A person moves$30\sqrt{3}\text{m}$ towards North and 20 m towards East and finally$10\sqrt{3}\text{m}$ towards South. The magnitude of displacement of the person from the origin is

Given that

A person started at origin

Travelled $\text{OE}=30\sqrt{3}$  towards North

From E to D i.e. ED = 20m towards East

From D to C i.e. $\text{DC}=10\sqrt{3}$  towards South

Now displacement of person from origin is OC.

By drawing $\perp$  CB on to OE

We get $\text{OB}=20\sqrt{3},\text{CB}=20\text{m}$

From right angle triangle OBC

Using Pythagerus property we can have

${\text{OC}}^{\text{2}}={\text{OB}}^{\text{2}}+{\text{CB}}^{\text{2}}$

${\text{OC}}^{\text{2}}={\left(20\sqrt{3}\right)}^2+{\left(20\right)}^2$

$=1200+400$

${\text{OC}}^{\text{2}}=1600$

$\therefore \text{OC}=\sqrt{1600}=40\text{m}$