Chapter 10: Heron's Formula

In earlier classes you learned that the area of a triangle is $\tfrac{1}{2} \times \text{base} \times \text{height}$ . But what if you only know the three side lengths and not the height? Heron of Alexandria solved this problem nearly two thousand years ago with a beautiful formula: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$ where $a, b, c$ are the side lengths and $s = \tfrac{a+b+c}{2}$ is the semi-perimeter. This chapter teaches the formula, proves a few results that motivate it, and applies it to find areas of triangles and quadrilaterals in the real world.

The chapter is short and very practical. There is just one big formula to memorise, plus a few tricks for special cases (equilateral, right triangle, isosceles) and a smart strategy for splitting quadrilaterals into triangles. The arithmetic can get messy , square roots, big multiplications , but the conceptual content is contained in one line.

Heron's formula is surprisingly powerful. It works for any triangle, regardless of whether it is acute, obtuse, or right. It works for irrational side lengths. It works for tiny triangles and huge ones. The only requirement is that the three sides actually form a triangle (triangle inequality).

By the end of the chapter you should be able to (1) recognise when Heron's formula is the right tool, (2) compute the semi-perimeter and apply the formula efficiently, (3) split a quadrilateral into triangles and add up areas, and (4) handle real-world questions like "how much area of grass needs to be mowed?" or "how much paint is needed for this triangular wall?"

What's inside

Area of a triangle by base and height , review and rationale.
Heron's formula , statement, simple proofs for special cases, and the general formula.
Worked applications to special triangles , equilateral, isosceles, right.
Area of a quadrilateral by triangulation , split, compute, add.
Real-world applications , fields, garden plots, banners, irregular plots.

Key results / Formula card

Result	Statement
Triangle area (base-height)	$\text{Area} = \tfrac{1}{2} \times b \times h$
Semi-perimeter	$s = \tfrac{a + b + c}{2}$
Heron's formula	$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
Equilateral triangle	$\text{Area} = \tfrac{\sqrt{3}}{4} a^2$ where $a$ is the side
Right triangle	$\text{Area} = \tfrac{1}{2} \times \text{leg}_1 \times \text{leg}_2$
Isosceles triangle	$\text{Area} = \tfrac{a}{4} \sqrt{4b^2 - a^2}$ , where $a$ is the base and $b$ is each equal side
Quadrilateral by triangulation	Split into two triangles using a diagonal; add the two areas

Memorise this card. Every problem in the chapter is one of these formulas applied carefully.

Chapter 10: Heron's Formula

What's inside

Key results / Formula card

Sub-topics

Practice quiz