Question

In a rectangular park, a man walks $2$ km from south to north and then makes a ${160}^{\circ }$ turn to right and walks $2$ km more. He then again makes another ${160}^{\circ }$ turn to right and walks $2$ km more. If he continues this pattern until he reaches his starting point again, then the distance travelled by the man during this journey is
(correct answer + 2, wrong answer - 0.50)

• A. $18$
• B. $36$
• C. $27$
• D. $24$

Answer 1

The correct option is A $18$


Let the man starts at $A,$ moves $2$ km to reach $B,$ then makes ${160}^{\circ }$ turn to right and moves $2$ km to reach $C.$

From diagram, $AB=BC$
$\therefore \angle OAB=\angle ABO=\angle OBC={10}^{\circ }$
$\Rightarrow \angle AOB={160}^{\circ }$

So, after each move of $2$ km the radius joining position of the man to centre $O$ rotates ${160}^{\circ }$ anticlockwise.
$\therefore$ Man reaches starting point again in $m$ moves iff
${160}^{\circ }\times m={360}^{\circ }\times n,(m,n\in N)$
$\Rightarrow 4m=9n$
$\Rightarrow \frac{m}{9}=\frac{n}{4}=k\left(\text{say}\right)$
$\Rightarrow m=9k,n=4k$
$\therefore$ Least value of $m=9$

Hence, total distance covered $=9\times 2=18$ km

Alternate Solution:
$\theta ={20}^{\circ }=\frac{\pi }{9}$
Since, man is moving in a star shap position.
Let there be $m$ sides.
$\therefore m\times \theta =n\pi ,(m,n\in N)$
$\Rightarrow m\left(\frac{\pi }{9}\right)=n\pi$
$\Rightarrow m=9n$
$\therefore$ Least value of $m=9$
Hence, total distance covered $=9\times 2=18$ km