ducem DE║BC
cu heron in tr. ADE
p=42, p-AD=12, p-DE=16, p-AE=14
aplicam formula si gasim aria AED
A=336
AE*H/2=336
H=24
aria trapez
At=(38+10)*24/2
At=576
ducem DF⊥AB
cu pitagora in ADF calculam AF
AF=√(AD^2-H^2)
AF=18
FB=38-18=20
cu pitagora in DFB calculam DB
DB=√(H^2+FB^2)
DB=4√61
ducem CG⊥AB
analog gasim:
GB=√(CB^2-H^2)
GB=10
AG=38-10=28
AC=√(H^2+AG^2)
AC=4√85
se observa ca EB=10 si GB=10 prin urmare desenul corect ar fi sa fie cu E≡G
dar nu aveam de unde sti
