// maths › Algebra
See how Heron's area formula is derived with algebra and the Pythagorean theorem: drop a height onto the base, use Pythagoras to find where it lands and how tall it is, then one half base times height — shown to equal Heron's square-root formula exactly.
d = (a²−b²+c²)/(2c), h = √(a²−d²), Area = ½·c·h = √(s(s−a)(s−b)(s−c))
A mind behind this: Hero of Alexandria c. 10 – c. 70 AD
Drop a height onto the base, splitting the triangle into two right triangles that share that height. The Pythagorean theorem locates where the height lands and gives its length, then Area = 1/2 base x height. Simplifying with the difference-of-squares identity yields Heron's formula.
It is the only tool used. Writing Pythagoras for each of the two right triangles and subtracting eliminates the height to find the foot of the altitude; applying it once more gives the height itself. No trigonometry is involved.
For sides a=4, b=3, base c=5: d = (9-16+25)/(2 x 5) = 1.8, height = sqrt(9 - 1.8 squared) = 2.4, so Area = 1/2 x 5 x 2.4 = 6 -- matching Heron's formula.
Because the proof is carried out entirely with algebraic manipulation and the Pythagorean theorem -- expanding, subtracting, and factoring -- rather than with angles or trigonometric ratios.
The base-times-height idea and the algebra behind it underpin area calculations in construction, land surveying, and any setting where a triangle's area is found from measured lengths rather than angles.