$\sin \alpha =\frac{3}{5}$,   $\sin \beta =\frac{24}{25}$,   $h=\frac{144}{25}$,   $\left|EB\right|=\frac{42}{25}$,   $\left|AD\right|=6$,   $\left|AB\right|=\left|CD\right|+2\cdot \left|EB\right|$

Korzystamy z definicji sinusa w trójkącie $AEC$:
$\sin \alpha =\frac{h}{\left|AC\right|}$    $\to$    $\left|AC\right|=\frac{h}{\sin \alpha }=\frac{144}{25}\cdot \frac{5}{3}=\frac{48}{5}$

Teraz korzystamy w tym trójkącie
z twierdzenia Pitagorasa:
${\left|AE\right|}^2+{\left(\frac{144}{25}\right)}^2={\left(\frac{48}{5}\right)}^2$
${\left|AE\right|}^2=\frac{2304}{25}-\frac{20736}{625}=\frac{57600-20736}{625}=\frac{36864}{625}$    $\to$    $\left|AE\right|=\frac{192}{25}$

$\left|AB\right|=\left|AE\right|+\left|EB\right|=\frac{192}{25}+\frac{42}{25}=\frac{234}{25}$,      $\left|CD\right|=\left|AB\right|-2\cdot \left|EB\right|=\frac{234}{25}-2\cdot \frac{42}{25}=\frac{150}{25}=6$

$Ob=\frac{234}{25}+6+6+6=\frac{684}{25}$ ,      $P=\frac{\left(\frac{234}{25}+6\right)\cdot \frac{144}{25}}{2}=\frac{\frac{384}{25}\cdot \frac{144}{25}}{2}=\frac{27648}{625}$

Odp.: $Ob=\frac{684}{25}$, $P=\frac{27648}{625}$.