Question

Ankit walks 4m from his house towards west. He then turns towards south and walks for 5m to reach a temple. Find the shortest distance between his house and the temple.

• A. $\sqrt{65}$ m
• B. 8 m
• C. $\sqrt{41}$ m
• D. 7 m

Answer 1

The correct option is C $\sqrt{41}$ m


Given, Ankit walks 4 m west followed by 5 m south. So the net displacement is AB (as shown in the figure).

The shortest distance between the starting point and the temple is the hypotenuse of the right-angled triangle formed.

Using Pythagoras' Theorem,
we get
$A{B}^2=4^2+5^2.$
$=16+25$
$=41$
$\therefore AB=\sqrt{41}m$.