I can find the area of a triangle with known vertices but the problem here is that the question is general: I have to use double integral to prove that the area of the triangle is:
$$A_{\text{triangle}}=\frac {\text{base}\times\text{height}}{2}$$
I assume that the width is $b$ and the height is $h$ so I have to integrate from $0$ to $b$ and from $0$ to $h$ but this would lead me to the area of a rectangle, that is, $bh$.
Any help is appreciated.
You can place one vertex at the origin, and another at $(b,0)$. The other vertex we assume is at $(t,h)$, where $t \in (0,b)$. The area of the triangle is then
$$\int_0^t dx \, \int_0^{h x/t} dy + \int_t^b dx \, \int_h^{h (b-x)/(b-t)} dy$$
which is
$$\begin{align}\frac{h}{t} \frac12 t^2 + h \int_t^b dx \frac{b-x}{b-t} &= \frac12 h t + \frac{h}{b-t}[b (b-t) -\frac12 (b^2-t^2)]\\ &= \frac12 h t + b h -\frac12 h(b+t)\\ &= \frac12 b h\end{align}$$
Here is an approach. Just assume it has the three vertices $(0,0),(0,b)$ and $(c,h)$ such that $b,c >0$. Now, you need to
1) find the equation of the line passing through points $(0,0),(c,h)$
2) find the line passing through the points $(0,b), (c,h) $
3) consider a horizontal strip; i.e. the double integral
$$ \iint_{D} dxdy .$$
