Răspuns:
a)DC=AE=8cm
BE=12-8=4cm
∆CEB-dreptunghic=>TP
[tex]ce = \sqrt{8 {}^{2} - {4}^{2} } \\ ce = \sqrt{48} \\ ce = 4 \sqrt{3} [/tex]
CE=AD=4√3 cm
b)∆ABM-dreptunghic
AB=12cm
in ∆BCE:
[tex]be = \frac{bc}{2} [/tex]
=> m<(BCE)=30 => m<(EBC)60
=> în∆ BMA: m<(BMA)=30
=> BM= 24cm
in ∆BMA: TP=>
[tex]am = \sqrt{ {24}^{2} - {12}^{2} } \\ am = \sqrt{432} \\ am = 12 \sqrt{3} [/tex]
Aria ∆ ABM =
[tex] \frac{12 \times 12 \sqrt{3} }{2} [/tex]
[tex]72 \sqrt{3} [/tex]
Sper că te-am ajutat!
