Question

Calculati aria unui trapez isoscel cu diagonalele perpendiculare si cu lungimea inaltimii de 12 cm

Answer 1

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=144 cm²

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