Heron's Formula - Definition, Proof, Theorem and Examples

Lesson Overview

• What Is Heron's Formula?
• Heron's Formula Definition
• Proof of Heron's Formula
• Heron's Formula for Triangle Area Calculation
• Applications of Heron's Formula
• Solved Examples On Heron's Formula

Heron's Formula is a mathematical method for finding the area of a triangle using only the lengths of its sides. It is useful when height is not available and works for all types of triangles. 

This ancient formula, developed by the Greek mathematician Heron, transforms basic side measurements into a straightforward area calculation.

What Is Heron's Formula?

Heron's Formula provides a way to calculate the area of a triangle without requiring the height. Instead, it uses the semi-perimeter, which is half the perimeter, along with the lengths of the three sides. 

This makes it a versatile tool for solving geometric problems where height measurements are impractical.

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Heron's Formula Definition

Heron's Formula states that the area of a triangle can be determined using the semi-perimeter, denoted as s, and the side lengths a, b, and c. 

The formula is:

Area = √[s(s-a)(s-b)(s-c)]  

where 's' is the semi-perimeter of the triangle, calculated as:

s = (a + b + c)/2

Example 

Given a triangle with side lengths 7, 8, and 9, we can calculate the area using Heron's Formula.

1. Calculate the semi-perimeter:
   s = (7 + 8 + 9) / 2 = 12
2. Apply the formula:
   Area = square root of 12 × (12 - 7) × (12 - 8) × (12 - 9)
   Area = square root of 12 × 5 × 4 × 3
   Area = square root of 720
   Area ≈ 26.83 square units.

So, the area of the triangle is approximately 26.83 square units.

Proof of Heron's Formula

Heron's Formula can be derived using two main approaches:

1. Using the Cosine Rule
2. Using Pythagoras' Theorem

The Law of Cosines

For any triangle with sides a, b, and c, and opposite angles α, β, and γ, the Law of Cosines states:

Again, using trig identity, we have 

Here, Base of triangle = a

Altitude = b sinγ

Now, 

Using Pythagoras Theorem

Area of a Triangle with 3 Sides

Area of ∆ABC is given by

A = 1/2 bh _ _ _ _ (i)

Draw a perpendicular BD on AC

Consider a ∆ADB

x2 + h2 = c2

x2 = c2 − h2-(ii)

⇒ x = √(c2−h2)−−−−−−-(iii)

Consider a ∆CDB,

(b−x)2 + h2 = a2

(b−x)2 = a2 − h2

b2 − 2bx + x2 = a2–h2

Substituting the value of x and x2 from equation (ii) and (iii), we get

b2 – 2b√(c2−h2)+ c2−h2 = a2 − h2

b2 + c2 − a2 = 2b√(c2 − h2)

Squaring on both sides, we get;

(b2+c2–a2)2 = 4b2(c2−h2)

The perimeter of a ∆ABC is given by:

P = a + b + c

Thus, substituting a + b + c = P in the above equation, we get;

Substituting the value of h in equation (i), we get;

Note: Heron's formula is applicable to all types of triangles and the formula can also be derived using the law of cosines and the law of Cotangents.

Heron's Formula for Triangle Area Calculation

Heron's formula for the area of a triangle is derived using the Pythagorean theorem, the area formula of a triangle, and algebraic identities. 

Consider a triangle with side lengths a, b, and c. 

Let s be the semi-perimeter, P be the perimeter, and A be the area of the triangle. 

Assume that side b is divided into two parts, p and q, by a perpendicular h dropped from vertex B onto side AC at point M.

