Question

plz ex 2 si 1 Dau funda repede

Answer 1

aplic teorema catetei
EF^2=DF*AF
DF=EF^2/AF=12*12/6=2*12=24
DA=DF-AF=24-6=18
teorema inaltimii spune ca
EA^2=AF*DA
EA^2=6*18   EA=6rad3
-u Pitagora
DE^2=DF^2-EF^2=24*24-12*12=3*12*12   DE=12rad3

1.aplic teorema inaltimii
CD^2=AD*DB   
DB=12*12/18=8   DB=2rad2
AB =18+2rad2
aplic teorema catetei
BC^2=BD*AB=2rad2*(18+2rad2)=36rad2+8
tot cu teorema catetei
AC^2=AD*AB=18*(18+2rad2)

Answer 2

1.
AD=18 cm
DC=12 cm
In ∆ABC dr =>T in
AD²= DC×BD
18²=12×BD
BD= 324:12
BD=27 cm
BC =BD+DC = 27+12= 39 cm
In ∆ ABC dr => T cat
AC²= BD×BC
AC²=12×39
AC² =468
AC=√468
AC=6√13 cm
In ∆ ABC dr =>TP
AB²+AC² =BC²
AB²= 39²-(6√13)²
AB² = 1521-468
AB²=1053
AB=√1053
AB=9√13. cm

2.
EF=12 cm
AF=6 cm
In ∆DEF dr =>T cat
EF²=AF×DF
DF= 12²:6
DF=144:6
DF=24 cm
DF =AF+ DA=>DA = DF-AF=24-6= 18 cm
In ∆ DEF dr =>TP
DE²+EF²=DF²
DE²= 18²-12²
DE² = 324-144
DE=√180
DE= 6√5 cm
In ∆ DEF dr =>T in
EA ²= DA×AF
EA²= 18×6
EA²=108/
EA=√108/
EA=6√3