04 // Interactive Formula Compendium // GEOMETRY-TRIGA = √[ s(s − a)(s − b)(s − c) ], where s = (a + b + c) / 2
Heron's Formula for Triangle Area — SSS Semi-Perimeter Sheet
Heron's formula calculates the area A of a triangle with side lengths a, b, c as A = √[s(s − a)(s − b)(s − c)], where s = (a + b + c)/2 is the semi-perimeter.
Formula Reference
Main Formula
A = √[ s(s − a)(s − b)(s − c) ], where s = (a + b + c) / 2
Rearranged Forms
r_inradius = A / s (Inscribed Circle Radius)
R_circumradius = (a · b · c) / (4A) (Circumscribed Circle Radius)
h_a = 2A / a (Altitude to Side a)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| A | Triangle Surface Area (A) | m² |
| a | Side Length a | m |
| b | Side Length b | m |
| c | Side Length c | m |
Inputs & Parameters
[m]
[m]
[m]
Calculated Result (Triangle Surface Area (A))
A =84m²
Formula Notes & Background
- Start with A = ½ab sin(C) and substitute sin(C) = √(1 − cos²C) = √[(1 − cos C)(1 + cos C)].
- Insert the Law of Cosines cos(C) = (a² + b² − c²) / (2ab) and factor the numerators as differences of squares.
- Introduce the semi-perimeter s = (a + b + c)/2 so the four factors become 2s, 2(s − a), 2(s − b), and 2(s − c), yielding A = √[s(s − a)(s − b)(s − c)].
Step-by-Step Examples
#1: Find the area and inradius of a classic 13-14-15 Heronian triangle.
a = 13 mb = 14 mc = 15 m
s = (13 + 14 + 15)/2 = 21; A = √[21 × (21−13) × (21−14) × (21−15)] = √(21 × 8 × 7 × 6) = √7056
→ A = 84 m² (inradius r = 84 / 21 = 4 m)
#2: Verify Heron's formula on a 3-4-5 right triangle.
a = 3 mb = 4 mc = 5 m
s = 6; A = √[6 × 3 × 2 × 1] = √36
→ A = 6 m² (matching ½ × 3 × 4 = 6 m²)
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Frequently Asked Questions
What causes Heron's radicand s(s−a)(s−b)(s−c) to become zero or negative?+
If the longest side is greater than or equal to the sum of the other two sides (violating the Triangle Inequality a + b > c), no non-degenerate triangle can exist.
How states of needle-thin triangles are computed accurately with Kahan's formula?+
Sort sides a ≥ b ≥ c and compute ¼√[(a+(b+c))(c−(a−b))(c+(a−b))(a+(b−c))] with parentheses preserved to avoid floating-point loss.
What is Brahmagupta's generalization of Heron's formula?+
For any cyclic quadrilateral inscribed in a circle with sides a, b, c, d, Area = √[(s−a)(s−b)(s−c)(s−d)]; setting d = 0 recovers Heron's formula.