a)
∡(DOA) = ∡DAC = 60°
ABCD - dreptunghi ⇒ BD = AC = 16cm
OD = OB = OA = OC = 16:2 = 8cm
Δ AOD - isoscel, cu ∡ (DOA) = 60° ⇒ Δ AOD -echilateral ⇒
⇒ AD= 8cm
ABCD - dreptunghi ⇒ BC = AD = 8cm ⇒ΔOBC -echilateral
[tex]\bf\mathcal{P}_{OBC}=3\cdot\ell = 3\cdot8=24\ cm[/tex]
b)
[tex]\bf \Delta ABC-de,\ \widehat{ABC}=90^o,\ \stackrel{T.P.}{\Longrightarrow}\ AB^2=AC^2-BC^2=16^2-8^2=\\ \\ =(16-8)(16+8)=8\cdot24=8\cdot8\cdot 3=8^2\cdot3 \RightarrowAB=8\sqrt3\ cm[/tex]
[tex]\bf AM=AB=8\sqrt3\ cm\\ \\ OM=AM-AO=8\sqrt3-8<8\cdot1,74-8 \Rightarrow OM<13,92 -8 \Rightarrow \\ \\ \Rightarrow OM<5,92<6\ cm[/tex]
