Stosujemy twierdzenie Pitagorasa w trójkącie $ADE$:
${\left|AE\right|}^2+{\left(11x\right)}^2={\left(14x\right)}^2$
${\left|AE\right|}^2+121x^2=196x^2$
${\left|AE\right|}^2=75x^2$
$\left|AE\right|=\sqrt{75}x=5\sqrt{3}x$

Stosujemy twierdzenie Pitagorasa w trójkącie $ACE$:
${\left|AC\right|}^2={\left|AE\right|}^2+{\left(3x\right)}^2$
${\left|AC\right|}^2=75x^2+9x^2$
${\left|AC\right|}^2=84x^2$

$P_p=\frac{{\left|AC\right|}^2\sqrt{3}}{4}=\frac{84x^2\cdot \sqrt{3}}{4}=21\sqrt{3}{x}^2$
$P_b=3P_{ADC}=3\cdot \frac{\left|CD\right|\cdot \left|AE\right|}{2}=3\cdot \frac{14x\cdot 5\sqrt{3}x}{2}=105\sqrt{3}{x}^2$

$\frac{P_c}{P_p}=\frac{P_p+P_b}{P_p}=\frac{21\sqrt{3}{x}^2+105\sqrt{3}{x}^2}{21\sqrt{3}{x}^2}=\frac{126\sqrt{3}{x}^2}{21\sqrt{3}{x}^2}=6$               Odp.: $\frac{P_c}{P_p}=6$