Răspuns:
[tex]\color{#CC0000}\Large\boxed{\bf A_{ABCD} = 60~cm}[/tex]
[tex]\color{blue}\Large \boxed{\bf Primetrul_{ABCD}=40~cm}[/tex]
Explicație pas cu pas:
TEORIE:
Rombul este paralelogramul cu 2 laturi consecutive congruente, așadar toate laturile sunt congruente.
[tex]\color{red}\large\boxed{\boxed{ \bf Arie~romb=\dfrac{d_{1}\cdot d_{2}}{2}}}[/tex]
[tex]\color{magenta}\large\boxed{\boxed{\bf Perimetru ~romb=4\cdot \ell}}[/tex]
Diagonale:
se înjumătățesc: [tex]\large \bf AO \equiv OC, \ BO \equiv OD[/tex]
sunt perpendiculare
sunt bisectoarele unghiurilor
Laturile rombului sunt egale si cele opuse sunt paralele două câte două [tex]\large\bf AB \parallel CD,\ AD \parallel BC[/tex]
[tex]\large \bf AB = BC=CD=DA= 10~cm[/tex]
[tex]\large \bf Primetrul_{ABCD}= 4\cdot 10 \implies \boxed{\bf Primetrul_{ABCD}=40~cm}[/tex]
[tex]\bf In~ triungiul~\Delta AOB ~avem:[/tex]
[tex]\large \bf AB = 10~cm[/tex]
[tex]\large\bf m(\measuredangle AOB)=90^{\circ}[/tex]
[tex]\large\bf m(\measuredangle ABO)=60^{\circ}[/tex]
[tex]\large\bf m(\measuredangle BAO)=30^{\circ}[/tex]
[tex]\large \bf BO = AB:2\implies BO = 10:2\implies\boxed{\bf BO=OD = 5~cm}[/tex]
[tex]\large \bf BO +OD = BD\implies\boxed{\bf BD = 10~cm}[/tex]
[tex]\large \bf Aplicam ~teorema ~lui~ Pitagora~ in~\Delta AOB\implies AO^{2}+ BO^{2}=AB^{2}[/tex]
[tex]\large \bf AO^{2}+ 5^{2}=10^{2}[/tex]
[tex]\large \bf AO^{2}+ 25=100[/tex]
[tex]\large \bf AO^{2}=100-25\implies AO^{2} = 36 \implies AO=\sqrt{36} \implies[/tex]
[tex]\large \bf \boxed{\bf AO = 6~cm}[/tex]
[tex]\large\bf AO +OC = AC\implies\boxed{\bf AC = 12~cm}[/tex]
[tex]\large \bf AC, BD~ diagonale~rombui\implies AC = d_{1}~si ~BD = d_{2}[/tex]
[tex]\large\bf A_{ABCD}=\dfrac{12\cdot 10}{2}=\dfrac{120}{2}\implies \large \boxed{\boxed{\bf A_{ABCD} = 60~cm}}[/tex]
==pav38==
