$\underline{\text{In }\triangle \text{ DEH}}$
$\sin 45=\dfrac{\text{opposite side}}{\text{hypotenuse}}=\dfrac{EH}{4}$
$\Rightarrow EH=2\sqrt{2}$
$\cos 45=\dfrac{\text{adjacent side}}{\text{hypotenuse}}=\dfrac{DE}{4}$
$\Rightarrow DE=2\sqrt{2}$
$\underline{\text{In }\triangle \text{ BJC}}$
$\sin 45=\dfrac{\text{opposite side}}{\text{hypotenuse}}=\dfrac{CJ}{10}$
$\Rightarrow CJ=5\sqrt{2}$
$\cos 45=\dfrac{\text{adjacent side}}{\text{hypotenuse}}=\dfrac{BJ}{10}$
$\Rightarrow BJ=5\sqrt{2}$
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: $C$