles.
  • The hexagon having the interior angles measure less than 180° is a Convex Hexagon.
  • The hexagon having at least one interior angle more than 180° is a Concave Hexagon.
  • Area of a Hexagon: 6 x (3\(\sqrt{3}\)) (a^2)/4 
  • The perimeter of a Hexagon: 6a

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Sample Questions 

Ques 1. Find the measure of interior angle if a regular polygon has 10 sides. (2 marks)

Ans.  For a regular polygon of “n” sides, the measure of interior angle: 

180 - {360/n} 

For n = 10, 

Interior angle = 180 - {360/10} = 144° 

Ques 2. If the interior angle of a regular polygon is 80°, what regular polygon will we get? (3 marks)

Ans. Interior angle = 180 - {360/n}

80° = 180 - {360/n}

360/n = 180° - 80°

360/n = 100°

n = 3.6 

Since “n” is not an integer, no such regular polygon is possible. 

Ques 3. If a regular polygon has 14 diagonals, which polygon is it? (3 marks)

Ans. No. of diagonals of a regular polygon = n(n-3)/2

14 = n(n-3)/2

28 = n^2 - 3n

n^2 - 3n - 28 = 0

n^2 - 7n + 4n - 28 = 0

n(n - 7) + 4(n - 7) = 0

Therefore, n = 7, -4 

Since the number of sides cannot be negative, n = 7. Hence, we have a regular heptagon. 

Ques 4. Find the sum of exterior angles of a regular nonagon. (2 marks)

Ans. Number of sides of a nonagon = n = 9

Exterior angle = 360/n = 360/9 = 40°

Therefore, the sum of exterior angles = 40° x 9 = 360°

Ques 5. Find out the area and perimeter of a hexagon with the length of the side equal to 5cm. (2 marks)

Ans. a = 5cm 

Area of Hexagon =  (3\(\sqrt{3}\)/2) a^2 =  (3\(\sqrt{3}\)/2) (5x5) = 75\(\sqrt{3}\)/2 sq. cm 

Perimeter of Hexagon = 6a = 6 x 5 = 30 cm

Ques 6. If the area of a hexagon is 60\(\sqrt{3}\) sq. cm, find the length of its side and use it to find its perimeter. (3 marks)

Ans. Area of Hexagon = (3\(\sqrt{3}\)/2) a^2

60\(\sqrt{3}\) = (3\(\sqrt{3}\)/2) a^2

40 = a^2

a = 2\(\sqrt{10 }\) cm

Perimeter of Hexagon = 6a = 6 x 2\(\sqrt{10 }\) = 12\(\sqrt{10 }\) cm

Ques 7. If the perimeter of a hexagon is 42cm, find its area. (2 marks)

Ans. Perimeter of a hexagon = 6a

42 = 6a

a = 42/6 = 7cm 

Area of Hexagon = (3\(\sqrt{3}\)/2) a^2 = (3\(\sqrt{3}\)/2) (7 x 7) = 127.305 sq. cm

CBSE X Related Questions

1.

Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.

(v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

      2.

      Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

          3.
          A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

              4.
              Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
              (i) 2, 4, 8, 16, . . . .
              (ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
              (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
              (iv) – 10, – 6, – 2, 2, . . .
              (v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
              (vi) 0.2, 0.22, 0.222, 0.2222, . . . .
              (vii) 0, – 4, – 8, –12, . . . .
              (viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
              (ix) 1, 3, 9, 27, . . . .
              (x) a, 2a, 3a, 4a, . . . .
              (xi) a, \(a^2, a^3, a^4,\)  . . . .
              (xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
              (xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
              (xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
              (xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

                  5.
                  Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) \(x + y = 5\),\( 2x + 2y = 10\) (ii)\( x – y = 8 , 3x – 3y = 16\) (iii) \(2x + y – 6 = 0\) , \(4x – 2y – 4 = 0\) (iv) \(2x – 2y – 2 = 0,\) \( 4x – 4y – 5 = 0\)

                      6.
                      The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

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