The recurrence is given by $A(n)=A(n-1)+n$. Each new $n^{th}$ line drawn is creating $n$ new partitions. While creating partitions,draw the new line in such a way that it cuts the all the previously drawn $n-1$ lines, then the $n^{th}$ line will create $n$ new partitions and previous $A(n-1)$ partitions will remain the same.
Then $A(3)=A(2)+3=4+3=7$
$A(4)=A(3)+4=7+4=11$
$\vdots$ $\vdots$ $\vdots$
$A(10)=A(9)+10=46+10=\textbf{56}$
P.S. Try dividing $\mathbb{R}^{2}$ by $n=3 ,n=4,n=5$ lines using the method above, you'll get the idea.