Question

Somu who is facing North turns to his right and walks 30 m. then he turns to his right and walks 14 m, then facing East he walks 30 m. how far is he from his original position?

Answer 1

Somu is 61.6 m away from his original position.

Step-by-step explanation:

Somu starts from A and reached to D

We are required to find out the distance AD

AB = CD = 30 m

In ΔABO and Δ CDO

AB = CD = 30 m

And ∠ABO = ∠CDO = 90°

Therefore, ΔABO ≅ Δ CDO

Thus, BO = OC = BC/2 = 14/2 = 7 m

And AO = OD

Thus, AD = 2AO

Now,

In ΔABO

By Pythagoras theorem

$AO^2=AB^2+BO^2$

$\implies AO^2=30^2+7^2$

$\implies AO^2=900+49$

$\implies AO=\sqrt{949}$

$\implies AO=30.8$

Therefore, AD = 2AO = 2 x 30.8 = 61.6 m

Hope this helps.

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