An Intuitive Derivation of Heron's Formula

From elementary geometry we learn that two triangles are congruent if their edges have the same lengths, so it should come as no surprise that the edge lengths of a triangle determine the area of that triangle. On the other hand, the explicit formula for the area of a triangle in terms of its edge lengths, named for Heron of Alexandria (although attributed to Archimedes [4]), seems to be less commonly remembered (as compared with, say, the formulas for the volume of a sphere or the area of a rectangle). One reason why Heron’s formula is so easily forgotten may be that proofs are usually presented as unwieldy verifications of an already known formula, rather than as expositions that derive a formula from scratch in a constructive and intuitive manner. Perhaps the derivation that follows, while not truly elementary, will render Heron’s formula more memorable for its symmetric and intuitive factorization. The first step of this derivation is to recall that the square of the area of a triangle is a polynomial in the edge lengths. More generally, suppose that T is a simplex in Rn with vertices x0, x1, . . . , xn , where x0 = 0, the origin. Let A denote the n × n matrix whose columns are given by the vectors x1, . . . , xn , and suppose that the xi are ordered so that A has positive determinant. The volume of T is then given by det(A) = n! V (T ), implying that