Question

Vă rog repede!!! Am nevoie mâine!!​

Answer 1

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

$A=\frac{(AB+CD)\times h}{2} =256\ cm^2$

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

$\frac{AO}{OC}=\frac{BO}{OD} =\frac{AB}{DC}$

$\frac{AO}{OC}=\frac{AB}{DC}\\\\\frac{AO}{OC}=\frac{12}{20}$

Facem proportie derivata si obtinem:

$\frac{AO}{AO+OC} =\frac{12}{12+20} \\\\\frac{AO}{AC} =\frac{12}{32} \\\\\frac{AO}{8\sqrt{2} } =\frac{12}{32}$

AO=3√2 cm=OB

OC=OD=5√2 cm

$P_{AOB}=AB+AO+OB=6\sqrt{2} +12\\\\P_{DOC}=DO+OC+CD=10\sqrt{2} +20$