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(D) Let the vertices of the triangle be $O(0,0)$,$A(x,0)$,and $B(0,y)$,where $x, y \in \mathbb{Z} \setminus \{0\}$.The area of the triangle is given b

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$36$

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(D) Let the vertices of the triangle be $O(0,0)$,$A(x,0)$,and $B(0,y)$,where $x, y \in \mathbb{Z} \setminus \{0\}$.The area of the triangle is given by $\frac{1}{2} |x| |y| = 50$,which implies $|xy| = 100$.Since $x$ and $y$ are non-zero integers,we need to find the number of pairs $(x, y)$ such that $|x| |y| = 100$.The number of divisors of $100 = 2^2 	imes 5^2$ is $(2+1)(2+1) = 3 	imes 3 = 9$.For each divisor $d$ of $100$,we have $|x| = d$ and $|y| = 100/d$. Since $x$ and $y$ can be positive or negative,there are $4$ possible sign combinations for each pair $(|x|, |y|)$ (i.e.,$(+,+), (+,-), (-,+), (-,-)$).Thus,the total number of triangles is $4 	imes 9 = 36$.

Let $S$ be the set of all triangles in the $xy$-plane,each having one vertex at the origin and the other two vertices lying on the coordinate axes with integral coordinates. If each triangle in $S$ has an area of $50$ sq. units,then the number of elements in the set $S$ is:

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