a)
Aria ABCD = AB*AD*sinA= 5*4*[tex] \sqrt{3} [/tex]/2=10[tex] \sqrt{3} [/tex]
Aria= DC*d(B, CD)= 5*d(B, CD)=10
[tex] \sqrt{3} [/tex]
d(B, CD)=2
[tex] \sqrt{3} [/tex]
Aria= d(B, AD)*AD=10[tex] \sqrt{3} [/tex] => d(B, AD)*4=
10[tex] \sqrt{3} [/tex]
d(B, AD)=5
[tex] \sqrt{3} [/tex]/2
b) Construim BM perpendicular pe Da. Cum m<A=60 => m<ABM=30
Conform R. T. 30 => AM=2,5 => MD=1,5 => BM=5[tex] \sqrt{3} [/tex] /2
Aplicand teorema lui Pitagora in BMD => BD= [tex] \sqrt{21} [/tex]
Fie AN perpendiucular pe BC (cade in exterior). Aplicand Teorema lui Pitagora in ANC => AC=[tex] \sqrt{61} [/tex]
