Question

ABCD trapez isoscel , AB= 24 cm , CD= 12 cm
AB||CD; m (Â= 45 grade) ; AD=BC
Aflati perimetrul si aria

Answer 1

Notam DE inaltimea (h) trapezului.

AE = (AB -DC) /2 = (24-12) / 2 = 6
In Δ ADE :

AE = 6
m(A) = 45 grade ⇒ m(D) = 45 grade ⇒ Δ ADE isoscel si DE = AE = 6

AD² = AE²+DE²
AD² = 36 +36
AD² = 72
AD = 6√2

P trapez= AD + BC + AB + DC= 6√2 + 6√2 +24 +12 = 12 (√2+3) cm

A= (B+b) h / 2 = (24+12) ×6 /2 = 108 cm²