Răspuns:
Explicație pas cu pas:
3
⇒∡(AE, BE)=45° ⇒b
4
d(C, (ADM))=CN , CN⊥AD
Teorema 3⊥ MD⊥CN ⇒CN este⊥pe 2 drepte ∈ (ADM) ⇒CN = d( C, planul (ADM)
Aabcd=ABsin 60°AB=12²√3/2 cm²
Aabcd=CN·AB ⇒ABsin 60°AB=CN·AB ⇒CN=AB sin 60°=12√3/2
CN=12√3/2=6√3cm raspuns: b
5
T Thales
AM/AC=1/4 AM/(AM+MC)=1/(3+1)=1/4
Ampbq=Aabc-Aamq-Amcp
Ampbq=Aabc-Aabc/16-Aabc·9/16
Ampbq=Aabc(1-1/16-9/16)=Aabc·6/16 Aabc=Aabcd/2
Ampbq=Aabcd·6/2·16 ⇒Ampbq/Aabcd=6/32=3/16 raspuns: d
