AO∧BC={M}
AM=2cm ⇒ OM=8cm
AC≡AB
CD≡BD
In ΔCMO, M=90*, aplicam Pitagora:
[tex]MC^{2} = OC^{2} - OM^{2}= 10^{2} - 8^{2} =100-64=36 \\ MC=6cm[/tex]
In ΔACM, M=90*, aplicam Pitagora:
[tex]AC^{2} = AM^{2}+ CM^{2}= 2^{2} +6^{2} =4+36=40 \\ MC=2 \sqrt{10} cm[/tex]
In ΔACD, C=90* (un triunghi inscris intr-un cerc cu o latura diametrul cercului este dreptunghic), aplicam Pitagora:
[tex]CD^{2} = AD^{2}-AC^{2}= 20^{2} -40 =400-40=360 \\ CD=6 \sqrt{10} cm[/tex]
[tex] P_{ACDB} =AC+CD+DB+AB=2*AC+2*CD= \\ =2*(2 \sqrt{10} +6 \sqrt{10})=2*8 \sqrt{10}=16 \sqrt{10}cm[/tex]
