Problema 7
Ducem diagonalele rombului AC si BD.
AC ∩ BD ={ O }
BD bisectoarea ∡ADC ⇒∡ADB≡∡BDC = 60°
BO≡OD =12√2
AO≡OC
In Δ dreptunghic AOD, cos ∡ODA = OD / AD
cos 60°=OD/AD
√3/ 2=12√2 / AD
AD=24√2 / √3 = 24√6 / 3 = 8√6
Perimetrul =4·AD = 4·8√6 = 32√6
In ΔAOD aplicam T.Pitagora
AD²=AO²+OD²
384=AO²+288
AO²=384-288
AO²=96
AO=4√6
AC= 2·AO=2·4√6=8√6
Problema 8
In ΔABC aplicam T.Pitagora
BC²=AC²+AB²
BC²=144+432
BC²=576
BC=24
sin ∡B = AC/BC = 12√3 / 24 = √3 / 2 ⇒∡B=60°
∡C = 90° -∡B = 90°-60°=30°
