A plane flies 320 km due west and then 240 km due north. Find the shortest distance covered by the plane to reach its original position.

Let the plane start from O and travels 320 km in the west to reach A and then travel 240 km in the north to reach B.

Now OA is the shortest distance to travel from O to B.

So, in △OAB, ∠OAB = 90^o

OB² = OA² + AB²   (By Pythagoras property)

⇒ OB² = 320² + 240²
⇒ OB² = 102400 + 57600
⇒ OB² = 160000
⇒ OB = √160000 = 400 km
Hence, the required shortest distance is 400 km.