Question

Mă poate ajuta cineva ?​

Answer 1

Răspuns:

Explicație pas cu pas:

a) ∡BAE=60°=∡AEB, ∡EAD=90°- ∡BAE=90°-60°=30°. AD=AE, ⇒ΔADE este isoscel cu baza DE, ⇒∡AED=∡ADE=(180°-∡EAD):2=150°:2=75°.

Atunci, ∡BED=∡BEA+∡AED=60°+75°=135°.

b)  Trasăm DN⊥AE, N∈AE. ΔADN dreptunghic, ∡DAN=30°, deci DN=(1/2)·AD=(1/2)·24=12cm=d(D,AE).

c) Aria(BEF)=???

Trasăm EP⊥AB, P∈AB. Din ΔBEP, sin(∠EBP)=EP/BE, ⇒sin60°=EP/24, ⇒

√3/2=EP/24, ⇒2·EP=24√3 |:2, ⇒EP=12√3. Atunci GE=24-12√3=12·(2-√3).

Atunci, CF=2·GE=24·(2-√3).

Aria(BEF)=Aria(BCGP)-Aria(CGEF)-Aria(BEP).

Aria(BCGP)=BP·BC=12·24

CGEF este trapez, deci Aria(CGEF)=CG·(CF+GE)/2=12·(24·(2-√3)+12·(2-√3))/2=12·36(2-√3)/2=12·18·(2-√3).

Aria(BEP)=(1/2)·BP·EP=(1/2)·12·12√3=6·12√3

Deci, Aria(BEF)=12·24-12·18·(2-√3)-6·12√3=12·(24-36+18√3-6√3)=12·(12√3-12)=12·12·(√3-1)=144·(√3-1)cm².