smc

Revised by Kinglogic and RJ5303707 (red shows lattice points within the triangle) If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is $16$ squares $(y=5.1\ast \pi )$, and the limit for the $x$-value is $5$ squares. First we count the $1\ast 1$ squares. In the rightmost column, there are $12$ squares with length $1$ because $y=4\ast \pi$ generates squares from $(4,0)$ to $(4,4\pi )$, and continuing on we have $9$, $6$, and $3$ for $x$-values for $1$, $2$, and $3$ in the equation $y=\pi x$. So there are $12+9+6+3=30$ squares with length $1$ in the figure. For $2\ast 2$ squares, each square takes up $2$ units left and $2$ units up. Squares can also overlap. For $2\ast 2$ squares, the rightmost column stretches from $(3,0)$ to $(3,3\pi )$, so there are $8$ squares with length $2$ in a $2$ by $9$ box. Repeating the process, the next column stretches from $(2,0)$ to $(2,2\pi )$, so there are $5$ squares. Continuing and adding up in the end, there are $8+5+2=15$ squares with length $2$ in the figure. Squares with length $3$ in the rightmost column start at $(2,0)$ and end at $(2,2\pi )$, so there are $4$ such squares in the right column. As the left row starts at $(1,0)$ and ends at $(1,\pi )$ there are $4+1=5$ squares with length $3$. As squares with length $4$ would not fit in the triangle, the answer is $30+15+5$ which is $\boxed{50}$.