Răspuns:
In ABCD- trapez isoscel
AC ⊥ BC
AB = 26 cm
CD= 10 cm
DN, CM- inaltimi
DC = NM= 10 cm
AN= MB= 26-10/2= 8 cm
In ABC- dreptunghic
CM inaltime =>T. catetei BC^2 = BM*AB
BC^2= 208 => BC= 4√13
In MCB- dreptumghic aplicam T.Pitagora
MC^2+MB^2= BC^2
MC^2+64= 208
MC= √144= 12 cm
Inaltimea trapezului are lungimea egala cu 12 cm.
Sper ca te-am ajutat!
