Area of a Triangle

$A = \tfrac{1}{2}bh$, and the proof takes one move: make a second copy of your triangle, turn it half a turn, and slide it against the first. The two fit together into a parallelogram of base $b$ and height $h$, with no gap and no overlap, so one triangle is half of it. A parallelogram is in turn a sheared rectangle — cut the triangle off one end and it fits exactly onto the other — so every triangle traces back to length times width, the definition of area itself. That is the entire content of the formula, and it is why a triangle is the atom of area measurement: any polygon whatever can be cut into triangles, and that is precisely how surveying software computes the area of an irregular lot.

A consequence worth having: since only the perpendicular height enters, sliding the apex sideways along a line parallel to the base changes the shape but not the area. A tall thin sliver and a tidy isosceles triangle on the same base, with the same height, cover exactly the same ground.

A worked instance. The gable end of a garage is 6.4 m wide at the plate and rises 2.1 m to the ridge, so $A = \tfrac{1}{2}(6.4)(2.1) = 6.72$ m² — about two and a half sheets of drywall, or one litre of paint per coat at typical coverage. Backwards, $h = 2A/b$ and $b = 2A/h$: a triangular flowerbed of 5 m² against a 4 m wall must run $2(5)/4 = 2.5$ m out from it.

Where it goes wrong is almost always the height. $h$ is the perpendicular distance from the base to the opposite vertex, not the length of the slanted side that runs up to it — and on a drawing the slant is the number that is written down, because it is the one somebody could measure. Using it inflates the answer, badly on a steep triangle. Two more. On an obtuse triangle the foot of that perpendicular lands outside the base, so the height has to be measured to an extension of the base line; the formula still holds, but the picture stops being reassuring. And $b$ and $h$ must be a matched pair — any of the three sides can serve as the base, but the height must be the one dropped to that side. If all you have is the three side lengths, this formula cannot help you and Heron's can; if you have two sides and the angle between them, use $A = \tfrac{1}{2}ab\sin C$, which is this same formula with $b\sin C$ quietly supplying the height.