Since the vertex of given triangle is $O$ (origin) and other two vertices $A$ and $B$ lies on the coordinate axes, hence triangle must be a right-angled triangle.

Area $=\frac{1}{2}x.y ( x, y\in I)$

As per given condition

$\frac{1}{2}x.y=50\Rightarrow x. y=100$

Possible ordered pairs of $\left(x, y\right)$ when $∆$ formed in $1^{st}$ quadrant are $9$ i. e. $\left(1,100\right), \left(2,50\right), \left(4,25\right), \left(5,20\right), \left(10,10\right), \left(25, 4\right), \left(50, 2\right), \left(100,1\right)$

$∆$ can be formed in any of the quadrants, hence total number of cases $9\times 4=36.$