a)[tex]P=3l=3*3 \sqrt{5} =9 \sqrt{5} [/tex]
b)[tex]P_{romb} =4l=60cm[/tex] rezulta [tex]l= \frac{60}{4} =15cm[/tex]
Notam cu O intersectia diagonalelor. Intr-un romb diagonalele sunt perpendiculare si intersectia lor formeaza segmente congruente. Deci DO=OB=12CM
In triunghiul AOB se aplica TP: [tex] AB^{2} = AO^{2} + OB^{2} \\ AO^{2}=AB^{2} -OB^{2}=15^{2} - 12^{2} =225-144=81 \\ AO=9cm \\ AC=AO+OC=2AO=18cm[/tex]
[tex] A_{romb} = \frac{ d_{1}* d_{2} }{2} = \frac{AC*DB}{2} = \frac{18*24}{2} = \frac{432}{2} =216cm ^{2} [/tex]
Deasemenea, intr-un romb avem:
[tex] A_{romb}=h*l \\ 216=h*15 \\ h= \frac{216}{15} 14,4cm [/tex]
