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5,955 views
15 votes
15 votes

$M$ and $N$ start from the same location. $M$ travels $10$ km East and then $10$ km North-East. $N$ travels $5$ km South and then $4$ km South-East. What is the shortest distance (in km) between $M$ and $N$ at the end of their travel?

  1. $18.60$
  2. $22.50$
  3. $20.61$
  4. $25.00$
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1 Answer

Best answer
28 votes
28 votes

$\underline{\text{In }\triangle \text{ DEH}}$

$\sin 45=\dfrac{\text{opposite side}}{\text{hypotenuse}}=\dfrac{EH}{4}$

$\Rightarrow EH=2\sqrt{2}$

$\cos 45=\dfrac{\text{adjacent side}}{\text{hypotenuse}}=\dfrac{DE}{4}$

$\Rightarrow DE=2\sqrt{2}$

$\underline{\text{In }\triangle \text{ BJC}}$

$\sin 45=\dfrac{\text{opposite side}}{\text{hypotenuse}}=\dfrac{CJ}{10}$

$\Rightarrow CJ=5\sqrt{2}$

$\cos 45=\dfrac{\text{adjacent side}}{\text{hypotenuse}}=\dfrac{BJ}{10}$

$\Rightarrow BJ=5\sqrt{2}$

Required Shortest Distance $HC$

Using Pythagoras's theorem in Triangle $\triangle HIC $

$HC=\sqrt{(HI)^{2}+(CI)^{2}}$

Where as ,

$HI=HL+LI$

From Figure, $HL=10-2\sqrt{2}$ and  $LI=5\sqrt{2}$

$HI= 10-2\sqrt{2}+5\sqrt{2}=10+3\sqrt{2}$

$CI=CJ+JG+GI$

Fom Figure ,  $CJ=5\sqrt{2}$ , $JG=5$ , $GI=2\sqrt{2}$

$CI=5\sqrt{2}+5+2\sqrt{2}=5+7\sqrt{2}$

Therefore ,

$HC=\sqrt{(10+3\sqrt{2})^{2}+(5+7\sqrt{2})^{2}} = 20.61$km

Correct Answer: $C$

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