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Base 10, height 6 → 30. Three sides 3-4-5 → Heron returns exactly 6. Sides 7 and 9 at 45° → 22.27. Every homework format covered.
Triangle area questions arrive in three disguises and each has its own formula: base-and-height when you have it, Heron's when you only know the sides, and the sine formula when you have two sides and the angle between them. This calculator runs all three, validates that three sides can actually form a triangle, and flags right triangles when the Pythagorean shortcut applies.
Geometry students use it three ways: ½bh, Heron for three sides (3-4-5 → 6), or two sides + angle — with the triangle inequality checked for you.
Compute the semi-perimeter s = (a+b+c)/2, then A = √(s(s−a)(s−b)(s−c)). For 3-4-5: s = 6, and A = √(6×3×2×1) = √36 = 6 — matching ½×3×4 as it must. Heron's is the tool when height is unknown: land parcels from surveyed sides, sail sails, triangular garden beds. The formula validates reality first — if the two shorter sides do not sum beyond the longest, no triangle exists (3, 4, and 8 cannot close). Watch for degenerate cases: sides 2, 3, 5 sum exactly, giving zero area — a flattened line, not a triangle. The formula dates to Heron of Alexandria around 60 AD and remains the fastest general method without trigonometry.
A = ½ab·sin(C) — the SAS formula, the most elegant of the three. Sides 7 and 9 with a 45° angle between them: ½×7×9×sin45° = 22.27. The sine term is doing geometric work: it converts the tilted side into an effective height — sin(90°) = 1 recovers the maximum area for given sides, which is why right angles maximize area and a collapsing angle (sin → 0) flattens the triangle to nothing. Critical detail: the angle must be the INCLUDED angle, the one between the two known sides. Given a non-included angle (SSA), the triangle may not even be unique — the infamous ambiguous case — which is why homework specifies "included" and this calculator requires it.
Because every triangle is exactly half of a parallelogram — duplicate the triangle, rotate the copy 180°, and the two form a parallelogram with the same base and height whose area is base × height. The triangle is half of that. The corollary students miss: any side can serve as the base, but the height must be measured PERPENDICULAR to it — the altitude, not a neighboring side. In an obtuse triangle the altitude can fall outside the triangle entirely, which is the classic homework trap. Practical confirmation from Heron: for right triangles the legs are perpendicular by definition, so ½ab always works — 3-4-5 gives 6 by both methods, which is how you know your Heron arithmetic is right.