construim:
DE⊥AB;CF⊥AB, E;F∈AB
notam:
x=AE
28-x=FB (EF=10)
H=DE=CF
h^2=DA^2-x^2=CB^2-(28-x^2)^2 din ultimele 2 egalitati rezulta:
900-x^2=676-784+56x-x^2
56x=1008
x=18
AE=18
FB=10
h^2=AD^2-AE^2=900-324=576
h=24=DE=CF
aria trapez At
At=(38+10)24/2=576
AC^2=CF^2+AF^2=576+784=1360
AC=4√85
DB^2=DE^2+BE^2=576+400=976
DB=4√61
scriem aria tr ACB in 2 moduri
AB x CF=CB x d unde d e distanta ceruta adica perpendiculara din A pe CB)
d=AB x CF/CB = 38 x 24/26 = 456/13
daca e de corectat, s-o faci !
