Question

A man has strayed from his path while on his way to the park.He moves 100km towards south,then another 40km towards west.He then travels 70km towards north and reaches the park.What is the distance of the shortest possible route?
a.50km
b.40km
c.60km
d.30km

Answer 1

The distance of the shortest possible route is 50 km.

Explanation:

Referring to the figure attached below we can see the given data as,

The at first he walks 100 km towards south i.e. AC = 100 km

Then he moves 40 km towards west i.e. CD = 40 km

Then again he moves 70 km towards North i.e. DE = 70 km

Let the distance of the shortest possible route be denoted as “AE”.(as shown in the figure)

Also, we can see that DE = BC = 70 km

∴ AB = AC - BC = 100 – 70 = 30 km

And, DC is parallel to EB ∴ DC = EB = 40 km

Now, applying Pythagoras theorem for ΔABE, we get

Hypotenuse² = Perpendicular² + Base²

⇒ AE² = AB² + EB²

⇒ AE = √[AB² + EB²]  

⇒ AE =√[30² + 40²]  

⇒ AE = √[900 + 1600]  

⇒ AE = √2500

⇒ AE = 50 km

Thus, the distance of the shortest possible route is option (a): 50 km.

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Difference between Distance & Displacement ?

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The given figure shows the distances covered by Ariv from place A to place H. His direction of movement from A toB, C to D, E to F and G to H is due North and that from B to C. D to E and F to G is due East. Find the aerial distance between A and H.

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Answer 2

The shortest distance from the starting point is 50 km.

Option (a).

Explanation:

Given that,

A man go to 100 km toward south then he turn 40 km toward west and after that he turn 70 km toward north.

We need to calculate the total distance

$D=100-70 = 30 km$

Let the shortest distance is x.

We need to calculate the shortest distance

Using Pythagorean theorem

$x^2=40^2+30^2$

$x=\sqrt{40^2+30^2}$

$x=50\ km$

Hence, The shortest distance from the starting point is 50 km.

#Learn more

Topic : Shortest path between two points

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