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


Heron's formula is a very useful formula in mathematics, especially when it comes to finding the area of a triangle when you know the length of all three sides. This formula is named after Hero of Alexandria, who was a Greek engineer and mathematician. It is particularly useful because it does not require you to calculate angles or heights, making it versatile and applicable to any triangle.

Breaking down Heron's formula

To use Heron's formula, you need to know the lengths of all three sides of the triangle. Once you know these, the formula allows you to calculate the area using only these side lengths. The formula is set up in two main steps:

Step 1: Calculate the semi-perimeter

The semi-perimeter of a triangle is half of its perimeter. If the sides of the triangle are a, b and c, then the semi-perimeter s is calculated as follows:

s = (a + b + c) / 2

Step 2: Apply Heron's formula

Once you find the semi-perimeter, you can find the area A of the triangle using Heron's formula:

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

Understand with an example

Let's look at an example to make this clearer:

Imagine that you have a triangle with sides 7 cm, 10 cm, and 5 cm. To find the area of this triangle using Heron's formula, follow these steps:

Calculate the semi-perimeter

First, calculate s:

s = (7 + 10 + 5) / 2 = 11 cm

Calculate the area

Now, substitute the values into Heron's formula:

A = √(11 * (11 - 7) * (11 - 10) * (11 - 5)) A = √(11 * 4 * 1 * 6) A = √(264) A ≈ 16.25 cm²

Therefore, the area of the triangle is approximately 16.25 square centimeters.

7 cm 10 cm 5 cm

Explore more examples

Consider another triangle with sides 8 cm, 15 cm and 17 cm.

Step 1: Find the semi-perimeter

s = (8 + 15 + 17) / 2 = 20

Step 2: Use Heron's formula to find the area

A = √(20 * (20 - 8) * (20 - 15) * (20 - 17)) A = √(20 * 12 * 5 * 3) A = √(3600) A = 60 cm²

The area of this triangle is exactly 60 square centimeters.

8 cm 15 cm 17 cm

Why is Heron's formula important?

Heron's formula is extremely important because it provides a way to find the area of any triangle, without needing height or angle measurements. This formula is especially useful in real-world problems where you may only know the lengths of the sides. It is also helpful in a variety of fields such as engineering, architecture, and land surveying, where the triangle is a fundamental shape used in design and measurement.

Mathematical proof

Although the proof of Heron's formula may be complicated to derive at a 9th grade level, it is useful to understand its background:

Heron's formula can be derived from the law of cosines and from the formula for the area of a triangle derived from trigonometry. Nevertheless, at this stage, it is more important to understand how to apply the formula rather than to deduce it.

Other examples

Consider a triangle whose sides are 9 cm, 12 cm and 15 cm.

Calculate the semi-perimeter

s = (9 + 12 + 15) / 2 = 18

Substitute into Heron's formula

A = √(18 * (18 - 9) * (18 - 12) * (18 - 15)) A = √(18 * 9 * 6 * 3) A = √(2916) A = 54 cm²

Thus, the area of this triangle is 54 square centimeters.

9 cm 12 cm 15 cm

Challenges and considerations

Sometimes you may come across special types of triangles such as right triangles or equilateral triangles. Although Heron's formula applies universally, you can simplify your calculations by using special properties of such triangles. For example, in right triangles, using the base height may be more straightforward.

Practice problems

Practice is a must to become proficient at using Heron's formula. Try solving these sample problems using what you've learned:

  1. Find the area of a triangle whose sides are 6 cm, 8 cm and 10 cm.
  2. Find the area of a triangle with sides 11 cm, 13 cm and 16 cm.
  3. Find the area of a triangle with sides 21 cm, 20 cm and 29 cm.

Conclusion

Heron's formula is a simple but powerful tool in geometry. It emphasizes the importance of understanding the relationship between different aspects of geometry through side lengths and gives a practical approach to calculating areas efficiently. As you continue learning, the skill of using Heron's formula will enhance your problem-solving abilities in a variety of geometric challenges.


Grade 9 → 12


U
username
0%
completed in Grade 9

← Prev (11.5)
Constructing Angles of Given Measure
Next (12.1) →
Area of a Triangle Using Heron’s Formula

Comments 



Heron's Formula

https://www.buddymath.com/g9/12?bmal=en