Question

Q1: 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.​

Answer 1

Answer:

400 km

Step-by-step explanation:

As plane firstly goes 320 km towards west and then to North 240 km

Clearly a right angle triangle is formed by plane so

=> shortest distance is A and B let say

now

(Ab)² = (ac)²+(cb)²

=> Ab = √(240)²+(320)²

=> ab = √57600+102400

=> ab =√160000

=> ab = 400

Answer 2

Given: A plane flies 320 km due west and then 240 km due north.

To find: The shortest distance covered by the plane to reach its original position.​

Solution:

The path followed by the plane and the line joining the initial and final positions of the plane forms a right-angled triangle. The distance of 320 km towards the west forms the perpendicular of the triangle. The distance of 240 km towards the north forms the base of the triangle. Now, the distance between initial and final positions is found using Pythagoras Theorem.

$distance = \sqrt{320^{2} + 240^{2}}$

              $= 400 m$

Therefore, the shortest distance covered by the plane to reach its original position. is 400 m.