Question

A person on his way home from work. He drives 12km due north and then 5km due west. Find the shortest distance he can cover to reach home early. ​

Answer 1

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Answer 2

Given :- A person drives 12km due north and then 5km due west .

To Find :- The shortest distance he can cover to reach home early. ?

Solution :-

Refer to image once .

from image let person home is at point A . So, from point A he drives 12km due north and reach at point B .

So,

→ AB = 12 km

now, from point B he drives 5 km due west and reach at point C .

So,

→ BC = 5 km .

then,

→ AC = Shortest distance from home to point C .

→ AC = √[AB² + BC²] { using pythagoras theorem in right angled ∆ABC }

→ AC = √(12² + 5²)

→ AC = √(144 + 25)

→ AC = √(169)

→ AC = 13 km (Ans.)

Hence, the shortest distance he can cover to reach home early is equal to 13 km .

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