Răspuns :

Răspuns:

Explicație pas cu pas:

CE⊥AB     ∡B=60°    CE=AD=6 cm

tg ∡B=CE/EB      √3=6/EB        EB=6√3/3=2√3 cm

AB=CD+EB=4√3+2√3cm=6√3cm

sin ∡B=CE/BC      BC=CE/sin 60=6/√3/2=12/√3=4√4cm

a)P=AB+BC+CD+AD=6√3+4√3+4√3+6=14√3+6 cm

b)BD²=AD²+AB²=36+108=144      BD=12 cm

AC²=AD²+CD²=36+48=84           AC=2√21 cm

Ignatovvlad201id
ABCD trapez dreptunghic
AB||CD
m(CD=4V3 cm
m(AD=6 cm
CE|AB-m(AECD dreptunghi
CD=AE=4V3 cm
AD=CE=6 cm

UnghiulCEB dreptunghic
CE=6 cm
m(sin60=V3/2; 6/CB=V3/2; CB V3=6*2;
CB V3=12 cm; CB=12/V3; CB=12 V3/3;
CB=4V3 cm.
cos B=EB/CB
cos 60=1/2
EB/4V3=1/2
EB*2=4V3
EB=2V3 cm.

AE=4V3; AB=6V3 cm
P abcd= AB+BC+CD+AD
6V3+4V3+4V3+6=14V3+6=2(7V3+3)

b) AC,BD deagonale
Unghiu ADC dreptunghic
AD=6 cm
DC= 4V3 cm; AC^2=AD^2+ DC^2
AC^2=6^2+(4V3)^2
AC^2=36+48
AC^2=84
AC=2V21 cm

BAD dreptunghic
AD=6 cm
AB=6V3; AD^2=AB^2+AD^2
BD^2=(6V3)^2+6^2 ; BD^2= 108+36;
BD^2= 144 cm; BD=12cm

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