Răspuns:
Explicație pas cu pas:
Construim BE⊥CD ,In Δ BEC ∡C=60° ; ∡CEB=30°
tg CEB=EC/AD=√3/3 (AD+CE) ⇒EC/4√3=√3/3 ⇒EC=4 cm
BC²=AD²+EC² ⇒BC²=16·3+16 =16·4 ⇒BC=8 cm
a) Pabcd=AD+AB+BC+CD=2·4√3+4√3+4+8=12+9√3 cm
b)BD²=2AB²=2·16·3 ⇒BD=4√6 cm
AC²=AD²+CD²=16·3+(4√3+4)²=48+16(3+2√3+1)=48+48+32√3+16=112+16√2=16(7+√2)
AC=4√(7+√2) cm
c) tgBAC=tg(90°+30°)=-ctg30°=-√3
