Question

Parvesh travels 15m toward north, turns right and travels 20m. Then turns
left and travels 5m. How far is he from the starting point and in which
direction?​

Answer 1

Given :- Parvesh travels 15m toward north, turns right and travels 20m. Then turns left and travels 5m.

To Find :- How far is he from the starting point and in which direction ?

Solution :-

from image let Pravesh travel from point A and walks towards north and after 15 m stops at point B .

So,

→ AB = 15 m

now, from point B he turns right and walks another 20 m and stops at point C .

So,

→ BC = 20 m .

now, from point B he turns left and walks another 5 m and stops at point D .

So,

→ CD = 5 m .

Now, we have,

→ A = Starting point .

→ D = End point .

Let us assume that, BE = x m .

In ∆ABE and ∆DCE we have,

→ ∠ABE = ∠DCE { Each 90° }

→ ∠BEA = CED { Vertically opposite angles }

So,

→ ∆ABE ~ ∆DCE { By AA similarity. }

then,

→ AB/DC = BE/CE { When two ∆'s are similar their corresponding sides are in same ratio . }

→ 15/5 = x/(20 - x)

→ 3(20 - x) = x

→ 60 - 3x = x

→ 60 = x + 3x

→ 4x = 60

→ x = 15 m

therefore,

→ BE = 15 m

→ CE = BC - BE = 20 - 15 = 5 m .

now, in right angled ∆ABC right angle at E we have,

→ AE = √(AB² + BE²) { By pythagoras theorem }

→ AE = √(15² + 15²)

→ AE = √(225 + 225)

→ AE = √(2 × 225)

→ AE = 15√2 m

similarly, in right angled ∆DCE right angle at E we have,

→ DE = √(DC² + CE²) { By pythagoras theorem }

→ DE = √(5² + 5²)

→ DE = √(25 + 25)

→ DE = √(2 × 25)

→ DE = 5√2 m

hence,

→ AD = AE + DE = AE + ED = 15√2 + 5√2 = 20√2 m (Ans.)

also,

→ Direction of point D from point A = North - East (Ans.)

∴ Parvesh is 20√2 m away from the starting point and in North - East direction .

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