The area of the triangle is given by the formula:

A = (1/2) * b * h

To calculate h, we start with the relationship:

b = p + q

From this, we have:

q = b - p (1)

Squaring both sides:

q^2 = b^2 + p^2 - 2bp (2)

Adding h^2 to both sides:

q^2 + h^2 = b^2 + p^2 - 2bp + h^2 (3)

Using the Pythagorean theorem in triangle BCM:

h^2 + q^2 = a^2 (4)

Applying the Pythagorean theorem in triangle MBA:

p^2 + h^2 = c^2 (5)

Substituting the values from equations (4) and (5) into equation (3):

q^2 + h^2 = b^2 + p^2 - 2bp + h^2

This simplifies to:

a^2 = b^2 + c^2 - 2bp

Solving for p:

p = (b^2 + c^2 - a^2) / 2b (6)

From equation (5):

p^2 + h^2 = c^2

Thus:

h^2 = c^2 - p^2

Substituting equation (6) into this expression for h^2:

h^2 = (c + (b^2 + c^2 - a^2) / 2b) * (c - (b^2 + c^2 - a^2) / 2b)

Simplifying further:

h^2 = ((b + c)^2 - a^2) / 2b * (a^2 - (b - c)^2) / 2b

This results in:

h^2 = ((b + c + a)(b + c - a)(a + b - c)(a - b + c)) / 4b^2 (7)

The perimeter of the triangle is P = a + b + c, and the semi-perimeter is s = P / 2, so:

2s = a + b + c (8)

Substituting equation (8) into equation (7):

h^2 = ((b + c + a)(b + c - a)(a + b - c)(a - b + c)) / 4b^2

This simplifies to:

h^2 = 16s(s - a)(s - b)(s - c) / 4b^2

Thus:

h = 2√(s(s - a)(s - b)(s - c)) / b (9)

The area A of the triangle is:

A = (1/2) * b * h

Substituting equation (9) into this formula:

A = (1/2) * b * 2√(s(s - a)(s - b)(s - c)) / b

Simplifying:

A = √(s(s - a)(s - b)(s - c))

Therefore, the area of triangle ABC is:

A = √(s(s - a)(s - b)(s - c))

This is the formula for the area of a triangle known as Heron's formula

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Applications of Heron's Formula

Heron's Formula has several practical uses:

• It is commonly used to calculate the area of various types of triangles when the lengths of all three sides are known.
• It can also help find the area of a quadrilateral when the lengths of its sides are given.

Solved Examples On Heron's Formula

Example 1: Find the area of a triangle with side lengths 6, 8, and 10.

Solution:

1. Semi-perimeter, s = (6 + 8 + 10) / 2 = 12
2. Area = square root of [12 × (12 - 6) × (12 - 8) × (12 - 10)]
   Area = square root of [12 × 6 × 4 × 2]
   Area = square root of 576
   Area = 24 square units.

Example 2: Find the area of a triangle with side lengths 5, 12, and 13.

Solution:

1. Semi-perimeter, s = (5 + 12 + 13) / 2 = 15
2. Area = square root of [15 × (15 - 5) × (15 - 12) × (15 - 13)]
   Area = square root of [15 × 10 × 3 × 2]
   Area = square root of 900
   Area = 30 square units.

Example 3: Find the area of a triangle with side lengths 7, 8, and 9.

Solution:

1. Semi-perimeter, s = (7 + 8 + 9) / 2 = 12
2. Area = square root of [12 × (12 - 7) × (12 - 8) × (12 - 9)]
   Area = square root of [12 × 5 × 4 × 3]
   Area = square root of 720
   Area ≈ 26.83 square units.

Example 4: Find the area of a triangle with side lengths 10, 24, and 26.

Solution:

1. Semi-perimeter, s = (10 + 24 + 26) / 2 = 30
2. Area = square root of [30 × (30 - 10) × (30 - 24) × (30 - 26)]
   Area = square root of [30 × 20 × 6 × 4]
   Area = square root of 14400
   Area = 120 square units.

Example 5: Find the area of a triangle with side lengths 11, 60, and 61.

Solution:

1. Semi-perimeter, s = (11 + 60 + 61) / 2 = 66

Area = square root of [66 × (66 - 11) × (66 - 60) × (66 - 61)]
Area = square root of [66 × 55 × 6 × 5]
Area = square root of 108900
Area = 330 square units.