a) fie CF⊥AB
CF=AD
FB=AB-CD=20-8=12
CB²=CF²+FB²=9²+12²=81+144=225 ⇒CB=√225=15
Perimetru=AB+BC+CD+AD
P=20+15+8+9=52
[tex] Aria~trapez= \frac{(Baza~Mare+Baza~mica)*Inaltimea}{2} [/tex]
[tex]A= \frac{(AB+CD)*AD}{2} = \frac{(20+8)*9}{2} =126[/tex]
b) distanta de la un punct la o dreapta este o perpendiculara dusa din cel punct pe dreapta
Notam cu "d" inaltimea din A pe BC
In ΔABC:
[tex]Aria~triunghi= \frac{baza +inaltimea}{2} [/tex]
[tex]Aria~ABC= \frac{AB*FC}{2} = \frac{20*9}{2} =90[/tex] (1)
Dar, daca consieram baza BC, atunci
[tex]Aria~ABC= \frac{BC*d}{2} = \frac{15*d}{2} [/tex] (2)
din (1) si (2) ⇒
[tex] \frac{15*d}{2} =90 =\ \textgreater \ d= \frac{90*2}{15} =12[/tex]
c) ΔEDC~ΔEAB =>
[tex] \frac{CD}{AB} = \frac{ED}{EA} [/tex]
[tex] \frac{CD}{AB} = \frac{ED}{ED+DA} [/tex]
[tex] \frac{8}{20} = \frac{ED}{ED+9} [/tex]
8E+72=20ED
72=12ED
ED=6
Aria ΔAEB= AB*EA/2=20*15/2=150
