Question

Alex walks 20m from his house towards East. He then turns towards North and walks for 10m to reach his school. Find the shortest distance between his house and the school.
(2 marks)

Answer 1

Given, Alex walks 20 m East followed by 10 m North. So the net displacement is AB (as shown in the figure).
(1 mark)
The shortest distance between the starting point and his school is the hypotenuse of the right-angled triangle formed.

Using Pythagoras' Theorem,
we get
$A{B}^2={20}^2+{10}^2.$
$=400+100$
$=500$
$\therefore AB=\sqrt{500}m$= 10$\sqrt{5}$ m
​​​​​​​ (1 mark)