Grade 9
  1. Number Systems  1.1. Real Numbers  1.2. Rational Numbers  1.3. Irrational Numbers  1.4. Properties of Real Numbers  1.5. Laws of Exponents and Surds  1.6. Representation on the Number Line  1.7. Decimal Representation of Real Numbers
  2. Polynomials  2.1. Definition and Types of Polynomials  2.2. Degree of a Polynomial  2.3. Zeroes of a Polynomial  2.4. Remainder Theorem  2.5. Factorization of Polynomials  2.6. Algebraic Identities  2.7. Polynomial Equations and Roots
  3. Coordinate Geometry  3.1. Cartesian System  3.2. Plotting Points on the Plane  3.3. Distance Formula  3.4. Section Formula  3.5. Area of a Triangle  3.6. Coordinates of Midpoint and Centroid
  4. Linear Equations in Two Variables  4.1. Solutions of a Linear Equation  4.2. Graphical Method  4.3. Algebraic Method  4.4. Applications of Linear Equations  4.5. Linear Inequalities
  5. Introduction to Euclidean Geometry  5.1. Basic Definitions and Terms  5.2. Axioms and Postulates  5.3. Theorems on Angles and Lines  5.4. Properties of Parallel Lines  5.5. Construction of Triangles  5.6. Congruence Axioms
  6. Lines and Angles  6.1. Angle Pair Relationships  6.2. Parallel Lines and Transversal  6.3. Properties of Angles Formed by Parallel Lines  6.4. Angle Sum Property of a Triangle  6.5. Types of Angles
  7. Triangles  7.1. Congruence of Triangles  7.2. Criteria for Congruence  7.3. Properties of Triangles  7.4. Inequalities in a Triangle  7.5. Similarity of Triangles  7.6. Pythagorean Theorem
  8. Quadrilaterals  8.1. Properties of Quadrilaterals  8.2. Types of Quadrilaterals  8.3. Midpoint Theorem  8.4. Properties of Parallelograms  8.5. Trapezium and Kite Properties  8.6. Diagonals and their Properties
  9. Areas of Parallelograms and Triangles  9.1. Area of a Parallelogram  9.2. Area of a Triangle  9.3. Proofs of Area Theorems  9.4. Applications of Area Theorems
  10. Circles  10.1. Definitions and Terms  10.2. Properties of a Circle  10.3. Chord Properties  10.4. Tangents to a Circle  10.5. Cyclic Quadrilaterals  10.6. Arcs and Angles Subtended
  11. Constructions  11.1. Basic Constructions  11.2. Constructing Bisectors  11.3. Constructing Triangles  11.4. Constructing Tangents to a Circle  11.5. Constructing Angles of Given Measure
  12. Heron's Formula  12.1. Area of a Triangle Using Heron’s Formula  12.2. Application to Different Triangles and Quadrilaterals  12.3. Proof of Heron’s Formula
  13. Surface Areas and Volumes  13.1. Surface Area of a Cube Cuboid and Cylinder  13.2. Volume of a Cube Cuboid and Cylinder  13.3. Surface Area and Volume of a Cone  13.4. Surface Area and Volume of a Sphere and Hemisphere  13.5. Conversion of Units  13.6. Volume of a Prism
  14. Statistics  14.1. Collection of Data  14.2. Presentation of Data  14.3. Measures of Central Tendency  14.4. Mean Median and Mode  14.5. Graphical Representation of Data  14.6. Range and Measures of Dispersion
  15. Probability  15.1. Introduction to Probability  15.2. Experimental Probability  15.3. Theoretical Probability  15.4. Simple Problems on Probability

Grade 9 → Heron's Formula ↓


Proof of Heron’s Formula


Heron's formula is a mathematical formula used to find the area of a triangle when you know the lengths of all three sides. Named after the Greek engineer and mathematician, Hero of Alexandria, this formula is useful because it avoids the need to find the height of the triangle, which isn't always easy.

