C

$\because \triangle ABC$∽$\triangle {A\’}{B\’}{C\’}$，$AD$和${A\’}{D\’}$是它们的对应高线，若$AD=4$，${A\’}{D\’}=2$，
$\therefore$两三角形的相似比为：$\frac{AD}{A′D′}=\frac{4}{2}=\frac{2}{1}$，
$\therefore \frac{{S}_{△ABC}}{{S}_{△A′B′C′}}=(\frac{AD}{A′D′})^{2}=(\frac{2}{1})^{2}=\frac{4}{1}$.
故选：$C$.