Heron's formula: Approximately 2000 years ago, Heron of...

Heron's formula: Approximately 2000 years ago, Heron of Alexandria derived a formula for the area of a triangle in terms of the lengths of the sides. A more modern derivation of Heron's formula is indicated in the steps that follow. (a) Use the expression for $\sin A$ in Exercise $56(b)$ to show that $\sin ^{2} A=\frac{(a-b+c)(a+b-c)(b+c-a)(b+c+a)}{4 b^{2} c^{2}}$ Hint: Use difference-of-squares factoring repeatedly. (b) Let $s$ denote one-half of the perimeter of $\triangle A B C$. That is, let $s=\frac{1}{2}(a+b+c) .$ Using this notation (which is due to Euler), verify that (i) $a+b+c=2 s$ (ii) $-a+b+c=2(s-a)$ (iii) $a-b+c=2(s-b)$ (iv) $a+b-c=2(s-c)$ Then, using this notation and the result in part (a), show that $$ \sin A=\frac{2 \sqrt{s}(s-a)(s-b)(s-c)}{b c} $$ Note: since $\sin A$ is positive (Why?), the positive root is appropriate here. (c) Use the result in part (b) and the formula area $\triangle A B C=\frac{1}{2} b c \sin A$ to conclude that $$\text { area } \triangle A B C=\sqrt{s(s-a)(s-b)(s-c)} $$ This is Heron's formula. For historical background and a purely geometric proof, see An Introduction to the History of Mathematics, 6 th ed., by Howard Eves (Philadelphia: Saunders College Publishing, 1990 ), pp. 178 and 194

Heron's formula: Approximately 2000 years ago, Heron of Alexandria derived a formula for the area of a triangle in terms of the lengths of the sides. A more modern derivation of Heron's formula is indicated in the steps that follow.
(a) Use the expression for $\sin A$ in Exercise $56(b)$ to show that $\sin ^{2} A=\frac{(a-b+c)(a+b-c)(b+c-a)(b+c+a)}{4 b^{2} c^{2}}$
Hint: Use difference-of-squares factoring repeatedly.
(b) Let $s$ denote one-half of the perimeter of $\triangle A B C$. That
is, let $s=\frac{1}{2}(a+b+c) .$ Using this notation (which is due to Euler), verify that
(i) $a+b+c=2 s$
(ii) $-a+b+c=2(s-a)$
(iii) $a-b+c=2(s-b)$
(iv) $a+b-c=2(s-c)$ Then, using this notation and the result in part (a), show that
$$
\sin A=\frac{2 \sqrt{s}(s-a)(s-b)(s-c)}{b c}
$$
Note: since $\sin A$ is positive (Why?), the positive root is appropriate here.
(c) Use the result in part (b) and the formula area $\triangle A B C=\frac{1}{2} b c \sin A$ to conclude that $$\text { area } \triangle A B C=\sqrt{s(s-a)(s-b)(s-c)} $$
This is Heron's formula. For historical background and a purely geometric proof, see An Introduction to the History of Mathematics, 6 th ed., by Howard Eves (Philadelphia: Saunders College Publishing, 1990 ), pp. 178 and 194

Key Concepts

Trigonometric Relationships in Triangle Area

Trigonometric functions, especially the sine function, are integral in deriving area formulas for triangles. By relating the area to the product of two sides and the sine of the included angle, these relationships enable the translation of angular measures into algebraic forms that can be manipulated to yield formulas like Heron’s.

Algebraic Factoring Techniques

Algebraic methods such as difference-of-squares factoring are essential for simplifying and transforming expressions during derivations. These techniques help in revealing symmetric patterns and reducing complex expressions, which is a key step in deriving compact formulas such as Heron’s for the area of a triangle.

Heron's Formula

Heron's formula is a method for finding the area of a triangle using only its side lengths. It encapsulates the geometric properties of the triangle into a single algebraic expression that involves the semi-perimeter, thereby eliminating the need to directly calculate angles or heights.

Semi-perimeter

The semi-perimeter of a triangle, defined as half the sum of its side lengths, serves as a crucial component in Heron's formula. It simplifies the expression of the area by providing a symmetric measure that relates directly to all three sides of the triangle.

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