Question

A rat walk root 2 x meter towards east, then he turn left and walk x meter again he turn towards north-east and walk x meter and finally he turn towards south-east and walk x meter. Then find the shortest distance from the starting to end point?​

Answer 1

Question:

A rat walk $\sqrt{2}x$ meter towards east, then he turn left and walk x meter again he turn towards north-east and walk x meter and finally he turn towards south-east and walk x meter. Then find the shortest distance from the starting to end point?

Answer:

See the attachment

In triangle BCD, we have $\angle CBD=45^{\circ}$,

$cos45 = \frac{Adjacent} {Hypotenuse} \\\\ cos45 = \frac{Adjacent}{x} \\\\ \frac{1}{\sqrt{2}} = \frac{Adjacent}{x} \\\\ Adjacent = \frac{x}{\sqrt{2}}$

$BD = \frac{x}{\sqrt{2}} + \frac{x}{\sqrt{2}} \\\\ BD = \frac{2x}{\sqrt{2}} \\\\ BD = \frac{2x}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\\\ BD = \sqrt{2} x = AE$

Now come to $\Delta ODE$

By Pythagoras theorem,

$OD^2 = OE^2 + DE^2 \\\\ OD^2 = (OA+AE)^2 + DE^2 \\\\ OD^2 = (\sqrt{2}x+\sqrt{2}x)^2 + DE^2 \\\\ OD^2 = (2\sqrt{2}x)^2 + DE^2 \\\\ OD^2 = 8x^2+x^2 \\\\ OD^2 = 9x^2 \\\\ OD = 3x$

Hence, the shortest distance from the starting to end point is 3x