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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. $\text{10 km, North}$

1 Answer

Best answer
16 16 votes

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 $

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

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

=> $ZP=8\:Km$.

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