Question

A man has strayed from his path while on his way to the park.he moves 100km towards the south, then another 40km towards west.he then travels 70km towards the north and reaches the park.what is the distance of the shortest possible route?

Answer 1

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Answer 2

Answer:

50 km

Step-by-step explanation:

Refer the attached figure .

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

Now we are supposed to find the distance of the shortest possible route i.e. AE

Now referring the figure we can see that DE = BC = 70 km

So, AB = AC - BC = 100-70=30

Thus AB = 30 km

DC is parallel to EB and DC = EB = 40

Now to find AE :

In ΔABE

Using Pythagoras theorem :

$Hypotenuse^{2} =Perpendicular^{2} +Base^{2}$

$AE^{2} =AB^{2} +BE^{2}$

$AE^{2} =30^{2} +40^{2}$

$AE^{2} =900+1600$

$AE^{2} =2500$

$AE =\sqrt{2500}$

$AE =50$

Thus the distance of the shortest possible route is 50 km.