Area of a triangle

Draw a rectangle ABCD and put a point X anywhere on the top side. Join X to D and C. Do the same in an identical rectangle with a different point Y. Which triangle has the greater area?

Neither: they are equal. Drop the altitude (the perpendicular line) from X, and from Y. In each case the triangle fills exactly half the rectangle. Sliding the top point along changes the triangle's shape but not its area, because it changes neither the base nor the height.

That gives a way to find any triangle's area: put it inside a rectangle. Given ∆ABC, draw a line l through A parallel to BC, and drop perpendiculars from B and C to that line. That makes a rectangle BCDE round the triangle.

Call the rectangle's sides the base and the height. The triangle fills half of it, so Area(∆ABC) = ½ × base × height. And because BXAE is a rectangle too, the rectangle's height is the same as the triangle's own height: the perpendicular distance from A to BC.

So: area of a triangle = ½ × base × height. Base and height are all you need; the other two sides do not matter.

But there is one case the picture does not cover. If the triangle leans so far that the foot of the altitude falls outside BC, you cannot draw that rectangle on BC. Does the formula still work?

Yes, by subtraction. Let D be the foot of the perpendicular from A, with B between D and C. Then ∆ABC is what is left when ∆ADB is taken away from ∆ADC, and each of those can be boxed in a rectangle.

So Area(∆ABC) = ½ × h × DC − ½ × h × DB = ½ × h × (DC − DB) = ½ × h × BC. The same formula. It works for every triangle: right-angled, acute or obtuse.