Try drawing a triangle with its base on the x-axis, and a vertex at the origin. You might then proceed by cases.
The simplest case will be a right triangle with the height line on the y-axis. Then the other side has equation y = -(h/b)x + h, where "h" is the height and "b" is the length of the base. Use a basic "integral as area" argument.
Then consider a triangle with its "peak" vertex above the segment on the x-axis which is the base. You can split this into two right triangles, and thus two integrals, similar to the first case.
And so forth.
Eliz.