In Heron’s Formula, the sides of the triangle are represented by \(\small a\), \(\small b\), and \(\small c\).
Chapter 10 · CBSE · Class IX
📖 Introduction
📘 Definition
💡 Basic Concepts Related to Heron’s Formula
🎨 Triangle Representation
📐 Derivation of Heron’s Formula
1. Setup
Consider a triangle with sides \(\small a\), \(\small b\), and \(\small c\). Let the altitude \(\small h\) drop to the base \(\small b\), dividing it into two segments: \(\small d\) and \(\small b - d\).
By the Pythagorean theorem:
\[\small h^2 = c^2 - d^2 \quad \text{--- (1)} \] \[\small h^2 = a^2 - (b - d)^2 \quad \text{--- (2)} \]2. Solving for \(\small d\)
Equating equations (1) and (2):
\[\small c^2 - d^2 = a^2 - (b^2 - 2bd + d^2) \] \[\small c^2 - d^2 = a^2 - b^2 + 2bd - d^2 \] \[\small c^2 = a^2 - b^2 + 2bd \]Rearranging to find \(\small d\):
\[\small d = \frac{b^2 + c^2 - a^2}{2b} \]3. Expressing Height \(\small h\) in terms of sides
Substitute \(\small d\) back into \(\small h^2 = c^2 - d^2\):
\[\small h^2 = c^2 - \left( \frac{b^2 + c^2 - a^2}{2b} \right)^2 \]Using the difference of squares identity \(\small X^2 - Y^2 = (X - Y)(X + Y)\):
\[\small h^2 = \frac{(2bc)^2 - (b^2 + c^2 - a^2)^2}{4b^2} \] \[\small h^2 = \frac{[2bc - (b^2 + c^2 - a^2)][2bc + (b^2 + c^2 - a^2)]}{4b^2} \]Grouping terms to form perfect squares:
\[\small h^2 = \frac{[a^2 - (b - c)^2][(b + c)^2 - a^2]}{4b^2} \] \[\small h^2 = \frac{(a - (b - c))(a + (b - c))((b + c) - a)((b + c) + a)}{4b^2} \] \[\small h^2 = \frac{(a - b + c)(a + b - c)(b + c - a)(a + b + c)}{4b^2} \]4. Introducing the Semi-perimeter (\(\small s\))
Let the semi-perimeter be \(\small s = \frac{a + b + c}{2}\). Then \(\small 2s = a + b + c\). We can rewrite the factors as:
- \(\small a + b + c = 2s\)
- \(\small b + c - a = 2s - 2a = 2(s - a)\)
- \(\small a + c - b = 2s - 2b = 2(s - b)\)
- \(\small a + b - c = 2s - 2c = 2(s - c)\)
Substituting these into the expression for \(\small h^2\):
\[\small h^2 = \frac{2(s-b) \cdot 2(s-c) \cdot 2(s-a) \cdot 2s}{4b^2} \] \[\small \begin{aligned} h^2 &= \frac{16s(s-a)(s-b)(s-c)}{4b^2}\\ &= \frac{4s(s-a)(s-b)(s-c)}{b^2} \end{aligned}\]5. Final Area Calculation
The area \(\small \Delta\) is given by \(\small \frac{1}{2}bh\):
\[\small \Delta^2 = \frac{1}{4}b^2h^2 \] \[\small \Delta^2 = \frac{1}{4}b^2 \left( \frac{4s(s-a)(s-b)(s-c)}{b^2} \right) \] \[\small \Delta^2 = s(s-a)(s-b)(s-c) \]\[\small \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]
🔢 Formula
✏️ Example
Solved Example
Find the area of a triangle whose sides are \(\small 13\text{ cm}\), \(\small 14\text{ cm}\), and \(\small 15\text{ cm}\).
- Heron’s Formula
- Semi-perimeter concept
-
1Calculate semi-perimeter
-
2Substitute values in Heron’s Formula
-
3Simplify carefully
-
Given:
\[\small a=13,\quad b=14,\quad c=15 \]
-
Semi-perimeter:
\[\small s=\frac{13+14+15}{2} \]
-
Applying Heron’s Formula:
\[\small \begin{aligned} \text{Area}&=\sqrt{21(21-13)(21-14)(21-15)}\\ &=\sqrt{21\times8\times7\times6}\\ &=\sqrt{7056}\\ &=84 \end{aligned} \]
Therefore, area of the triangle is:
\[\small
84\text{ cm}^2
\]
🛠️ Applications of Heron’s Formula
- Finding area of triangular parks and fields
- Land surveying
- Construction engineering
- Architecture and map designing
- Navigation and geographical calculations
- Computer graphics and animation
🌟 Importance
❌ Common Mistakes
- Using perimeter instead of semi-perimeter
- Incorrect subtraction in \(\small (s-a)\), \(\small (s-b)\), or \(\small (s-c)\)
- Calculation errors while simplifying square roots
- Ignoring units in final answer
- Applying formula to invalid triangles
⚡ Exam Tip
📋 Case Study
A farmer owns a triangular field with sides \(\small 50\text{ m}\), \(\small 52\text{ m}\), and \(\small 26\text{ m}\). He wants to calculate the area for irrigation planning.
Questions
- Find the semi-perimeter of the field.
- Calculate the area using Heron’s Formula.
- If irrigation costs ₹12 per square metre, find total cost.
Solution
\[\small s=\frac{50+52+26}{2}=64 \]
\[\small \text{Area}=\sqrt{64(64-50)(64-52)(64-26)} \]
\[\small =\sqrt{64\times14\times12\times38} \]
\[\small =\sqrt{408576} \]
\[\small =639.2\text{ m}^2\ (\text{approx}) \]
Irrigation cost:
\[\small 639.2\times12=7670.4 \]
\[\small \text{Total Cost}=\text{₹ } 7670.4 \]