Răspuns:
Explicație pas cu pas:
BD bisectoare, deci α=β. AB║CD, deci α=y, (unghiuri alterne interne)
⇒ΔBCD isoscel, deci BC=CD.
a) din ΔDAB, T.P. ⇒BD²=AB²+AD²=60²+(20√3)²=(20·3)²+20²·3=20²·(9+3)=20²·4·3, ⇒BD=√(20²·4·3)=20·2√3=40√3 cm.
In ΔBCD, fie x=BC=CD. CE⊥AB, ⇒CD=EA=x.
in ΔBCE, dreptunghic, BE=AB-AE=60-x, CE=AD. T.P. ⇒ CE²=CB²-EB², ⇒
(20√3)²=x²-(60-x)², ⇒20²·3=x²-60²+120x-x², ⇒120x=60²+20²·3, ⇒
x=40=CD=CB.
atunci, din ΔADC, AC²=AD²+CD²=(20√3)²+40²=20²·3+20²·2²=20²·(3+4)=20²·7. Atunci AC=√(20²·7)=20√7.
b) P(ABCD)=AB+BC+CD+DA=60+40+40+20√3=140+20√3 cm
c) Aria(ABCD)=(AB+CD)·AD/2=(60+40)·20√3/2=1000√3 cm²
Aria(BCD)=(1/2)·DC·AD=(1/2)·40·20√3=400√3 cm²
