Question

Please tell the Derivation of heron's formula?

Answer 1

In ΔABC, the lengths of the segments from vertices to the points of tangency of the incircle are found to be

x = s - a, y = s - b, z = s - c,

so that Heron's formula can be also written as S² = sxyz.

Let r be the inradius of ΔABC. The rearrangement of the six triangles of the dissection as done at the bottom of the applet, shows immediately that S = rs.

Let I be the incenter and denote w = AI. From the diagram in the right portion of the applet,

xyz = r²(x + y + z) = r²s.

It then follows that sxyz = r²s² = S², which completes the proof.

Note: let the angles of the triangle be 2α, 2β, 2γ so that α + β + γ = 90°. The identity xyz = r²(x + y + z) is equivalent to the following trigonometric formula:

cotα + cotβ + cotγ = cotα cotβ cotγ,

where "cot" denotes the standard cotangent function.

Note: Heron's formula is an immediate consequence of that of Brahmagupta which is stated for cyclic quadrilaterals. Letting one of the sides vanish leads to Heron's formula. Curiously, Brahmagupta's formula can be derived from Heron's.