Răspuns:
Explicație pas cu pas:
Fie CP⊥AB
In ΔCPB dr avem ∡B=60°⇒ m(∡BCP)=30°⇒ 2BP=BC
AP=CD=24
AB=AP+BP=36
BP=36-24=12⇒BC=12×2=24 cm
Aplicam Pitagora in ΔCPB
BC²=BP²+CP²
576=144+CP²
CP=√432=12√3=AD
P=CD+BC+AB+AD=24+24+36+12√3=84+12√3
[tex]A=\frac{(36+24)\cdot 12\sqrt{3} }{2} =360\sqrt{3}\ cm^2[/tex]
In ΔDAC avem pitagora
AC²=AD²+CD²
AC²=432+576
AC=√1008=12√7
In ΔDAB avem pitagora
BD²=AD²+AB²
BD²=432+1296
BD=√1728=24√3
Fie AM⊥BC
In ΔAMB ∡ ABM=60° si ∡MAB=30°⇒ 2BM=AB
BM=18
Aplicam pitagora
AB²=AM²+BM²
AM²=1296-324
AM=√972=18√3cm
