Răspunsuri

  • Utilizator Brainly
2014-09-12T17:46:42+03:00
a) AD=6√3
Vezi desenul atasat.
Ducem CM_|_ AB
CM=AD=6√3, pt ca AD_|_AB
inΔ BCM, CM=6√3
tg <CBM=CM/MB
tg 60=√3=6√3/MB
MB=6
sin <CBM= CM/BC
sin 60=√3/2=6√3/BC
BC=12

in Δ dreptunghic CAB, avem teorema inaltimii:
CM²=AM*MB
(6√3)²=AM*6
AM=108/6=18
dar, AM=DC
DC=18
AB=AM+MB=18+6
AB=24

Perimetru trapez=AB+BC+CD+DA=24+12+6+6√3=42+6√3
Aria trapez=(AB+DC)*AD:2=(24+18)*6√6:2=126√3

b) AC²=AM*AB=18*24=432
AC=12√3
DB²=AB²+AD²=24²+(6√3)²=576+108=684
DB=6√19