Fie AC ∩ BD = {O}
[tex]\mathit ABCD - romb \Rightarrow \begin{cases} A O=DO=\dfrac{AC}{2}=8\,cm\\ \\\ BO=DO=\dfrac{BD}{2}=6\,cm \end{cases}[/tex]
[tex]\mathit ABCD - romb \Rightarrow AC \perp BD[/tex]
[tex]\triangle AOB : \sphericalangle O=90^{\circ} \xrightarrow{TP} AB^2=AO^2+BO^2[/tex]
[tex]\mathcal AB^2=8^2+6^2=64+36 \Rightarrow AB=\sqrt{100} = 10\,cm[/tex]
[tex]\mathcal P_{ABCD}=4l=4 \cdot 10=40\,cm \, (a)[/tex]
[tex]\mathcal A_{ABCD}=\dfrac{d_1 \cdot d_2}{2} =l \cdot h \Rightarrow \dfrac{AC \cdot BD}{2}=AD \cdot d(B,AD)[/tex]
[tex]\dfrac{16 \cdot 12}{2} = 10 \cdot d(B,AD) \Rightarrow d(B,AD)=\dfrac{48}{5}\, cm[/tex]
#copaceibrainly
