A person moves $30 \mathrm{~m}$ North and then $20 \mathrm{~m}$ towards East and finally $30 \sqrt{2} \mathrm{~m}$ in South-West direction. The displacement of the person from the origin will be

Let us use the conventional directions, as shown in figure, to solve this problem.
According to question, let us draw the diagram.
There are two possible diagrams. Clearly, $\quad C E=A B=30 \mathrm{~m} \Rightarrow C D=30 \sqrt{2} \mathrm{~m}$ Using Pythagoras theorem, $(C E)^2+(D E)^2=(C D)^2$ $\Rightarrow(D E)^2=(30 \sqrt{2})^2-(30)^2=1800-900=900 \mathrm{~m}$ $\Rightarrow \quad D E=30 \mathrm{~m}$ Now, it is clear that Case I is not possible. Using Case II, Clearly, displacement of the person is End point - Start point $=D A$ (along West) and $\quad D A=D E-A E$ $=30 \mathrm{~m}-20 \mathrm{~m}=10 \mathrm{~m} \quad$ (along West)