Easy derivation for area of obtuse triangle

Question

I am looking for an easy derivation for the area of an obtuse triangle. As everybody knows the formula for any triangle is the same: half the product of the base and its height.

The derivation for an acute triangle is quite neat and purely geometrical (you draw a rectangle twice the size of the triangle and are done).

I am looking for an equally simple derivation for the area of an obtuse triangle.

Just to be clear I don't have a problem deriving this in all kinds of ways (including using integrals). However the simplest I found would still require either to solve a set of linear equations or the distributive law.

Is it possible to do it without these?

Answer 1

Suppose $\triangle ABC$ is obtuse at $A$, and orient the plane so $AB$ is our base. Drop a perpendicular from $C$ to $AB$ extended through $A$, say with foot $F$. Define $G,\,H$ so $FBGC,\,FAHC$ are rectangles. Then $$[\triangle ABC]=[\triangle FBC]-[\triangle FAC]=\frac{[\square FBGC]-[\square FAHC]}{2}=\frac{[\square ABGH]}{2}.$$(Sorry for using a square symbol for rectangles.) But $ABGH$ has the same base and height as $\triangle ABC$.

Answer 2

The area of an obtuse triangle with base $b$ and a point $p$ that is a distance $h$ above the base but a distance $c$ outside it, is equal to the area of the right-angled triangle with base $b+c$ and height $h$, minus the area of the right-angled triangle with base $c$ and height $h$.

Answer 3

You can use a similar argument for obtuse triangles Hint

$$[\Delta ABC]=\frac{[EBDC]}{2}-[\Delta EAC]$$