How to derive the area of triangle without using quadrilaterals?

Question

I want to derive A = 1/2bh. please tell me how to derive without using parallelogram and rectangle.

Answer 1

Well if you were to use the example of a right triangle with the 90 degree angle at the lower left corner then you could derive this using an integral. You would just integrate the equation of the hypotenuse from 0 to b
The equation of the hypotenuse would be y=(-h/b)x + h and when integrated it would look like

(-h/2b)x^2 + hx

Then plug in your limits and you would end up with:

(-bh/2 ) + bh

This simplifies to bh/2 or 1/2*b*h

Answer 2

This can be done by means of an integral. First, let's consider a right triangle. Suppose one of the vertexes is at (0,0) and the 2 other vertexes are at (0, b) and (b, h), so that (0,b) is the vertex of the righ angle. Then, the hypotenuse has equation y = h/b x. If we integrate this equation from 0 to b we get the area of the triangle, that is

A = Integral 0^b h/b x dx = h/b x2/2 (0^b) = h/b b^2/2 = bh/2

So, this formula holds for right triangles. For general triangles ABC, observe that, if you trace it's height AH= h relative to side BC, the your triangle is split into 2 right triangles, with areas AH h/2 and Hb h/2. If H is outside the triangle, one of the areas is negative. So, you have A = AH h/2 + HB h/2 = (AH + HB)h/2 = ah/2, showing this is true for any triangle.