AB=12 cm
CD=20 cm
∡DAC=90°=∡DBC
Intr-un trapez isoscel ortodiagonal, inaltimea este egala cu linia mijlocie, adica cu jumatate din suma bazelor
a) h=(12+20):2=16 cm
[tex]A=\frac{(AB+CD)\times h}{2} =256\ cm^2[/tex]
b) Ducem inaltimile AE si BF
EF=AB=12 cm
DE=FC=(20-12):2=4 cm
Calculam AD din ΔAED, aplicam Pitagora (suma catetelor la patrat este egala cu ipotenuza la patrat)
AD²=AE²+DE²
AD²=256+16
AD=4√17 cm=BC
In ΔADC aplicam Pitagora
DC²=AD²+AC²
400=272+AC²
AC=8√2 cm=BD
ΔAOB~ΔCOD
[tex]\frac{AO}{OC}=\frac{BO}{OD} =\frac{AB}{DC}[/tex]
[tex]\frac{AO}{OC}=\frac{AB}{DC}\\\\\frac{AO}{OC}=\frac{12}{20}\\\\[/tex]
Facem proportie derivata si obtinem:
[tex]\frac{AO}{AO+OC} =\frac{12}{12+20} \\\\\frac{AO}{AC} =\frac{12}{32} \\\\\frac{AO}{8\sqrt{2} } =\frac{12}{32}[/tex]
AO=3√2 cm=OB
OC=OD=5√2 cm
[tex]P_{AOB}=AB+AO+OB=6\sqrt{2} +12\\\\P_{DOC}=DO+OC+CD=10\sqrt{2} +20[/tex]
