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Fatima starts from point $P$, goes North for $3$ km, and then East for $4$ km to reach point $Q$. She then turns to face point $P$ and goes $15$ km in that direction. She then goes North for $6$ km. How far is she from point $P$, and in which direction should she go to reach point $P$?

  1. $\text{8 km, East}$
  2. $\text{12 km, North}$
  3. $\text{6k m, East}$
  4. $\text{10 km, North}$
in Numerical Ability by Active (2.4k points)
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Migrated from GO Electronics 6 months ago by Arjun
0
8 km towards east.
0
can anyone please answer this question????

1 Answer

+3 votes
Best answer

Fatima traverses like this:-

 

Making right angle triangles, $\unicode{0x25FA} \: PXQ$ & then $ \unicode{0x25FA} \: PZY$. 

So, to reach point $P$ from $Y$ she has to travel =$\sqrt{(10^2-6^2)} = 8 \: km$. (in East direction)

Hence, Correct Answer: $ A. \: 8 \: km, \: East $

by Boss (15.2k points)
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+4
After travelling 6km from y to z how can we assume that z is on straight line from p to z??

Iam stuck here
0

@Naveen Kumar 3 plz reply chiru's comment

+2

@chirudeepnamini @Shubhm

 In $\Delta QPX \:\text{&}\:\Delta PYZ$,

$\frac{QP}{PY}=\frac{PX}{YZ}=\frac{1}{2}$

&, $\angle QPX=\angle PYZ$ {$\because$ corresponding angles of two parallel lines XP & ZY}

$\therefore$ $\Delta QPX \:\text{&}\:\Delta PYZ$ are similar (from $\text{SAS}$ similarity).

thus, angles of similar triangles are same.

So,  $\angle YZP=\angle PXQ = 90^\circ$


Also, similar triangles have proportional sides,

so, $\frac{QP}{PY}=\frac{PX}{YZ}=\frac{XQ}{ZP}$

=>$\frac{XQ}{ZP}=\frac{1}{2}$

=> $ZP=8\:Km$.

0

@Naveen Kumar 3 thank you.. got it now..

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