Question

A child walks $6m$ towards the west and then $20m$ in towards south. He then turns east and walks another $12m.$ Then he walks $12m$ towards the north. How far is he from his initial position ?

Answer 1

From the given details we note that

$AB=6,BC=20,CD=12,DE=12$

From the diagram, we note that

$\angle ABC=\angle BCD=\angle CDE=\angle ABC=90°$

Construction: Draw the lines $AF$and $EF$which are the extensions of $BA$and $DE$respectively such that they meet at point $F$at the right-angle also.

So, in rectangle $FBCD$we have four angles equal to $90°$

We get $BC=FD=20,BF=CD=12$. (opposite sides are equal in rectangle)

In triangle $OED$,

$EF=FD-DE=20-12=8$

$AF=BF-AB=12-6=6$
On applying Pythagoras’ theorem For right-angle triangle ΔAEF, we get

$$\begin{gathered} A{F}^2+F{E}^2=A{E}^2 \\ =6^2+8^2=36+64 \end{gathered}$$

$A{E}^2=100$

Hence, distance from initial point to terminal point $\sqrt{100}=10m$.