Herons Formula

Herons Formula (Hero's Formula):
Heron gave the famous formula for finding the area of triangle in terms of its three sides.Herons Formula is named after
Hero of Alexandria. It is used when all the three sides are given and there is no height or altitude.Hero's Formula is given below :



Where,
is the semi-perimeter of ΔABC.

Note : It is applicable to all types of triangles whether it is right angled or equilateral or isosceles triangle.

Some solved examples :

1) Find the area of a triangle whose sides are 13 cm, 14 cm and 15 cm .
Solution :
a =13 cm , b = 14 cm and c = 15 cm.
∴ s = (a + b + c )/ 2
s = (13 + 14 + 15) / 2 = 21 cm.
By Heron’s formula


Area = √( 21 x 8 x 7 x 6 )

Area = √ ( 7 x 3 x 4 x 2 x 7 x 2 x 3 )

Area = √ ( 7
2 x 2 2 x 2 2 x 3 2 )

Area = 7 x 2 x 2 x 3

Area = 84 cm
2
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2) The perimeter of triangular field is 450 m and its sides are in the ratio of 13 : 12: 5. Find the area of the triangle.
Solution:
It is given that the sides a, b, c of the triangle are in the ratio of 13 :12: 5
⇒ a: b : c = 13 : 12 : 5 ⇒ a= 13x , b = 12x and c = 5x
The perimeter of triangular field = 450 m
⇒ 13 x +12x + 5x = 450
⇒ 30x = 450
∴ x = 15
∴ each sides are, a= 13 x 15 =195 m ; b = 12 x 15 = 180 m
and c = 15 x 5 = 75 cm.
By Heron’s formula


Area = √ ( 225 x 30 x 45 x 150 )
Area = √( 45,562,500)
Area = 6750 m
2
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3) A traffic signal board, indicating “SCHOOL AHEAD is an equilateral triangle with sides ‘a’. Find the area of the signal board using Heron’s formula. If its perimeter is 180 cm. What will be the area of the signal board?
Solution :
Perimeter = 180
a + a + a = 180
3a = 180
a = 60 cm
s = semi-perimeter = Perimeter /2
s= 180 / 2 = 90 cm By Heron’s formula


Area = √( 90 x 30 x 30 x 30 )
Area = √( 2,430,000 )
Area = 1558.85 cm
2
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Some Applications of Heron’s Formula

Heron’s formula can be used to find the area of irregular quadrilateral if its all sides are given. For that we can convert that quadrilateral into triangles and then apply Heron’s formula.

Solved Examples :

1) Find the area of quadrilateral ABCD, in which AB = 7 cm, BC = 6 cm,
CD = 12 cm , DA = 15 cm and AC = 9 cm.
Solution :

For ΔABC
s = ( a + b + c ) / 2 =( 6 + 7 + 9 ) / 2 = 22 / 2 = 11 cm
By Heron’s formula


Area of ΔABC = √( 11 x 5 x 4 x 2 )
⇒ Area of ΔABC = √(440)
⇒ Area of ΔABC = 20.98 cm
2
For ΔACD,
s = ( 9 + 12 + 15 ) / 2 = 36 / 2 = 18 cm
By Herons formula


∴ Area of ΔACD = √( 18 x 9 x 6 x 3)
⇒ Area of ΔACD = √ (2916 )
⇒ Area of ΔACD = 54 cm
2
Hence, the area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD
= 20.98 + 54
∴ Area of quadrilateral = 74.98 cm
2
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2) An umbrella is made by stitching 10 triangular pieces of cloth of two different colors red and blue. Each piece is measuring 20 cm ,50 cm and 50 cm. How much cloth of each color is required for the umbrella?

Solution :
Sides of triangular cloth are a = 20 cm, b = 50 cm and c = 50 cm
s = ( 20 + 50 + 50 ) /2 = 60 cm
By Herons formula,


∴ Area of triangular piece = √( 60 x 40 x 10 x 10 )
⇒ Area = √( 240,000)
⇒ Area = 489.89 com
2
∴ The area of each color = 5 x 489.89 = 2,449 .48 cm
2

Mensuration : Area and Perimeter

Perimeter and Area of Irregular Shape
Area and Perimeter of the Rectangle
Area of Square (perimeter of square)
Perimeter of Parallelogram(Area of Parallelogram)
Area of Rhombus(Perimeter of rhombus)
Area of Trapezoid (Trapezium)
Triangle Area (Perimeter of triangle)
Herons Formula

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