Question

A baby crawls $8$ feet towards the west and then $8$ feet towards the north. It then moves $14$ feet towards the east. How far and in which direction is the baby from the starting point?

Answer 1

Determine the distance and direction of the baby from the starting point.

In the figure,$\mathrm{AB}=8$ is the path baby crawls toward the west,

$\mathrm{BC}=8$ is the path baby crawls toward the north after turning.

$\mathrm{CE}=14$is the path baby crawls toward east finally.

So, the path covered is $\mathrm{A}\stackrel{\mathrm{west}}{\to }\mathrm{B},\mathrm{B}\stackrel{\mathrm{north}}{\to }\mathrm{C}\mathrm{and}\mathrm{C}\stackrel{\mathrm{East}}{\to \mathrm{E}}$ as shown below:

$$\begin{gathered} \mathrm{C}-----\mathrm{D}---->\overline{)\mathrm{E}} \\ \uparrow ⋮⋰ \\ \mathrm{I}⋮⋰ \\ \mathrm{I}⋮⋰ \\ \mathrm{B}<----\overline{)\mathrm{A}}⋰ \end{gathered}$$

Distance between the initial and final point is $\mathrm{AE}$ toward North east direction.

Let $\mathrm{D}$ be any point in $\mathrm{CE}$ such that $\mathrm{CD}=8\mathrm{ft}$.

Since with respect to the direction of paths, $\mathrm{AB}\parallel \mathrm{CD}\mathrm{and}\mathrm{AD}\parallel \mathrm{BC}$ and also $\mathrm{AB}=\mathrm{CD}=\mathrm{BC}=\mathrm{AD}=8\mathrm{ft}$.

Thus a quadrilateral $\mathrm{ABCD}$ is obtained which is square.

And also a right-angled triangle $∆\mathrm{ADE}$ formed.

Apply Pythagoras theorem in right angle to find the distance from point $A$ to point $E$:

$\begin{array}{rcl}{\mathrm{AE}}^2& =& {\mathrm{AD}}^2+{\mathrm{DE}}^2\\ & \Rightarrow & {\mathrm{AE}}^2={\mathrm{AD}}^2+{\left(\mathrm{CE}-\mathrm{CD}\right)}^2\\ & \Rightarrow & {\mathrm{AE}}^2=8^2+{\left(14-8\right)}^2\\ & & \Rightarrow {\mathrm{AE}}^2=64+36\\ & \Rightarrow & {\mathrm{AE}}^2=100\\ & \Rightarrow & \mathrm{AE}=\sqrt{100}\\ & \Rightarrow & \mathrm{AE}=10\end{array}$

Therefore, distance between the starting point and ending point is $10\mathrm{m}$ in the north east direction.