AD=BC
CD=4√2 cm
AD=8 cm
∡ABC=45°
Ducem CM⊥AB
CM=MB (ΔCMB dreptunghic isoscel)
Aplicam Pitagora (suma catetelor la patrat este egala cu ipotenuza la patrat)
BC²=CM²+MB²
64=2CM²
32=CM²
CM=MB=4√2 cm
AB=2MB+DC
AB=8√2+4√2=12√2 cm
ΔAEB dreptunghic isoscel in E
Pitagora:
AB²=AE²+EB²
288=2AE²
AE²=144
AE=EB=12 cm
DC║AB⇒
[tex]\frac{ED}{AE}=\frac{CD}{AB} \\\\\frac{ED}{12} =\frac{4\sqrt{2} }{12\sqrt{2} }[/tex]
ED=4 cm=EC
In ΔBED dreptunghic in E aplicam Pitagora:
BD²=BE²+DE²
BD²=144+16
BD²=160
BD=4√10 cm
[tex]A_{ABCD}=\frac{(CD+AB)\cdot CM}{2} =\frac{16\sqrt{2}\cdot 4\sqrt{2} }{2}=64\ cm^2[/tex]
[tex]A_{ABE}=\frac{AE\cdot BE}{2} =\frac{144}{2} =72\ cm^2[/tex]
[tex]64=\frac{8}{9}\cdot 72[/tex]
Un alt exercitiu de geometrie gasesti aici: https://brainly.ro/tema/8601313
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