Question

UN TRAPEZ ABCD ARE ARIA 528 CM PATRATI SI INALTIMEA=22
AFLATI LINIA MIJLOCIE

Answer 1

ABCD-trapez
AriaABCD=528 cm^2
h=22
Lm=??

*AABCD=(B+b)×h/2 (baza mare plus baza mica ori inaltimea supra doi)*



AABCD=Lm•h
➡528=Lm•22
Lm=
$= \frac{528}{22} = 24$

Lm=24

Baftă!!!✌✌

Answer 2

[tex]\it Linia\ mijlocie\ este\ :\ \ \ \ \ell_m = \dfrac{\mathcal{B}+b}{2} \ \ \ (*) \\\;\\ \mathcal{A} = \dfrac{\mathcal{B}+b}{2}\cdot h \stackrel{(*)}{\Longrightarrow} \mathcal{A} = \ell_m\cdot h \Rightarrow \ell_m = \cfrac{\mathcal{A}}{h}[/tex]