Question

Am nevoie urgent de rezolvare! Va rog! Dau 10 puncte!

Answer 1

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

$Aria~trapez= \frac{(Baza~Mare+Baza~mica)*Inaltimea}{2}$

$A= \frac{(AB+CD)*AD}{2} = \frac{(20+8)*9}{2} =126$

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:
$Aria~triunghi= \frac{baza +inaltimea}{2}$

 $Aria~ABC= \frac{AB*FC}{2} = \frac{20*9}{2} =90$       (1)

Dar, daca consieram baza BC, atunci
 $Aria~ABC= \frac{BC*d}{2} = \frac{15*d}{2}$                (2)

din (1) si (2) ⇒
$\frac{15*d}{2} =90 =\ \textgreater \ d= \frac{90*2}{15} =12$

c) ΔEDC~ΔEAB =>
$\frac{CD}{AB} = \frac{ED}{EA}$

$\frac{CD}{AB} = \frac{ED}{ED+DA}$

$\frac{8}{20} = \frac{ED}{ED+9}$

8E+72=20ED
72=12ED
ED=6

Aria ΔAEB= AB*EA/2=20*15/2=150