263. Heron's Formula Unveiled · Area from sides alone

Try to break it

Drag A to make the triangle very tall, very wide, or nearly flat against BC. The three side lengths shift, the semi-perimeter s = (a + b + c)/2 updates, and √(s(s − a)(s − b)(s − c)) always matches the geometric ½·base·height. Try to find a shape where it fails; impossible.

How you build it

Construct a triangle from its three sides and find its area.

• Point tool: mark vertex B near the left.
• Ray tool: click B, then click to the right — the base ray BC will lie along.
• Arc tool: open the compass on the ruler — click the 0 mark, then the 14 mark — the arc, centred on B, cuts the base ray at C (BC = 14).
• Point tool: mark C where the arc cuts the base ray.
• Arc tool: open the compass on the ruler — 0 then 15 — centre B, and swing an arc above the base (this is side BA = 15).
• Arc tool: open the compass on the ruler — 0 then 13 — centre C, and swing an arc that crosses the first arc (side CA = 13).
• Point tool: mark A where the two arcs cross — its position is fixed by the three sides.
• Segment tool: join B to A (15 cm).
• Segment tool: join C to A (13 cm). Sides 13, 14, 15 → s = 21, area = √(21·7·6·8) = 84 by Heron.

The proof, step by step

Prove that the Heron formula gives the true area of a triangle.

• Let ABC be a triangle with sides a, b, c. Draw altitude AD perpendicular to BC.
• Let BD = x. Then CD = a - x.
• In right triangles ABD and ACD, apply Pythagoras theorem to express h².
• Equate the expressions for h² and solve for x.
• Substitute x back to find h, then calculate Area = 0.5 * a * h.
• Simplify the algebraic expression to arrive at √[s(s-a)(s-b)(s-c)].

Worked example

A triangle has sides of length 5 cm, 12 cm, and 13 cm. What is the area of the triangle?

s = (5+12+13)/2 = 15. Area = √[15(15-5)(15-12)(15-13)] = √[15×10×3×2] = √900 = 30 cm².

• 30 cm² — correct
• 60 cm²
• 25 cm²
• 15 cm²