Understanding the formula

Heron's formula states that if the lengths of the sides of a triangle are a, b, and c, and s is the semi-perimeter of the triangle (which is half of the perimeter), then the area A of the triangle is given by:

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

Visualizing a triangle

To better understand Heron's formula, let's first imagine a triangle. Below is a simple representation of a triangle with its sides labeled:

C A B

A triangle with sides a, b, and c.

Analysis of the formula

To understand how Heron's formula works, we need to explore its derivation and see how it relates to the basic principles of geometry. Let's explore the components one by one.

Step 1: Calculate the semi-perimeter

First, we calculate the semi-perimeter of the triangle. The semi-perimeter s is half the sum of the three sides of the triangle:

    s = (a + b + c) / 2

This gives us a measurement that is directly related to the perimeter of the triangle, but is easier to calculate by halving it.

Step 2: Calculate the area

The next part of Heron's formula involves adding square roots and several subtractions and multiplications:

    A = √(s * (s - a) * (s - b) * (s - c))

Here, (s - a), (s - b) and (s - c) are the terms we subtract from the semi-perimeter. This is because we are essentially balancing the contribution of each triangle's side to the area.

Proof of Heron's formula

Use of algebra and geometric identities

Let's derive Heron's formula using geometry and algebra. We'll start with some known formulas and manipulate them to get Heron's result.

Consider a triangle whose sides are a, b and c

The standard formula for finding the area of a triangle is:

    A = 1/2 * base * height

However, calculating the height without specific angles or additional information can be cumbersome. Instead, we will use the cosine rule to express the angles in terms of the sides.

Deriving the derivative using the cosine rule

The cosine rule is:

    c^2 = a^2 + b^2 - 2ab * cos(C)

This can be rearranged to solve for angle C:

    cos(C) = (a^2 + b^2 - c^2) / (2ab)

Involving trigonometry

Using half of the side lengths and the angle between them, we calculate the area:

    A = 1/2 * ab * sin(C)

According to the Pythagorean identity, sin^2(C) = 1 - cos^2(C) Substituting and simplifying these with the earlier expression for cos(C) gives a formula similar to Heron's.

Connecting it all together

Ultimately, by expanding and simplifying the expressions obtained using the cosine and sine identities, we arrive at an expression that can be simplified to match the form of Heron's formula. This involves a few steps of algebraic manipulation and is the result of integrating all of these inputs into a single expression.

After some laborious but simple algebra (which involves expanding the terms, removing parentheses, and recombining them), we get Heron's formula:

    A = √(s * (sa) * (sb) * (sc))

Examples of Heron's formula

To see Heron's formula in action, let's consider practical examples using typical triangle measurements.

Example 1

Suppose we have a triangle with sides a = 7, b = 8 and c = 9 Calculate the area using Heron's formula.

First, find the half-perimeter s:

    s = (7 + 8 + 9) / 2 = 12

Then calculate the area A:

    A = √(12 * (12 - 7) * (12 - 8) * (12 - 9)) = √(12 * 5 * 4 * 3) = √720 ≈ 26.83

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

Example 2

Consider another triangle with sides a = 5, b = 6, and c = 7.

Calculate s:

    s = (5 + 6 + 7) / 2 = 9

Now find the area A:

    A = √(9 * (9 - 5) * (9 - 6) * (9 - 7)) = √(9 * 4 * 3 * 2) = √216 ≈ 14.70

So, the area of this triangle is about 14.70 square units.

Why is Heron's formula important?

Heron's formula allows us to find the area of a triangle using only the lengths of its sides. This is especially helpful in situations where it is difficult to measure the height directly or when working with coordinates.

Since it only requires basic arithmetic operations and square roots, it can be performed without advanced equipment or technology, making it a valuable tool in both the classroom and the field.

Heron's formula is a beautiful example of mathematical beauty, turning a geometric problem into a simple algebraic calculation. The formula shows how different areas of mathematics can come together to solve real-world problems efficiently and reliably.


Grade 9 → 12.3


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Application to Different Triangles and Quadrilaterals
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Proof of Heron’s Formula

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