Vezi desenul atasat. Avem trapezul ABCD, cu bazele AB|| CD , si AD=BC
Ducem CE_|_ AB , CD=inaltimea trapezului =12 cm
Fie o intersectia diagonalelor AC si BD, mai stim ca ca AC_|_BD
Putem rezolva problema in 2 moduri:
Mod 1.
in ΔOAB, daca <AOB=90° ⇒<OAC=<CAB=45°
in ΔCAB, avem sin <CAB=CE/AC
√2/2=12/AC
AC=24/√2=24√2/2=12√2 cm
DB=AC=12√2 cm
Arie trapez ABCD=(AC*BD/2)*sin <DOA=(12√2*12√2/2)*sin 90°=144*2/2*1=144cm²
Mod 2.
in ΔOAB, daca <AOB=90° ⇒<OAC=<CAB=45°
in ΔCAE, daca <CEA=90°, si <CAE=45⇒ACE=45°⇒ ΔCEA=isoscel, si CE=AE=12 cm
dar CD=AE-EB
AB=AE+EB
Arie trapez ABCD = (AB+CD)*CE/2= [(AE+EB)+(AE-EB)]*CE/2 = =(12+12)*12/2=24*6=144
cm²
