GATE2016 EC-2: GA-9

Question

$M$ 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?

• A. $18.60$
• B. $22.50$
• C. $20.61$
• D. $25.00$

Comments

I'm getting B as the answer

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Anyone please explain this solution easy way

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To solve this type of distance related questions only way is to use the properties of right angle triangle and if there’re lengths available then we can directly solve using Pythagoras theorem or else have to use trigonometric ratios

Answer 1

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

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

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

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

$\Rightarrow DE=2\sqrt{2}$

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

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

$\Rightarrow CJ=5\sqrt{2}$

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

$\Rightarrow BJ=5\sqrt{2}$

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: $C$

Comments

Great Explanation .And if we applied Pythagoras theorem while finding DE,E then we’ll be  getting complex numbers.