Heron's Formula

Heron's formula is a popular method for calculating the area of a triangle when the lengths of its three sides are known. It was introduced by Heron of Alexandria in his book "Metrica". This formula applies to all types of triangles, including right-angled, equilateral, and isosceles.

According to this formula, the area of a triangle is directly proportional to the square root of the semi-perimeter of the triangle. The area "A" of a triangle with sides of lengths a, b, and c is given by:

A = \sqrt{s(s - a)(s - b)(s - c)}

Where, 

• A is area of Triangle ABC, 
• a, b, c are lengths of the sides of the triangle, and 
• s is semi-perimeter = (a + b + c)/2.

Below is the demonstration of Heorn's Formula for finding the area of the triangle with sides of length a, b, c:

History of Heron's Formula

In 60 CE, Heron of Alexandria gave a formula which is known as the Heron's Formula. Heron was a Greek Mathematician who gave the formula for finding the area of a triangle when the length of all sides of the triangle is given. He used this formula for solving various problems in triangles and trigonometry. He proved the Law of Cosine and the Law of Cotangent using this formula.

Derivation of Heron’s Formula

There are two primary methods to derive Heron’s formula: using the Pythagorean Theorem or the Cosine Rule.

Heron's Formula Derivation Using Pythagoras Theorem

To begin the derivation, recall that the area of a triangle can be expressed as:

Area of a Triangle = (1/2) × b × h 

Where,

• b is the base, and 
• h is the height.

The image shown below shows the triangle ABC.
Start by breaking the triangle into two smaller right-angled triangles and applying the Pythagorean Theorem to each, we can solve for the area in terms of the sides a, b, and c.

Draw a perpendicular AD on BC

From the ∆ ABD,

a² = x² + h²
⇒ x² = (a² − h²)....(I)
⇒ x = √(a² − h²)....(ii)

Consider the  ∆ACD,

(b − x)² + h² = c²
⇒ (b − x)² = c² − h²
⇒ b² − 2bx + x² = c² – h²

Putting the value of x and x² from equations (i) and (ii) in the above equation, we get

b² – 2b√(a² − h²)+ a² − h² = c² − h²
⇒ b² + a² − c² = 2b√(a² − h²)

Squaring on both sides, we get;

(b² + a² – c²)² = 4b²(a² − h²)
⇒ {(b² + a² – c²)²) / 4b² = (a² − h²)
⇒ a² + {(b² + a² – c²)²) / 4b² = h²

Simplifying, we get
h² = (a + b + c)(b + c - a)(a + c - b)(a + b - c) / 4b² 

Now, 2s = a + b + c, where s is the semi-perimeter of the triangle.

h² = 2s(2s - 2a)(2s - 2b)(2s - 2c) / 4b²
⇒ h = √[2s(2s - 2a)(2s - 2b)(2s - 2c)] / 2b
⇒ h = 2 × √[s(s - a)(s - b)(s - c)] / b...(iii)

From, area of triangle = 1/2 × b × h

Now, area of triangle = 1/2 × {b × 2 × √[s(s - a)(s - b)(s - c)]} / b, cancelling out the similar terms we get:

Area of Triangle (A) = \sqrt{s(s-a)(s-b)(s-c)}

Heron's Formula Derivation Using Cosine Rule

Heron's Formula can also be easily solved using the Cosine Rule. Now for any triangle ABC if the sides of the triangle are a, b, and c and their opposite angles are, α, β, and γ.

The law of cosine states, cos γ = (a² + b² - c²)/2ab

Using Trigonometric identites

cos² γ + sin² γ = 1
⇒ sin γ = √(1 - cos² γ)
⇒ sin γ = √[1 - {(a² + b² - c²)/2ab}²]
⇒ sin γ = √[(4a²b² - (a² + b² + c²)²]/2ab

If the base of the triangle is a then its altitude is b sin γ

Area of Triangle = 1/2 base × height
⇒ Area of Triangle = 1/2 × a × b sin γ
⇒ Area of Triangle = 1/2 ab × √[(4a²b² - (a² + b² + c²)²]/2ab
⇒ Area of Triangle = 1/4 √[c² - (a - b)²][(a + b)² - c²]
⇒ Area of Triangle = √(b + c - a)(a + c - b)(a + b - c)(a + b + c)/16

Area of Triangle = √s(s - a)(s - a)(s - b)

Where s = (a + b + c)/2 is the semi perimeter.

Heron’s Formula for Equilateral Triangle

For an equilateral triangle, all sides are equal. Now, the semi-perimeter of the equilateral triangle is (s) = (a + a + a) / 2
⇒ s = 3a / 2
where a is the length of the side.

