Inaltimea DF_|_AB
AF=(AB-DC)/2=(24-18)/2=3cm
In triunghiul DFA:
tg A=DF/AF
√3=DF/3 deci DF=3√3cm
[tex]A= \frac{(AB+CD)*DF}{2}= \frac{42*3\sqrt{3}}{2}=63\sqrt{3}~cm^2[/tex]
FB=AB-AF=24-3=21cm
Pitagora in triunghiul DFB
DF²+FB²=BD²
(3√3)²+21²=BD²
BD²=441+27 deci BD=√468=√4·117 deci BD=2√117cm
d=CE
Prelungesti AD si BC si notezi intersectia cu T, astfel obtinand un triunghi isoscel TAB.
TDC si TAB:
<T comun
<D=<A corespondente
deci ΔTDC≈ΔTAB
TD/AT=DC/AB=TC/TB
In triunghiul DFA: F=90 si A=60 deci F=30 grade
Cateta opusa unghiului de 30 de grade este ipotenuza/2 deci AD=2*AF=2*3=6cm
TD/AT=DC/AB
DC/AB=18/24=3/4
AT=TD+AD=TD+6
inlocuiesti in raportul TD/AT=DC/AB
TD/TD+6=3/4
4TD=3*(TD+6)
TD=18cm=TC=CD deci triunghiul TDC echilateral
h=EC
EC=[tex] \frac{DC \sqrt{3}}{2}= \frac{18\sqrt{3}}{2}=9\sqrt{3} [/tex] cm