Now, using Heron’s Formula,
Area of Equilateral Triangle = √(s(s – a)(s – a)(s – a), upon multiplication, the formula becomes:

Area of Equilateral Triangle = √3 / 4 × a²

Heron's Formula for Isosceles Triangle

Isosceles Triangle is a triangle that has two equal sides, their area can be easily calculated using Heron's Formula. For any isosceles triangle △ABC where sides AB = a, and BC = a are equal, and the third side is CA = b. The formula for its area is,

Area of iscosceles triangle ABC(A) = √s(s - a)(s - a)(s - b)

Where s is the semi-perimeter i.e., s = (a + a + c)/2.
⇒ s = (a + a + b)/2
⇒ s = (2a + b)/2

Simplifying, A = √s(s - a)(s - a)(s - b)

Substituting "s" = (2a + b)/2:

⇒ s − a = (2a + b)/2 ​− a = b/2​
⇒ s − b = (2a + b)/2 ​− b = (2a − b)/2​

Thus: a =\sqrt{(\frac{2a + b}{2})(\frac{b}{2})(\frac{b}{2})(\frac{2a - b}{2})}
Simplyfy further: A = \sqrt{\frac{(2a+b)(2a-b)(b^2)}{2} }
A = 1/4√(4a² - b²)(b²)

Final Formula:

A = \frac{b}{2}\sqrt{a^2 - \frac{b^2}{4}}

Note: Heron's formula for a Scalene triangle remains the same as the default formula, as it applies to all types of triangles, including scalene, isosceles, and equilateral, provided the lengths of the three sides are known.

Heron's Formula for Area of Quadrilateral

Heron's formula is used to determine the formula for the area of the quadrilateral. We can divide the quadrilateral into two separate triangles using any one of its diagonals and then the area of the two separate triangles is calculated using Heron's Formula.

The area of the quad ABCD is calculated by dividing it into two triangles using its diagonal. Let's say we join the vertices A and C to form the diagonal AC then we divide it into two triangles △ABC and △ADC. If we take the length of sides of the quadrilateral as,

AB = b, BC = c, CD = d, and DA = a, and the length of diagonal AC is e then its area is calculated using,

Area of quad ABCD = Area of △ABC + Area of △ADC...(i)

Area of triangle ABC

Area of △ ABC = √(s₁(s₁ - b)(s₁ - c)(s₁ - e))

Where s₁ = (b + c + e)/2.

Area of triangle ADC

Area of △ ADC = √(s₂(s₂ - d)(s₂ - a)(s₂ - e))

Where s₂ = (d + a + e)/2.

Thus,
form eq(i)

Area of quad ABCD = √(s₁(s₁ - b)(s₁ - c)(s₁ - e)) + √(s₂(s₂ - d)(s₂ - a)(s₂ - e))

Where,

• s₁ = (a + b + e)/2, and
• s₂ = (a + d + e)/2

Applications of Heron's Formula

Heron's formula has various applications and some of the important applications of Heron's Formula are,

• For finding the area of the triangle if the sides of the triangle are given
• For finding the area of the quadrilateral the length of all the sides and the length of the diagonal are given.
• For finding the area of any polygon its sides and the length of all the principal diagonals are given.

Read in Detail: Applications of Heron’s Formula

Heron's Formula Examples

Example 1: Calculate the area of a triangle whose lengths of sides a, b, and c are 14cm,13cm, and 15 cm respectively.
Solution:

Given:  
a = 14cm
b = 13cm
c = 15cm

Firstly, we will determine semi-perimeter(s) s = (a + b + c)/2
⇒ s = (14 + 13 + 15)/2
⇒ s = 21 cm

Thus, A = √(s(s – a)(s – a)(s – a)
⇒ A = √(21(21 – 14)(21 – 13)(21 – 15)
⇒ A = 84 cm²

Example 2: Find the area of the triangle if the length of two sides is 11cm and 13cm and the perimeter is 32cm.
Solution:

Let a, b and c be the three sides of the triangle.
a = 11cm
b= 13 cm
c = ?

Perimeter = 32cm

As we know, Perimeter equals to the sum of the length of three sides of a triangle.
Perimeter = (a + b + c)
⇒ 32 = 11 + 13 + c
⇒ c = 32 - 24
⇒ c = 8 cm

Now as we already know the value of perimeter,
s = perimeter / 2
⇒ s = 32 / 2
⇒ s =16 cm

As, a = 11cm, b = 13 cm, c = 8 cm, s = 16 cm
Thus,  A = √(s(s – a)(s – a)(s – a)
⇒ A = √(16(16 – 11)(16 – 13)(16 – 8)
⇒ A = 43.8 cm²

Example 3: Find the area of an equilateral triangle with a side of 8 cm.
Solution: 

Given,
Side = 8 cm

Area of Equilateral Triangle = √3 / 4 × a²
⇒ Area of Equilateral Triangle = √3 / 4 × (8)²
⇒ Area of Equilateral Triangle = 16 √3 cm²

Practice Questions: Heron's Formula Questions with solutions.

Related Reads

• Perimeter of the triangle
• Three Dimensional Geometry
• Hypotenuse Side Theorem
• Acute Angle Triangle
• Right Angled Triangle